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Posted by Mariner (Member # 1618) on :
 
I seem to recall the Monty Hall problem appeared here in the past, so now I've got another one for y'all. This appeared on a blog I read a while back and I guarantee the answer given was wrong. Well, not entirely guarantee. But I'm pretty darn sure. Not only that, but it makes me think that the common "counterintuitive" answer to a similar problem is also wrong, and the supposedly wrong intuitive answer to the problem is correct. So here goes.

John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?

Now, the intuitive answer is that the odds are 1/2, since the sex of one child does not influence the sex of the other (obviously, we're simplifying biology here to a pure 50-50 chance of male vs female and no twins or any other confounding factors), and the whole thing about Tuesday is irrelevant. A clever man would then say it is actually 1/3, since there are 4 possibilities of a two child family: GG, GB, BG, and BB. Since GG is eliminated, there's only 1/3 chance. (He would also put the poison into his own goblet, because he knows only a great fool would reach for what he was given)

But the even more clever person would say it is 13/27. That's because out of all the possibilities of two child families when the day of the week they were born is included happens to be 196 (2 sexes * 7 days * 2 sexes * 7 days). Of those, 27 have at least one boy born on a Tuesday. And 13 of those have two boys. Mathematically, that's sound. (Also, he would build up an immunity to iocane and put the poison in both goblets)

But logically, that's stupid.

So do we trust the mathematics or do we trust our instincts? In the Monty Hall problem, we trusted the math, because our intuition was being deceitful. But here, I think the math is being deceitful.

The clearest explanation for that is that the whole Tuesday thing appears to be random. If John Doe had said that he had two kids and one was a boy, then the probably would be (supposedly) 1/3. Yet you know that boy was born on a day of the week. And the probability supposedly jumps to 13/27 as soon as that day of the week is mentioned? That's crazy!

Note that that is exactly why the Monty Hall problem works. The probability of your initial pick didn't change once you got more information. Yet here, we are saying it does.

But that brings me back to my initial point. Remember the clever man who said the probability was 1/3? I've never questioned this answer until I started questioning the Tuesday thing. And now I think that answer is wrong too.

After all, if Tuesday is just a random happenstance and shouldn't be relevant, then the fact that one's a boy shouldn't be relevant either. And in fact, you can create a game scenario to cheat the clever mathematicians out of money.

So, if anyone thinks the probability is still 1/3 (ignore the Tuesday bit entirely, we're simplifying the game a bit), please step forward. We're going to play a game.

I shall present to you the father of a two-child family. He will mention the sex of one of his kids. You then predict whether he has two boys, a boy and a girl, or two girls. If you're correct, you win $4. If you're wrong, you pay $6. You can play as often as you like, and I assure you I am not rigging the selection of parents. Should you pay?

If you believe you're clever, you'd say yes. After all, as soon as John Doe mentions he has a boy, you can immediately eliminate the two girls scenario. And then there's a 2/3 chance that it's a boy and a girl. So you go with that one. Likewise, if John Doe says he has a girl, then you can eliminate the BB option, and then there's a 2/3 chance it's a boy and a girl.

So you always say one boy, one girl. And then you win 2/3 of the time. And thus you should win, on average, $0.67 per game.

Except, from the way I described it, you should clearly see the problem. In the general population, boy-girl is only 50% of the two child families. So in actuality, you'd be losing $1 per game.

The difference is obvious. When I presented the father, I chose from a random audience. Thus, that is when the probability is chosen and locked in. The extra information he gives does not change that fact. Like the Monty Hall problem, the extra information transfers all of its probability to the other pick (ie, BB goes from 25% to 50% if John Doe says he has a boy). It may not seem like it should happen mathematically, but the way the game is played its true. You can program it yourself and try it.

So let's ignore the game for a moment and go back to the original John Doe. When he gives the information that he has a son born on a Tuesday, it does not change the random pool that he was chosen from. To your knowledge, his random pool is still all two child families. Thus, the fact that one of his kids is a boy is irrelevant to the sex of the other one, as is the whole born on a Tuesday thing. To your mind, he was still chosen from the universal pool, and thus the odds should still be 50-50.

Am I clear? And more importantly, am I right?

Interestingly enough, I'd say that there's still other information that might be relevant. For example, if you met John Doe at a function for fathers of Boy Scouts, then the probability should go back down to 1/3 (the Tuesday is still irrelevant). After all, then you ARE selecting from a smaller pool, since there are no fathers of two daughters at a fathers of Boy Scouts event. If the game we were playing only allowed fathers to say (truthfully) that they had at least one boy, then we would be self selecting only BG GB BB families, and you would win money.

But a random guy off the street? Who cares if his kid was born on a Tuesday. Who cares if one is a guy. Go with the original odds.

And does anyone's head hurt yet?
 
Posted by Brian (Member # 588) on :
 
My first question for the mathmeticians out there is: Does the original problen devolve into 'what is the sex of my second child'? This would be analogous to Mariner's first option of 50:50.
My thinking is: the 1:4 (or 1:3) ratio in the second option depends on there being a difference between boy-girl and girl-boy. Would it change the odds if he said 'my oldest child is a boy'?
 
Posted by JoshuaD (Member # 1420) on :
 
Brian: Yes. If we knew that the oldest child was a boy, then the sex of the 2nd child is 50/50.

[ July 23, 2010, 01:40 PM: Message edited by: JoshuaD ]
 
Posted by JoshuaD (Member # 1420) on :
 
quote:
But the even more clever person would say it is 13/27. That's because out of all the possibilities of two child families when the day of the week they were born is included happens to be 196 (2 sexes * 7 days * 2 sexes * 7 days). Of those, 27 have at least one boy born on a Tuesday. And 13 of those have two boys. Mathematically, that's sound. (Also, he would build up an immunity to iocane and put the poison in both goblets)
Something's definitely wrong with this reasoning. There's some sneaky mathematical slight of hand going on somewhere. I'm working on figuring out what it is now.
 
Posted by Brian (Member # 588) on :
 
See, that's what I don't get.

If the oldest child is a boy, the odds are 50:50.

If the youngest child is a boy, the odds are 50:50.

If there are only two children, and one of them is a boy, then one of those two scenarios must be true, therefor the odds are 50:50.

But simply because we don't know which one is true (even though we know one of them must be) then the odds drop to 1:3, and thus neither of them is true?
 
Posted by TheOtter (Member # 2917) on :
 
quote:
Originally posted by JoshuaD:
quote:
But the even more clever person would say it is 13/27. That's because out of all the possibilities of two child families when the day of the week they were born is included happens to be 196 (2 sexes * 7 days * 2 sexes * 7 days). Of those, 27 have at least one boy born on a Tuesday. And 13 of those have two boys. Mathematically, that's sound. (Also, he would build up an immunity to iocane and put the poison in both goblets)
Something's definitely wrong with this reasoning. There's some sneaky mathematical slight of hand going on somewhere. I'm working on figuring out what it is now.
It seems to me that the slight of hand goes like this. You're taking numbers from the general population, and then applying them to a situation that has already been narrowed down to a subset of the general population. The numbers 13, 27, and 196 have no bearing whatsoever on the gender of this guy's other kid, because we've already specified that the guy has one boy.

An equivalent question would be - I flipped two coins. The first one was on Tuesday and came up heads. What are the chances the second one also came up heads? No matter what day of the week it is, or which hand you used, the chances are 50/50 for that second coin. Same with the kids in this sort of a simplified puzzle.
 
Posted by JWatts (Member # 6523) on :
 
"John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys? "

It seems like an obvious masking problem. He has two kids one is a boy. What is the sex of the other?
Obvious answer = 50%.

Now rephrase: What are the odds he has two boys? Since we know the sex of one child is a boy, this becomes the equivalent of saying:
What are the odds my remaining child is a boy?

So the answer is 50%.
 
Posted by JoshuaD (Member # 1420) on :
 
JWatts: No. If you're a programmer, look at this and imagine the output:

code:
#DEFINE BOY = 1
#DEFINE GIRL = 2

$BoyGirl = 0;
$BoyBoy = 0;

for 1..100000 {

children[0] = Random() %2;
children[1] = Random() %2;

if ( children[0] == BOY && children[1] == BOY) {
$boyBoy++;
} elsif (children[0] == BOY && children[1] == GIRL) {
$boyGirl++;
} elsif (children[0] == GIRL && children[1] == BOY) {
$boyGirl++;
} elsif (children[0] == GIRL && children[1] == GIRL) {
#do nothing
}

}

print $BoyBoy / ($BoyGirl + $BoyBoy)



[ July 23, 2010, 03:04 PM: Message edited by: JoshuaD ]
 
Posted by hobsen (Member # 2923) on :
 
The laws of probability may not apply to questions involving human psychology. For example, if a man taken at random anywhere in the world says he had a child born last night, the odds are probably that the child was a son rather than a daughter - for some men are proud of producing sons and ashamed of producing daughters. The same applies to many women for that matter.

And the probability - as a matter of human reproduction - is not 50% in the United States. The observed ratio as I remember is 104 boys born for every 100 girls. That ratio is different for every country in the world, for reasons which are largely unknown. And in regions where sex selection is widely practiced, as in India, there may be 120 boys born for every 100 girls.

Otherwise I agree that, as a matter of logic, saying one of a man's two children is a boy says nothing about the other. But people do not engage in conversation for the most part to set logical puzzles for others, so their motives for speaking may change the probability that such a puzzle will be proposed by mere chance in the normal course of events.
 
Posted by TomDavidson (Member # 99) on :
 
Why is it mathematically sound that 13 of the 27 families with a boy born on Tuesday will have two boys? As 1/3 of all two-child families with one boy have two boys, there are thus 9 families in that result set.

9/27 = 1/3.

In other words, the specific day of the week is irrelevant.
 
Posted by vegimo (Member # 6023) on :
 
Let's say a blind man has a drawer full of blue and green socks, has an equal likelyhood of pulling either color, and pulls one out with each hand. He asks someone else if he is holding at least one blue sock, and the answer is yes. The probability that he is holding 2 blue socks is 1/3. This is analogous to the original problem (without the Tuesday portion). If he were to ask whether he was holding a blue sock in his left hand and got a positive response, then the probability he was holding a pair would be 1/2.

I'll have to think about the iocane in both goblets a bit more.
 
Posted by DonaldD (Member # 1052) on :
 
quote:
An equivalent question would be - I flipped two coins. The first one was on Tuesday and came up heads.
This is not an equivalent statement. It is not known whether it was the first or the second coin (in your example) that came up heads on Tuesday (not that Tuesday matters anyway)
 
Posted by Jordan (Member # 2159) on :
 
The reason you're having trouble articulating the flaw, Joshua, is because there isn't one, and you're way too smart to accept a fallacy just to make things match your intuition. [Smile] The reasoning is sound, if a little subtle; this is a much trickier one to explain than the Monty Hall problem, but I'll give it a go.

The key to understanding this sort of question is spotting the special cases. Start with the first question, and look at what you're actually being asked:

quote:
I have two children, and they're not both girls. What's the chance that they're two boys?
In other words, what's the probability of one special case (two boys) being true after eliminating another special case (two girls)? And because there's not so many possibilities to begin with, those special cases really skew the possibilities A LOT, which is why you turn out with the probability being a third instead of a half.

In the second question, there aren't so many special cases. In fact, there's only ONE special case: the one in which both of the boys were born on Tuesday. And because that special case is pretty unlikely, you end up with the probability being only a little bit less than a half, instead of way less like it was in the first question.


There's another way to look at the question that might make it clearer:

quote:
You have access to a database of all families in America. You do a filter to get all the families with two children. Now, you ask yourself: what proportion of those families have two boys, GIVEN that the family has one boy who was born on a Tuesday?
Now it might be a bit clearer why the special case is important. Your GIVEN is actually introducing the day of the week that one of the children was born on as an additional filtering factor. The few families in which both children are born on a Tuesday are counted only once in your filtered set of results, even though there are actually two boys in that family who meet your GIVEN, and that's why the probability of the other child being a boy is just a bit less than 50%.

Please tell me this is making a bit of sense? [Smile]

[ July 23, 2010, 07:35 PM: Message edited by: Jordan ]
 
Posted by JoshuaD (Member # 1420) on :
 
Jordon: You're definitely making sense. I was headed in that direction when I was working on it earlier today, but you put it in more concrete language than I had achieved.

I'm still a little skeptical, but I'm starting to think it's not broken after all.
 
Posted by seekingprometheus (Member # 3043) on :
 
Haven't bothered to read the whole thread so I don't know if someone already explained this, but the answer is 50%, because there is nothing in the way the question is formed that changes a variable to modify the probability.

They did this problem with the catch built correctly in the movie 21:

3 doors, 1 with a prize behind it. Contestant chooses door 1, host tells contestant door 3 does not have prize and asks if contestant would like to change mind. Contestant should change choice to 2 because of the new information that was contingent on his original choice changes the probability.

Sticking with original choice will still only be right 1 out of 3 times. Choosing to switch doors will result in a prize 2 out of 3 times because it's effectively the same as choosing both doors that weren't originally chosen and allowing the host to remove the incorrect one.

The information "one of them is a boy who was born on a Tuesday" doesn't modify anything about the probability of the 2nd second child. There isn't a probability, then a modification of the data set. To use the gameshow problem, this is equivalent to the host commenting that the prize was behind door number two yesterday. Interesting trivia, but irrelevant to the probability today.

The trick I'd that it's not a trick question.
 
Posted by Jordan (Member # 2159) on :
 
I'm sick of blanching pistachios for tonight, so I took some time out to put a summary of all the possibilities into a spreadsheet.


Edit: sp, you'd be right if the probabilities here were independent; instead, we're asking that given at least one child is a boy who was born on Tuesday, how likely is it that both children are boys. In other words, it's a Bayesian probability.

Starting with the set of all possibilities (i.e. all families of two children), we first filter it to obtain the set of families in which there is at least one boy who was born on Tuesday (the precondition). We then see how many of those families have two boys. The reason it's slightly less than half is because in one case, both boys were born on Tuesday, but those two only count for one family; so in the set of all possible families, there is one more possible scenario in which there's both a girl and a boy than there is in which there's two boys.

It's exactly the same situation you ended up in with Mariner's second question (one more outcome with both a girl and a boy than with two boys), except that in this case there are a lot more cases where we have two boys.

[ July 23, 2010, 09:06 PM: Message edited by: Jordan ]
 
Posted by Aris Katsaris (Member # 888) on :
 
There is indeed a sleight of hand, and it's that the guy selects what information he gives you.

For example - Let me give you the following possibilities: Either I have a brother born on January, or I have a sister born on any other date of the year. What are the chances I have a sister?

If you think the chances of a sister are greater, you err. I know what date my brother was born, and I selected the date I mentioned specifically for that reason.

The guy says "one of my children is a boy, born on Tuesday." But why did he give that information? If he was always going to give the gender/day information of the *other* child (e.g: one of my children is a girl, born on Wednesday), then this doesn't affect the probability because it's selected information, not a mathematically excluded possibility.

But *if* we asked him that information, not knowing what answer he'd give, then the information becomes relevant. IF not knowing anything about any dates you ask "do you have a brother born between July and December?" and I say NO, that indeed increases the probability of a sister. But if with my biased information, I CHOOSE to give you that knowledge unprompted, then I may just be trying to mislead you into thinking the opposite of what the numbers would tell you.

So in short, this isn't a pure math problem -- it involves psychological reasoning about the motivation of the guy to give you this information.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Originally posted by Jordan:
I'm sick of blanching pistachios for tonight, so I took some time out to put a summary of all the possibilities into a spreadsheet.

The summary of those possibilities still leads you astray. Because the problem doesn't explain the *reason* that the guy mentions Tuesday, or why he mentions one of the children is a boy -- and this is very very relevant to determing how it affects the probability.
 
Posted by Jordan (Member # 2159) on :
 
I think the real problem is more a matter of ambiguous wording. If you phrase the easier question differently, the answer becomes more obvious:

quote:
Given that I have two children, and at least one of them is a boy, what is the probability that I have two boys?
Unfortunately, the initial formulation is just ambiguous enough to be misread, which is why so many people go for the 50:50 option by mistake when the actual answer is one-in-three.

Similarly, if you phrase the other question:

quote:
Given that I have two children, and that at least one of them is a boy who was born on a Tuesday, what is the probability that I have two boys?
If the day of the week really were irrelevant information, this would essentially reduce to the first question and the probability of the other child being a boy would be a third. In actuality, you're now restricting the sample to those in which there are boys who were born on Tuesday, and if you do that you get a result which is much closer to a half simply because there are now more possibilities, and more of them in which there are two boys.

[ July 23, 2010, 09:17 PM: Message edited by: Jordan ]
 
Posted by Jordan (Member # 2159) on :
 
(Stupid sodding nuts… Surely you must be able to buy them with those silly little skins removed already?)

quote:
[T]his isn't a pure math problem -- it involves psychological reasoning about the motivation of the guy to give you this information.
I'm going to wait till tomorrow to see if you're referring of an á la Monty Hall "the host knows something" situation, some sort of unavoidable initial conditions, or simply assuming that the man might be trying to trick us somehow by randomly selecting the information he wants to tell us. (You should know that I tend to approach problems like this assuming point children and perfectly spherical quiz-show hosts, so the latter possibility just isn't going to cross my mind. [Smile] )
 
Posted by Aris Katsaris (Member # 888) on :
 
Let me put it in another way.
Possibilities for two children:
G/G 1/4
G/B 1/4
B/G 1/4
B/B 1/4
--
If guy decides to reveal one of these genders randomly, possibilities now become
G/G (reveals G) 1/4
G/B (reveals 1st: G) 1/8
G/B (reveals 2nd: B) 1/8
B/G (reveals 1st: B) 1/8
B/G (reveals 2nd: G) 1/8
B/B (reveals B) 1/4

Now if we know "B" was revealed, this corresponds to 1/8 + 1/8 + 1/4 = 50%. Out of these, it's even odds that the other one was a boy, or that the other one was a girl.

So if the guy randomly selected the kid whose gender he'd reveal, it's even odds that the other kid is either gender. 50% says common sense, and 50% it indeed is.
--
HOWEVER if the guy thinks: I will NOT mention there's a girl, but I will only reveal if there exists a boy. The possibilities become:
G/G (mentions no information) 1/4
G/B (mentions there's a B) 1/4
B/G (mentions there's a B) 1/4
B/B (mentions there's a B) 1/4

Now, knowing he revealed it was a boy, there only 33.3% chances that the other kid is a boy too, and 66.7% chances that the other kid is a girl.
--
AND if the guy thinks: I will mention ALL my boys, but none of my girls. The possibilities become:
G/G (mentions no information) 1/4
G/B (mentions there's a B) 1/4
B/G (mentions there's a B) 1/4
B/B (mentions there's two boys) 1/4

Now, with the knowledge he mentioned only *one* B for certain, we can be 100% sure that the other kid is a girl -- because he'd have mentioned two boys if B/B was the reality.
--
That's what I mean when I say motivation matters. WHY did he reveal the particular gender? Was he randomly picking a kid, or was he choosing that gender for some reason?

[ July 23, 2010, 10:34 PM: Message edited by: Aris Katsaris ]
 
Posted by Aris Katsaris (Member # 888) on :
 
To put it in yet another way: A sexist dad is more likely to have daughters even though he isn't mentioning any.
 
Posted by seekingprometheus (Member # 3043) on :
 
Jordan,

Whoops.

You're right, it is 1/3. The probability of an mm combination goes from 1/4 to 1/3 when we remove the possibility of ff. I got thrown by the Tuesday fluff, and somehow read the question as asking the sex of the other child.

[sheepish]
 
Posted by seekingprometheus (Member # 3043) on :
 
...which is funny, cause mariner's post explains it.

Teach me to scan and post.
 
Posted by TomDavidson (Member # 99) on :
 
Interestingly, though, Mariner's post explains the "problem" here incorrectly, or at least in a "mathemagician" sort of way. It is not the case that, of the 27 two-child families who have had a boy on a Tuesday, 13 of them can be assumed to have a second boy. At no point is the probability ever 13/27. (Heck, even if it were 50%, it still wouldn't be 13/27.)

Rather, the odds that a two-child family with one boy has a second boy are 1:3. Of 27 families, then, 9 of them can be assumed to have a second boy. Ergo, the actual odds are 9:27, or 1:3. Which is exactly what you would expect (assuming you understand the original Monty Hall example).

[ July 24, 2010, 12:19 AM: Message edited by: TomDavidson ]
 
Posted by TheOtter (Member # 2917) on :
 
I still say it's 50%.

It really is equivalent to tossing a coin. Forget the day of the week for a moment. If I tell you I tossed a coin twice and it came up heads the first time. Wouldn't you agree that there's a 50/50 chance that the second toss was also heads? However, if you toss two coins many times, you'll come up with the same distribution we've been talking about:

H/H
H/T
T/H
T/T

In a single run of two coin tosses, how can knowledge of the result of the first toss possibly affect the probability of the second toss? It can't. Also, if I tell you the first coin was tossed on a Tuesday, that would still have no bearing on the second toss's result. Any other conclusion must have a logical fallacy, since it's physically impossible for one coin toss to affect another.

I'm not 100% sure what the fallacy is, but I think it's false to continue to apply a percentage taken from the general population to a specific scenario that has already been nailed down in one way or another.
 
Posted by TomDavidson (Member # 99) on :
 
quote:
If I tell you I tossed a coin twice and it came up heads the first time. Wouldn't you agree that there's a 50/50 chance that the second toss was also heads?
Yes. But that's not the question. The question, rather, is: I tossed a coin twice, and it came up heads at least once. What are the chances that both tosses came up heads?

This is a very important distinction.

In your version, the result set is only (H/H, H/T). So there is in fact a 50% chance of another Heads result.

In the original, though, the result set is (H/H, T/H, and H/T), leaving a 33% chance of a second Heads result.

[ July 24, 2010, 01:16 AM: Message edited by: TomDavidson ]
 
Posted by Aris Katsaris (Member # 888) on :
 
"In the original, though, the result set is (H/H, T/H, and H/T), leaving a 33% chance of a second Heads result."

That's assuming that these three possibilities are equal in chance. However if the choice of revealing there exists a head was random (i.e. if the guy could have equally well have revealed the existence of a tails), then this increases the possibility of H/H, making the chance of a second Heads result indeed 50%.

If on the other hand *we* (not knowing the results) chose the question: "Is there at least one Heads?" then you're indeed correct that a positive answer makes the chance of a second Heads result 33%.

--
Think of it like this: If one randomly chooses a continent, and then one randomly chooses to reveal the race of a random person in that continent, then that person being black increases the probability that the continent randomly chosen was Africa.

That's how the revelation of a randomly chosen H, means there's a higher probability it was H/H than T/H. These two probabilities are no longer equal the same way that the probability of Africa is no longer equal to the probability of Europe, if a randomly chosen person from that continent happens to be black.

However if the guy chooses to reveal the race of the person chosen *because* it was an atypical result for the continent, then the probability of the continent being Africa actually diminishes.

This is the bias of selectively revealed information.
 
Posted by seekingprometheus (Member # 3043) on :
 
Otter:

Yeah, I missed it too. It's a probability problem in a riddle-ish form. The math is simple--it's just worded in the same riddle type as the "I have two coins with a summed value of 35 cents. One of them is not a quarter..."
quote:
In a single run of two coin tosses, how can knowledge of the result of the first toss possibly affect the probability of the second toss?
The point is that you don't know whether the result you have comes from the first or second toss.
 
Posted by DonaldD (Member # 1052) on :
 
Aris, what you are missing is that the fact of the sexes of the children is independent of any choice being made by the person making the statement. Unless you are positing that some misogynist fathers would exclude themselves by being unable not to say they have two boys when they do, but in that case, you would be better off factoring in female birth rates and twin factors into the equation.
 
Posted by TomDavidson (Member # 99) on :
 
Aris is making the question more complicated than it needs to be. Yes, the odds of a boy birth vs. a girl birth (as hobsen noted) are not actually 50% in any given instance. Yes, unless the problem explicitly rules out psychological factors, psychological factors will probably affect the outcome.

But I think it's pretty safe to say that the "puzzle" here fairly means us to assume both an equal chance of boy/girl births and no psychological factors. To quibble over these, on this particular question, is to miss the forest for the trees.
 
Posted by DonaldD (Member # 1052) on :
 
Heck, psycholgical factors could lead a father of two girls to claim that he had a boy born on Tuesday...
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
But I think it's pretty safe to say that the "puzzle" here fairly means us to assume both an equal chance of boy/girl births and no psychological factors.
I don't know what "no psychological factors" means: does it mean the guy randomly selects a kid's gender to reveal?

If so, then all you guys are wrong. If he randomly selected a kid whose gender he wanted to reveal, then the fact he mentioned it was a boy, leaves the other one's chance at 50%, not 33%.

If you want it to make it so that the chance is 33% then you rephrase the riddle, so that the other guy asks "Is at least one kid a boy?" and the father replies "Yes."

--
Some people found it weird that knowing the *older* kid is a boy, leaves the other one's chance of boyhood at 50%, but knowing that *a* kid is a boy, makes the other one's chance at boyhood at 33% - but this makes sense because the former is essentially logic A (declare gender of a kid) while the latter is logic B (declare existence of a boy).

quote:
Aris, what you are missing is that the fact of the sexes of the children is independent of any choice being made by the person making the statement. Unless you are positing that some misogynist fathers would exclude themselves by being unable not to say they have two boys when they do, but in that case, you would be better off factoring in female birth rates and twin factors into the equation.
This selective reporting can be done with coin-flips if you don't want to cloud the issue with gender politics. I can easily construct a program simulating coin-flips and using any of the 3 logics I listed, I could completely confuse your attempts to figure out the probability.
 
Posted by TomDavidson (Member # 99) on :
 
quote:
If he randomly selected a kid whose gender he wanted to reveal, then the fact he mentioned it was a boy, leaves the other one's chance at 50%, not 33%.
Um....As far as I can tell, you're confusing this with an actual Monty Hall problem.

In the original Monty Hall problem, it matters whether Monty knows which doors have goats behind them because he is required to open a door that contains a goat. If he were not required to do so, and in fact selected doors at random, then it's true that his opening a door has no effect on the probability that the original door selected contains the car.

But in this problem, John Doe walks up to you and says, "I have two kids, one of whom is a boy born on Tuesday," and you are then asked to compute the odds that he has two boys.

quote:
I don't know what "no psychological factors" means: does it mean the guy randomly selects a kid's gender to reveal?
No. It means that we are invited to assume the guy's motivations and methods are irrelevant. He has two kids, one of whom is a boy born on Tuesday. Full stop.

You can complicate things by speculating about whether or not he knows you're about to put odds down on the sex of his children, but that's outside the scope of the question as asked.

We are not asked to determine whether he has two sons until after we've already learned he has at least one son; there is no scope in which the possibility of two daughters exists.

[ July 24, 2010, 12:11 PM: Message edited by: TomDavidson ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
As far as I can tell, you're confusing this with an actual Monty Hall problem.

No, I think you confuse it for such. You believe the information he gives affects the chances of the remaining options, like the showhost removing one door, increasing the possibility of the remaining door containing the prize.

quote:
But in this problem, John Doe walks up to you and says, "I have two kids, one of whom is a boy born on Tuesday," and you are then asked to compute the odds that he has two boys.
Yes. Assuming he was being random with his choice of kid whose birth/gender information he revealed, the odds he has two boys are 50%. All you guys saying it's 33% are wrong. My math coincides with common sense in this: No matter how many further info he gives, (hey, it's a Capricorn redhaired boy born on Tuesday, on a moonless night, while the wolves howled and the omens looked favourably down upon him) it won't budge this from 50%.

quote:
"He has two kids, one of whom is a boy born on Tuesday. Full stop."
Okay. 50% then. The kids were born before he gave *any* information about them, so ANY information he gives about the gender, day of birth, zodiac signs whatever, won't budge this percentage from 50%.
 
Posted by PSRT (Member # 6454) on :
 
quote:
A clever man would then say it is actually 1/3, since there are 4 possibilities of a two child family: GG, GB, BG, and BB. Since GG is eliminated, there's only 1/3 chance
We're looking at combinations, not permutations. There are only three combinations of two children: GG, BG, BB. We've eliminated one. There are only two combinations left.
 
Posted by JWatts (Member # 6523) on :
 
Mariner , thanks for posting this. It's made for a good thinking exercise.
 
Posted by vegimo (Member # 6023) on :
 
The guy changes the probability when he says that he has two children, and that one is a Tuesday-born boy. This means that we are no longer dealing with the entire population. It also means that any other possibility should be equally likely. There is no motive for his revelation in the statement of the problem. Motive would make it impossible to calculate the true probabilities, so it should be left out. Thus we are left with pure mathematics, and the answer is 13/27.


Treatment A:

Consider the problem first with the Tuesday condition removed to make it easier to understand. In this case, the person only states the gender and you are supposed to calculate the odds of the gender of the other child. It could be a man or a woman, the child could be a boy or a girl, and you could be tasked with calculating whether the other child is a boy or a girl. It does not matter. The theory behind the calculation remains the same. You have to look at how the population has been reduced, and then consider the remaining portion of that subset which would satisfy the conditions of the riddle.

In this case, the population has been reduced by the conditions that there are only two children and that one of them is a boy. Accepting that boys and girls are equally likely (yeah it is math, not genetics, so that is reasonable), there should only be four ways to get two-child families – BB, BG, GB, and GG. The man said that one of the children is a boy, and this eliminates the GG combination. Now the entire population of concern has been reduced to the BB, BG, and GB combinations. Yes, for the sake of argument, BG and GB families are essentially the same, but their combination is twice as large as the BB group. Thus, the probability of the BB result, in this reduced and reduced again population, is 1/3.


Treatment B:

Now consider the problem with the Tuesday condition included. Again, it does not matter that it was a man, that he said one child was a boy, or that he specified Tuesday. The relevancy of this information is that it serves to reduce the sample size. If you include the birth-day in the matrix of possible permutations of two-child families, you will find 196 different possibilities (starting with B on Sunday-BS, BS-GS, GS-BS, GS-GS, BS-BM, BS-GM, GS-BM, GS-GM, and so forth).

In this case, the population has been reduced by the conditions that there are only two children, one of them is a boy, and that boy was born on a Tuesday. This eliminates all possible combinations without those conditions having been met. You are left with 27 possible cases out of the original 196, and this is the new population from which you can determine the probability. Of those, the probability that there are two boys is 13/27.


Additional thoughts:

If you are told birth order, as in, “Of my two children, my oldest child is a boy born on a Tuesday,” you would then find that the probability is always ½ that the other child is a boy. This stays the same no matter how many conditions get thrown in.

(This one is just my theory – I haven’t found any proof of it.)
If you further restrict the population, you change the odds again. Your original sample matrix grows and the subset from which you determine the answer shrinks to a much smaller portion of the total population. The resultant probability gets closer to ½ with each definite (countable) restriction. For example, if the guy were to further restrict the population by saying the one Tuesday-born son was also delivered in January, the probability of the other child being a boy would then be 167/335. If he added that the child was born in the noon hour, the probability would be 4031/8063.
 
Posted by Jordan (Member # 2159) on :
 
quote:
TomDavidson:
Why is it mathematically sound that 13 of the 27 families with a boy born on Tuesday will have two boys? As 1/3 of all two-child families with one boy have two boys, there are thus 9 families in that result set.

9/27 = 1/3.

In other words, the specific day of the week is irrelevant.

Tom, you're starting from the assumption that the day of the week is irrelevant; instead, take it as another piece of information.

There are two sexes, each with probability ½.

There are seven days of birth, each with probability 1/7.

This yields (2 × 7)² = 196 possible combinations of two-child families, each with probability 1/196.

You are given the information that at least one of the children is a) a boy; and b) was born on a Tuesday.

Of the 196 combinations of two-child families, there are 27 combinations in which at least one of the boys was born on a Tuesday. This is our initial restriction, the conditional in our conditional probability.

Of those 27 combinations, there are 13 in which there are two boys.
 
Posted by Jordan (Member # 2159) on :
 
OK, I'm going to take my best shot at explaining why the day of the week is NOT irrelevant.

If you're one of the people who thinks it is, then stop right there for a minute and let go of your preconceptions. Forget for a minute what you think you already know about the problem, and approach it from the perspective that, even though what I've been saying doesn't make sense to you (yet), you might be wrong nevertheless. Please? [Smile]


I think that a lot of people are thinking that the less intuitive answer assumes that there's something magical about the day of the week. There really isn't; it's just another piece of mundane information, just the same as the piece of information that tells us that one of the children is a boy. The important thing about this information is the same thing that's important about every piece of information, which is that it restricts the set of possibilities you're allowed to consider.

What a lot of people do is assume that the day is irrelevant, and just continue with the problem as though every possible set of families is still included in our consideration.

But this is absolutely, completely and explicitly contradicted by the premises set out in the question.

You're not considering every possible combination of boys or girls. You're considering only those combinations in which at least one of the children is a boy born on a Tuesday, and unless all boys are born on Tuesdays, that automatically eliminates an awful lot of configurations.

Not all boys are born on a Tuesday. Only a seventh of boys are born on a Tuesday. And not all combinations of two children with at least one boy will have at least one boy born on a Tuesday. There are only 27 combinations of two children in which one of the children is a boy who was born on a Tuesday, and only 13 of those combinations have two boys. Since all of those combinations are equally likely, we get the answer: 13/27.


Looking at the problem in terms of how each piece of information restricts the possible combinations under consideration might be helpful.

Starting no information about the two children, the probability that both of them were boys is 1/4.

The extra piece of information that one of the two children is a boy restricts the set of possibilities under consideration, and the probability that both of them are boys increases to 1/3.

The extra piece of information that the boy was born on a Tuesday restricts the set of possibilities even further, such that the probability that both of them are boys increases to 13/27.

And the most powerful piece of information yet, that the eldest child is a boy, eliminates a huge selection of combinations from consideration and increases the probability to 1/2.

[ July 24, 2010, 03:36 PM: Message edited by: Jordan ]
 
Posted by Jordan (Member # 2159) on :
 
Aris isn't getting confused; he is legitimately and correctly observing that the question, as phrased, is ambiguous, and thus subject to two interpretations.

Most of us are interpreting the question as, essentially, this:

quote:
There exists a family with exactly two children, at least one of whom is a boy. What is the probability that both of the children are boys?
However, as it was actually expressed, you're left wondering if the guy decided to tell us about a particular child before asking us, instead of intending to tell us if he does (or doesn't) have at least one boy. In that case, the probability is 1/2.

That's sort of what I thought was up last night, thus my remark about assuming point children. [Smile] We're taking a real scenario expressed in conversational English, and reformulating it into a strictly mathematical interpretation. Aris, quite rightly, observes that the question may easily be understood as asking:

quote:
I have two children. The particular child I'm thinking of right now is a boy. What is the probability that the other one is a boy?
As I said last night, I suspect more people would arrive at the "correct" answer if the question were phrased right.
 
Posted by TomDavidson (Member # 99) on :
 
Again; why is it the case that of the 27 families with two children who've had a boy on a Tuesday, 13 have two boys? That's simply not true.
 
Posted by Jordan (Member # 2159) on :
 
You are correct, Vegimo. As you further and further fix the known boy from the sample combinations with additional information, the probability of the other child being a boy gets closer to a half.
 
Posted by TomDavidson (Member # 99) on :
 
Of the 27 families with one boy born on a Tuesday, 9 have a second boy. Why do you think otherwise?
 
Posted by Jordan (Member # 2159) on :
 
quote:
Originally posted by TomDavidson:
Again; why is it the case that of the 27 families with two children who've had a boy on a Tuesday, 13 have two boys? That's simply not true.

Here is the complete list of possibilities:

  1. BMon BTue
  2. BTue BMon
  3. BTue BTue
  4. BTue BWed
  5. BTue BThu
  6. BTue BFri
  7. BTue BSat
  8. BTue BSun
  9. BWed BTue
  10. BThu BTue
  11. BFri BTue
  12. BSat BTue
  13. BSun BTue

Thirteen in total. Seven combinations in which the first boy is born on Tuesday, seven combinations in which the second boy is born on Tuesday, with one shared combination in which both boys were born on Tuesday.

I promise that I'm not trying to confuse or trick you in any way, shape or form, Tom. [Smile]


Edit: if there are only nine, then you should be able to identify the four combinations in my list of thirteen in which one of the children is a girl, or neither of them is born on a Tuesday.

Further edit: you are just too smart not to get this, Tom!! [Smile]

[ July 24, 2010, 04:16 PM: Message edited by: Jordan ]
 
Posted by DonaldD (Member # 1052) on :
 
Maybe a visualization is in order (and start with Jordan's plea regarding setting aside assumptions): imagine a world where the likelihood of any birth being a boy or a girl is exactly 50%. Imagine a world where the chance of any particular child being born on any given day is exactly 1 in 7.

Now imagine each of the two above statistics are completely independent such that boys and girls are equally likely to be born on any given day.

We then get to Mariner's 196 possibilities (on average). Imagine that one sample of each equally likely combination (Sunday Girl/Sunday Girl, Sunday Girl/Monday Girl... Saturday Boy/Friday Boy, Saturday Boy/Saturday Boy) is used.

Now here is the visualization: imagine a gatekeeper that only allows certain fathers into your area (the filter alluded to by Jordan): Two girls? Nope. Fathers of a boy or boys, but not or neither born on a Tuesday? Nope.

The gatekeeper allows exactly 27 fathers through in orde to speak to you. Of these 27, 7 are firstborn Girl, younger Boy (Girl born on any of 7 days, Boy on Tuesday) 7 are firstborn Boy, younger Girl, 6 are firstborn Boy (not Tuesday) and younger Boy (Tuesday), 6 are firstborn Boy (Tuesday) younger Boy (not Tuesday) and 1 is firstborn and younger Boy (both Tuesday). 13 of the 27 are boy-boy combinations.

If you change the filter to be 'boy born Sunday through Friday) you end up with 48/132 (approx 36%)
 
Posted by Jordan (Member # 2159) on :
 
Just a little more help, don't forget the spreadsheet of all the possibilities that I generated earlier in the thread
 
Posted by vegimo (Member # 6023) on :
 
Tom, list the matrix to show the possibilities, then restrict the population and count the results. Probability theory is a predictor for the results, but actual populations are what prove the theory.

Jordan, my guess was about the numbers (167/335 and 4031/8063). I arrived at those numbers through:

denominator = one less than the product of all the possibilities of each restrictor
numerator = one less than half of the original denominator

for no restrictor:
((1/2 * 2 * 2) - 1) / ((2 * 2) - 1) = 1/3

including month, day, and hour:
((1/2 * 2 * 2 * 12 * 7 * 24) - 1) / ((2 * 2 * 12 * 7 * 24) - 1) = 4031/8063
 
Posted by DonaldD (Member # 1052) on :
 
quote:
And the most powerful piece of information yet, that the eldest child is a boy, eliminates a huge selection of combinations from consideration and increases the probability to 1/2
Just checking - you realize this was not part of the original question, right?
 
Posted by Jordan (Member # 2159) on :
 
OK Tom, how about this. I've uploaded a spreadsheet with some lists so you can see how I generated the combinations, and what those combinations are. I'm not asking you to go that far, just try listing what your list of 27 possibilities is, and the nine possibilities in which there are two boys. I'm not sure where we're missing each other. [Smile]
 
Posted by Jordan (Member # 2159) on :
 
quote:
Originally posted by DonaldD:
quote:
And the most powerful piece of information yet, that the eldest child is a boy, eliminates a huge selection of combinations from consideration and increases the probability to 1/2
Just checking - you realize this was not part of the original question, right?
Sorry, I should have quoted or linked. [Smile] I was referring to Brian's modified question in the second post:

quote:
Brian:
Would it change the odds if he said 'my oldest child is a boy'?


 
Posted by Jordan (Member # 2159) on :
 
(Gosh, I don't think I've been this animated since the Monty Hall thread!)
 
Posted by Pyrtolin (Member # 2638) on :
 
Ah- finally spotted what's been twitching me here looking at that spreadsheet. We have to fully account for each child being independent of the other- so {BTue, BTue} isn't complete:

Both boys being born on Tuesday needs to be counted twice, because you don't know whether you're talking about the first child or the second child. So this really brings it back to 14 / 28 or 50%

But we've also accidentally eliminated a bunch of other cases by not considering the 1st child and 2nd child separately. I'll bet that if we step back a stage and account for that with all boy/girl combinations, we'll find another 14 possibilities to bring us back to 1/3 (while adding back in all of the girl/girl possibilities brings us to a 25% chance of a boy/boy set)
 
Posted by Pyrtolin (Member # 2638) on :
 
Actually, I don't have to bet:
b1 Tue-> 7 b2 possibilities, 7 g2 possibilities
b2 Tue-> 7 b1 possibilities, 7 g1 possibilities
g1 (any day)-> 7 b2 Tue possibilities
g2 (any day)-> 7 b1 Tue possibilities

So now we have 14/42 combinations, or 1/3
 
Posted by vegimo (Member # 6023) on :
 
No, each possibility is distinct, and none should be counted twice. There are only 27 possibilities where either child is a boy born on Tuesday. 13 of those have 2 boys.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Originally posted by Pyrtolin:
Both boys being born on Tuesday needs to be counted twice, because you don't know whether you're talking about the first child or the second child. So this really brings it back to 14 / 28 or 50%.

First, I'm really glad you're looking at the data carefully. [Smile]

Second, you're right (and quite perceptive) to pick up on the fact that {BTue BTue} is a special case that isn't being counted twice; your only mistake is in thinking that it should. Look back at the original problem which doesn't involve days of the week: {B B} only appears once. {B B} and {BTue BTue} being special cases that account for two children meeting our precondition, yet are counted only once, is important to understanding the problem fully.

More directly, in the list of the possibilities we have, it doesn't appear twice because there is only one case in which both children are boys born on Tuesday, not two. {BTue BTue} only happens if the first and second children are both boys who are born on Tuesday, and there's only one case in which this happens.

Aside from anything else, the fact that this stood out to you means you're well on the way to getting the answer. [Smile]
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
there's only one case in which this happens.
Only one absolute case? Sure. But that case is twice as likely to occur because there are two ways that it could be arrived at- one for each boy that the original asker might be referring to.

The probability has to account for the fact that he could be referring to either child, so it means that you have to count each possibility once for each child, rather than for the absolute distribution of cases.
 
Posted by Pyrtolin (Member # 2638) on :
 
Further- The actual equivalent from the original problem is the way that Boy/Girl is "counted twice".

It gets double weight becuase B/G and G/B are actually distinct, independent sets, in the same way That B1T/B2T and B2T/B1T are completely independent sets here.

If you can't count BT/BT twice here, you can't count B/G twice in the original set and are limited to just BB, BG, GG as your possibilities, which isn't accurate. The same principle that applied there applies here.
 
Posted by Pyrtolin (Member # 2638) on :
 
Actually, I'm wrong there, but leaving it for thought process. The problem is actually that we're using the wrong initial set because it accidentally happens to reduce properly.

The right basic set is:
B1 -> 1 G2, 1 B2
B2 -> 1 G1, 1 B1
G1 -> 1 G2, 1 B2
G2 -> 1 G1, 1 B1

Because we don't know which child he's referring to. The distribution ends up the same 2/8 BB, 4/8 GB, 2/8 GG, but the actual set is secretly larger.

Once he says that one child is a boy, then we have

lose the 2 GG scenarios, and are left with 2/6 BB and 4/6 GB. Again, those reduce properly so it's easy to accidentally undercount he full set of possibilities.

As Aris has noted- what he's talking about matters, not just the basic distribution.
 
Posted by Pyrtolin (Member # 2638) on :
 
The initial mistake enters because we're incidentally using 2 things from a 2 item set of possibilities. (If there were 3 sexes and he had two kids, the mistake wouldn't slip by so easily) Once you have 2 and 7, they don't reduce so easily, so the same shortcut doesn't work, and trying to use it accidentally factors out cases that need to be accounted for.

[ July 24, 2010, 07:02 PM: Message edited by: Pyrtolin ]
 
Posted by vegimo (Member # 6023) on :
 
No, the reason the B-Tue,B-Tue should only get counted once is because it is only one family. The father in question is not twice as likely to show up as any other father. There are only 196 distinct families, thus only 196 distinct fathers. Each one has the same possibility of being the father you happen to meet. Once he limits the population by specifying the restricting factors, he is still only one father with only one family, and is not counted twice. None of them are. There are only 27 families with a B-Tue included, and 13 of those have two boys.
 
Posted by TomDavidson (Member # 99) on :
 
Hm. In working this out for myself, it turns out to be quite amusing to substitute "born on Tuesday" with "prefers red to blue." And, indeed, I misunderstood. *laugh* Once you have any distinguishing characteristic beyond sex, this breaks down the same way. As Jordan observed on the first page (which I completely failed to understand), the addition of another distinguishing characteristic produces more situations in which both parties might share the same characteristic, thus skewing the totals further away from an expected percentage.

[ July 24, 2010, 07:43 PM: Message edited by: TomDavidson ]
 
Posted by Jordan (Member # 2159) on :
 
quote:
Pyrtolin:
Only one absolute case? Sure. But that case is twice as likely to occur because there are two ways that it could be arrived at- one for each boy that the original asker might be referring to.

I'm afraid not. What we're counting are families, not how many ways he could be asking the question.

quote:
As Aris has noted- what he's talking about matters, not just the basic distribution.
If you go by Aris' interpretation of the question, the conditional (i.e. the whole thing that makes the question mathematically interesting) disappears; every question we've examined so far reduces to, "What's the probability that a child is a boy?" and the answer in every case is ½.

The alternative interpretation, and the one you're "supposed" to take (in order to get the more interesting answers of 1/3 for the first question and 13/27 for the second) is easier to understand if you rephrase the question more precisely:

quote:
A family has two children. Given that at least one of them is a boy (or, given that one of them is a boy who was born on a Tuesday), what is the probability that both of the children are boys?
The subtlety in Aris' interpretation is that we aren't told how the man in question came to be telling us that he had one son, so the question may be "fixed" before you even start; the stricter formulation isn't ambiguous and is the more interesting interpretation that I (and others) are going by to get our figures.

[ July 24, 2010, 07:43 PM: Message edited by: Jordan ]
 
Posted by Pyrtolin (Member # 2638) on :
 
You have to explicitly account for all combinations of "Is he talking about the first child" and "Is he talking about the second child" or you're not getting all the possible sets. The fact that you don't know which one he's referring doubles the problem space. When the problem space is a multiple of two, that factors out, but when it's not, you can't factor it out.

What you're saying would be true if he said "My oldest child is a boy" , because that resolves a specific, relevant element the problem that was otherwise unknown, in the same way that if you were just looking at the B/G matrix, saying that the first child is a boy would resolve the probability to 50/50.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Tom:
Hm. In working this out for myself, it turns out to be quite amusing to substitute "born on Tuesday" with "prefers red to blue." And, indeed, I misunderstood. *laugh* Once you have any distinguishing characteristic beyond sex, this breaks down the same way.

Oh, sweet! I knew you'd get it soon. [Smile]

It's quite delightful, actually. Even something like, "is named after a famous comedian" affects the probabilities, bringing them yet closer to a half; the more specific the conditional, the more families are eliminated and the closer we come to just asking, "what's the chance that this specific child is a boy".

Or, the more specific you are about one of the children, the more likely it is that your precondition only applies to one of the two children. Questions like "the eldest child" can only apply to one of them off the bat, which is why the probability is exactly ½.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Pyrtolin:
What you're saying would be true if he said "My oldest child is a boy" , because that resolves a specific, relevant element the problem that was otherwise unknown, in the same way that if you were just looking at the B/G matrix, saying that the first child is a boy would resolve the probability to 50/50.

You're quite close here to grasping it intuitively. In the same way that saying that one of the children is both male and the eldest fixes the first column (by definition of "eldest"), saying that one of the children is both male and born on a Tuesday fixes most of the first column, except for the one case where both children were boys born on Tuesday.

Consider the question using the "at least one child" phrasing instead of asking how he "selected" one particular child might help.
 
Posted by edgmatt (Member # 6449) on :
 
Jordan - I was talking about this thread with a friend, and I couldn't get my head wrapped around how it's NOT 1/2, but when I thought about it this way it made sense: We are not asking 'what are the chances that this second child will be born a boy' we are asking the question "what are the chances that the man fathered two boys (since we know that he already fathered one)." Which is exactly the same as asking "what are the odds of flipping a coin twice and getting heads each time?" and not "I flipped a coin and got heads, what are the chances I'll get heads again." because the child has already been born.

When I saw it that way, I saw the math: 1/2 x 1/2
which is 1/4. There are 4 out comes to flipping a coin twice, and this has already been stated in this thread: HH, TT, HT, TH. since we can eliminate TT (since we know the first flip was heads) there are only three other possibilities, making it 1/3.

It is definitely not 1/2 though. That's just math, it really can't be argued as a matter of opinion.

I haven't quite got the "Tuesday" part of it yet, but I think it works from the same principle. Something about subtracting the time where BOTH are boys and BOTH are born on Tuesday figures in, but I'm not that strong on this sort of thing.
 
Posted by edgmatt (Member # 6449) on :
 
Or maybe re-phrase to "what are the chances of someone giving birth to a boy and then another boy?"
Which is 1/4, but then we say "ok we had two kids, one was a boy, what are the chances they are both boys?"

Eh I'm having trouble articulating this, yet I can see it in my head without all these damn words getting in the way. I guess that's the issue with the whole thread.
 
Posted by JoshuaD (Member # 1420) on :
 
Jordan: I suspect the following two cases are distinct. Do you agree?

1) A man walks up to you and tells you he has a son and another child. You then ask him what day of the week his son was born and he tells you it was Tuesday. Finally, he asks you the probability for the sex of his other child.

2) A man walks up to you during a meeting of "men who have sons who were born on Tuesday". He informs you that he has two children. Finally, he asks you the probability for the sex of his other child.

Do you believe this situations have identical answers or different answers?

[ July 24, 2010, 10:23 PM: Message edited by: JoshuaD ]
 
Posted by Pete at Home (Member # 429) on :
 
I can't wrap my brain around boy Tuesday.
 
Posted by Jordan (Member # 2159) on :
 
quote:
edgmatt:
We are not asking 'what are the chances that this second child will be born a boy' we are asking the question "what are the chances that the man fathered two boys (since we know that he already fathered one)."

Yes. [Smile] The crucial thing is, he's not asking, "What's the chance that this particular child is a boy?" but "What's the chance that both my children are boys?"

quote:
I haven't quite got the "Tuesday" part of it yet, but I think it works from the same principle. Something about subtracting the time where BOTH are boys and BOTH are born on Tuesday figures in, but I'm not that strong on this sort of thing.
You're dangerously close. If the numbers game is getting in the way, you can get quite close to understanding the why of it intuitively instead of mathematically.


Let's try an easier one. We're talking about coins again. You get a friend to toss two coins one after the other (without watching), and you need to work out what the chances are that he got two heads. You get to ask him one question about his toss, and that's it.

The key is that the more specific you are about which coin you're asking about, the closer the question becomes to, "what's the chances that one particular coin came up heads?" which is always ½.


—

To start with, you don't know anything about the coins. So the chance that both came up heads is 1/4.

—

You ask: "Did at least one of the coins come up heads?" He says yes. So now you know that at least one is heads, but not if it was the first one or the last one. So the possibilities now are:


So the chances are 1/3 that both his coins came up tails.

—

You ask: "Did the first coin come up heads?" He says yes. Now, that's pretty darn specific! It basically eliminates one of the coins completely. So now you're looking at:


In other words, there's a ½ probability that the toss came up {Heads Heads}.

—

You ask: "Did a silver coin come up heads?" He says yes.

Let's assume that coins are either silver or copper, and that both are equally common. Your question is a bit more specific than just asking if one coin came up heads, but not as specific as narrowing it down to the first coin. The possibilities here are:


Count them: there are seven combination of heads and tails, and each is equally likely; however, there are only three cases where both coins came up heads. So the probability is 3/7.

The reason that it's not ½ is that there's a small chance that both coins came up heads and both were silver, and in that case he could have been talking about either of the two coins. The coin being silver gives us more information, but not enough to distinguish between them when both of the coins are silver.

—

Just a final one to amuse you. Let's say that you know that in a tiny number of coins, Roosevelt is clearly wearing a feather boa. You ask: "Did a coin with Roosevelt wearing a feather boa come up heads?" He says yes. Now, you know that it's pretty darn unlikely that he has two of those coins, so you can almost be sure that he was talking about a specific coin and there's only one coin left to guess at. But, in the improbable event that two of his coins had the beboaed Roosevelt, he could have been talking about either of the two.


The idea is that the more likely it is that you've narrowed it down to just one of the coins, the closer you get to only having to guess at what one of the coins is.

[ July 25, 2010, 10:26 AM: Message edited by: Jordan ]
 
Posted by Jordan (Member # 2159) on :
 
quote:
JoshuaD:
I suspect the following two cases are distinct. Do you agree?

No, don't pollute my beautiful maths with tricksy semantics! [Smile]

The first one is very ambiguous. My first reading of it is that the man has to actually select a son to talk about in order to answer your direct question; in this case, it reduces to Aris' interpretation and the probability is 50%.

The second one is very clear, and it comes to 13/27 as per the dictates of conditional probability. [Smile]

So my first reading is that they're distinct; but if you had asked, "tell me the day that at least one of your sons was born on" in the first question, then I'd say the scenarios were identical. The key thing is that he's narrowed down which of his children he might be talking about a bit, but not all the way.
 
Posted by scifibum (Member # 945) on :
 
LOL@beboaed.
 
Posted by JoshuaD (Member # 1420) on :
 
Jordan: My point with those two scenarios is that in the first case the day the person was born on didn't matter. The man and I would be having that conversation regardless. In the second scenario, we'd only be having the conversation iff he had at least one son born on Tuesday.

The first case doesn't filter, the second case does. Do you think this matters?
 
Posted by Jordan (Member # 2159) on :
 
I see. No, it wouldn't matter. If you replied, "Tell me a day of the week on which at least one of your sons was born," the number of eventualities comes out the same regardless of what day of the week he replied with (and there would be less ambiguity that he was selecting the son to tell you about beforehand).

It doesn't matter how you come by the additional information, it's the fact that you have that additional information that counts. As soon as he tells you what day of the week one of his sons was born on, it increases the amount of information you have about his children.


Edited to add: looking at the "friend flipped two coins" question might help to clarify. [Smile]

[ July 25, 2010, 01:38 PM: Message edited by: Jordan ]
 
Posted by Slander Monkey (Member # 1999) on :
 
I didn't really write this in the flow of the conversation, so I apologize if I'm reiterating anthing that's been said already... I just needed to get it out, though it looks like it fits in well with Jordan and JoshuaD's discussion.

There's something to this problem that I think is being missed. It's extremely subtle, and it's bugging the heck out of me.

First, while I would agree that the math is correct for a particular interpretation of the statement made by the father, there remains some significant ambiguity in how the statement should, in fact, be interpreted. Consider the following generic form of the father's statement: "I have two kids. One of them is a boy. He was born on [day of the week]." This is trivially true for any boy, as all boys are born on some day of the week. So let's consider the original problem in which the day of the week is excluded: "I have two kids. One of them is a boy." The probability that the other is a boy is 1/3. Now let's say that after the fact, before revealing the identities of his children the father states: "he was born on [day of the week]." Does the probability suddenly change? He has to have been born on some day of the week, so it's not as if any new information is being offered. So where's the difference coming from?

Here's the deal: when considering two boys with different birth days of the week, the father must choose which boy's birthday to use in statement. What's to stop the father from using Monday in the statement (assuming it's not a lie), when the other boy's birth day of the week is Tuesday? If the father is the one choosing the day, we're left with no new information that would lead to a new probability (unless we know something more about how the father would choose). To illustrate this, imagine that the father uses the following selection criteria: if the father has two sons he will only use Tuesday in the statement if both boys were born on Tuesday (he'll use the other son's birthday in every other case). The father has not lied in any way here, or in my estimation been unfaithful to his statement, and the resulting probability would have to be, if my math is correct, 1/15 (subtract all the cases with a second boy not born on a Tuesday). Of course, if the father were to make the same statement with a different day of the week which would have to have different selection criteria the probability would be different. Just to give another example, if the father always stated the birth day of the week of the older brother we'd be back to 1/3 regardless of the day of the week stated, and if he always states Tuesday if any of the boys are born on Tuesday we get 13/27. I suspect, but haven't worked it out, that we would get 1/3 if he chose at random which boy's birthday to use. Furthermore, there doesn't appear to be any way to constrain the father's original statement in a practical way (without it taking on the flavor of a legal disclaimer) that would give the appropriate interpretation for the original mathematically derived result. On the other hand, if Disney World opened its doors one day only to families with two children, one of whom was a boy born on a Tuesday, we'd be in business. To be more explicit about it, if the father were somehow constrained to acknowledge Tuesday birthdays above all others, or the population has already been filtered then the math is right. Otherwise, I don't think it is.
 
Posted by Brian (Member # 588) on :
 
If I understand Jordan correctly, the more information we have, the more likely it is to be 50%. That we start out with no extra info, so we have to go with pure probability about a theoretical family, but that as we get more info about this particular family, we can narrow it down.
Okay. That makes sense.

But no-one has answered my last question yet. (Or if you did, it went completely over my head)

If you have exactly two kids, and one is a boy, then either the elder is a boy, or the younger is a boy. And everyone agrees that in both of those cases, it becomes 50:50 as to the sex of the other.

So why are the actual odds 1:3 with that level of information?
Does deduction count as information?
 
Posted by Badvok (Member # 1085) on :
 
OMG! What on earth is this thread on about? I can't believe no one has simply pointed out that if the outcomes of two events are not linked then you can not mathematically combine the probabilities!
The probability that John has two boys is 50:50 = the probability the unknown child is a boy. The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).
 
Posted by TomDavidson (Member # 99) on :
 
quote:
The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).
Hee. Badvok, Google the problem we discuss at the beginning of this scenario, the Monty Hall problem. [Smile]
 
Posted by TomDavidson (Member # 99) on :
 
quote:
If you have exactly two kids, and one is a boy, then either the elder is a boy, or the younger is a boy. And everyone agrees that in both of those cases, it becomes 50:50 as to the sex of the other.

So why are the actual odds 1:3 with that level of information?

Elder Boy
Younger Girl

Elder Girl
Younger Boy

Elder Boy
Younger Boy

As you can see, there is one chance in three, because the case of "Elder Boy/Younger Boy" actually collapses the two potential cases "Boy 1 older/Boy 2 younger" and "Boy 2 older/Boy 1 younger." When you know which one is older, you can specifically say "Boy 1 is older" and thus eliminate the special case.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by TomDavidson:
quote:
The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).
Hee. Badvok, Google the problem we discuss at the beginning of this scenario, the Monty Hall problem. [Smile]
The Monty Hall problem is totally different because there is a direct causal link.
 
Posted by JoshuaD (Member # 1420) on :
 
Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

It's not the asking of the question that matters. It's whether the question filtered the input.

Imagine the following scenario: I have a room of 1,000,000 dads who have two children and at least one son. I ask each of them "Is your other child a boy?" How many will answer yes? Approximately 333,333.

Now, same scenario. A room full of 1,000,000 dads who each have two children and at least one is a son. Instead I first ask, because I'm trying to kill time, "What day of the week was your son born on?" and receive an answer. I then continue to ask "Is your other child a boy?" How many will answer yes? Approximately 333,333. Nothing has changed from the first scenario.

Now, the final scenario. I have a room of 1,000,000 dads who each have two children, one of which is a boy who was born on Tuesday. I ask each of these men "Is your other child a boy?" How many will answer yes? Approximately 481,481.

See below for some more interesting contrasted scenarios. I'm about 99% sure this is correct. What do you think Jordan?



[ July 26, 2010, 10:50 AM: Message edited by: JoshuaD ]
 
Posted by TomDavidson (Member # 99) on :
 
quote:
The Monty Hall problem is totally different because there is a direct causal link.
What is the causal link? There is either a goat behind the door or there is not. Opening a door does not "cause" a goat to be placed; it simply removes one possibility from the set of unopened doors.

In the same way, telling someone that one of your two children is a boy removes one possibility from the set of possible children (namely, that you have two girls).
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by TomDavidson:
quote:
The Monty Hall problem is totally different because there is a direct causal link.
What is the causal link? There is either a goat behind the door or there is not. Opening a door does not "cause" a goat to be placed; it simply removes one possibility from the set of unopened doors.

Nope, it doesn't "cause" the goat to appear but the contestant's choice directly affects Monty's choice - which in turn affects the odds faced by the contestant (edit to clarify: because Monty will always pick a goat).
quote:
In the same way, telling someone that one of your two children is a boy removes one possibility from the set of possible children (namely, that you have two girls).

True, but that isn't the probability space. John has at least one son, so he can either have two sons or a son and a daughter - 50:50. The gender of one child doesn't alter the gender of the other child.

[ July 26, 2010, 11:16 AM: Message edited by: Badvok ]
 
Posted by JoshuaD (Member # 1420) on :
 
Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question. If you walks up to him and ask him if he has at least one son and he says no, you're going to simply walk away. It's only in the other 3 scenarios [BG, GB and BB] that you stay and take a guess at the gender of his other child.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by JoshuaD:
Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question.

What question? John states that he has one son! So the probability space only concerns the other child.
 
Posted by Pete at Home (Member # 429) on :
 
quote:
Originally posted by TomDavidson:
quote:
If he randomly selected a kid whose gender he wanted to reveal, then the fact he mentioned it was a boy, leaves the other one's chance at 50%, not 33%.
Um....As far as I can tell, you're confusing this with an actual Monty Hall problem.

In the original Monty Hall problem, it matters whether Monty knows which doors have goats behind them because he is required to open a door that contains a goat. If he were not required to do so, and in fact selected doors at random, then it's true that his opening a door has no effect on the probability that the original door selected contains the car.

And how exactly does the goat transform into a car? Higher math? [Wink]
 
Posted by JoshuaD (Member # 1420) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by JoshuaD:
Badvok: It does alter the probability space. You are going to be disregarding 1/4 of all fathers of two children as you ask this question.

What question? John states that he has one son! So the probability space only concerns the other child.
Read my long post above.

The unspoken assumption of this probability question is that you wouldn't be having this conversation with John at all if he had two daughters.
 
Posted by DonaldD (Member # 1052) on :
 
quote:
What question? John states that he has one son! So the probability space only concerns the other child
Badvok, you are ignoring that birth 'order' matters. You don't know whether 'John' is talking about his elder child or his younger child. There are 3 possible sequences of birth in this case: first born is Girl, second is Boy; first born is Boy, second is Girl; first born is Boy, second is also Boy.

That's 2 ways for one girl to be born and only one way for both children to be boys.

Or think about it this way: in a two-child family, the chance of having one boy and one girl is 50%, two boys is 25% and 2 girls is 25%.

By removing the 2-girl option, you are left with the other 3 options, where having a mix is still a higher proportion.

Or, imagine 4 doors: one with 2 girls, one with 2 boys, one with a boy and a girl, one with a girl and a boy.

Monty opens the door with 2 girls. How many doors are left with 2 boys, and how many are left with only one?
 
Posted by Badvok (Member # 1085) on :
 
Apologies to all - I seem to have gone down the wrong track here, I thought this long meandering thread was talking about:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
The answer to that question is simply 50:50. There is no other answer, we are simply asking for the probability a child is a boy or a girl.
 
Posted by DonaldD (Member # 1052) on :
 
Err, no,it's not [Smile]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Badvok, you are ignoring that birth 'order' matters. You don't know whether 'John' is talking about his elder child or his younger child. There are 3 possible sequences of birth in this case: first born is Girl, second is Boy; first born is Boy, second is Girl; first born is Boy, second is also Boy.[/QB]

Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.
 
Posted by DonaldD (Member # 1052) on :
 
No, you really shouldn't.

Unless you are suggesting that your first born can be younger than your second born child.

But even so, in such a time space you first born daughter could also be younger than your second born son, so that also cancels out.

Think of it in real-world terms to clarify the situation. You have two children: an elder child and a younger child. In how many ways can they be born?
 
Posted by JoshuaD (Member # 1420) on :
 
Badvok: No. If birth order cancels out, then the odds of me having two children of the same sex is 2/3: [B G], [B B], [G G]. This is incorrect. The odds of me having two children of the same sex is 1/2: [B G], [G B], [B B], [G G].

Please try to refrain from knee-jerk responses. Some of this mathematics is unintuitive. Take some time to think about the responses you receive before plowing forward with your insistence; this will save un-necessary repetition in the thread and it will also give you more time to formulate your objections in a coherent and complete way rather than shooting them from the hip as they come to mind.

I was suspicious of the 13/27th reasoning for nearly 4 days before I finally posted my reasoning on why I think it's not exactly correct. The thread will be here when you're done thinking, I promise. :-)
 
Posted by Badvok (Member # 1085) on :
 
OK, to clarify how birth order is irrelevant:

Child1:Child2
Eldest+Boy:Youngest+Boy
Youngest+Boy:Eldest+Boy
Eldest+Girl:Youngest+Boy
Youngest+Girl:Eldest+Boy

= 4 combinations

[ July 26, 2010, 12:05 PM: Message edited by: Badvok ]
 
Posted by DonaldD (Member # 1052) on :
 
You forgot about:

Youngest+Boy:Eldest+Girl
Eldest+Boy:Youngest+Girl

= 6 combinations
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by JoshuaD:
The odds of me having two children of the same sex is 1/2: [B G], [G B], [B B], [G G].

Yes, and?
quote:
Originally posted by JoshuaD:
Please try to refrain from knee-jerk responses.

I'll throw the same back at you!

I SAID birth order cancels out and is irrelevant NOT that the probability of same sex offspring is 2/3 - I don't know where you got that from!
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
You forgot about:

Youngest+Boy:Eldest+Girl
Eldest+Boy:Youngest+Girl

= 6 combinations

Eh? I think I listed those.
Edit: Ooops sorry, no I didn't - need to go home now!

[ July 26, 2010, 12:19 PM: Message edited by: Badvok ]
 
Posted by DonaldD (Member # 1052) on :
 
If birth order is irrelevant, then the only possibilities of child combinations are, as JD mentions, [B B], [G G] and [B G]. If order is truly irrelevant, then each of these options would be accorded the same probability, no? In which case, each situation would occur 1/3 of the time. [B B] + [G G] = 1/3 + 1/3 = 2/3.

I believe JD was attempting reductio ad absurdum.
 
Posted by DonaldD (Member # 1052) on :
 
quote:
Eh? I think I listed those
Nope, you listed their inverse cases only.
 
Posted by DonaldD (Member # 1052) on :
 
I think you may be getting caught up in Aris' option from the previous pages - specifically, that you are thinking of a particular child. Yes, if you rephrase the question as "John Doe comes up to you and says: "I have two kids. My son John is a boy..." then you end up with a 50% answer (unless George mentions his child George, in which case all bets are off.)
 
Posted by DonaldD (Member # 1052) on :
 
quote:
Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.
To focus on this just a little further - if Boy1 and Boy2 are not temporal distinctions, but rather actual names, think of it like this:
Boy1 = John
Boy2 = Sam
But then, you would need to name the girls as well:
Girl1 = May
Girl2 = Jan

In which case, you are more obviously missing some combinations. If you can have [B1 B2] (John/Sam) as well as [B2 B1] (Sam/John) then you should also accept both [B1 G1] (John/May) as well as [G1 B1] (May/John), [B1 G2] (John/Jan) and [G2 B1] (Jan/John)
 
Posted by vegimo (Member # 6023) on :
 
Badvok:
The reason birth order does not cancel out and remains relevant is the same reason you can't count the boys twice. The probability that any particular child will be a boy is 1/2. In a large 2-child-family population though, there will be 1/4 of the families with 2 boys, 1/4 of the families with 2 girls, and 1/2 the families with a boy and a girl. If we exclude the 2-girl families (because of the condition of the problem), there are now 1/3 of the families with 2 boys and 2/3 of the families with one of each. Just because a father has 2 boys does not mean he is twice as likely to be the particular father who approaches you. Any of the fathers could be the proud papa, and when he shticks his boast on you, there is a 1/3 chance that he is one of the fathers with 2 sons and a 2/3 chance that he is the father of a son and a daughter.
 
Posted by Brian (Member # 588) on :
 
Tom:
quote:
As you can see, there is one chance in three, because the case of "Elder Boy/Younger Boy" actually collapses the two potential cases "Boy 1 older/Boy 2 younger" and "Boy 2 older/Boy 1 younger." When you know which one is older, you can specifically say "Boy 1 is older" and thus eliminate the special case.
That was kinda my point. If the specific boy mentioned is older, then the odds are 50:50. If the specific boy mentioned is younger, then the odds are 50:50. Do I have that right so far?

If so, then those are the only two possibilities. That specific boy is either younger or older. Unless his last name is Schroedinger. [Smile]
If the only two choices lead to a 50:50 chance, why would the intermediate odds be different just because we don't know which one it is yet?
 
Posted by scifibum (Member # 945) on :
 
My thought after reading this thread: it's a lot harder to figure out how probability rules apply to informally described questions than it is to do the math.
 
Posted by DonaldD (Member # 1052) on :
 
Brian, the range of truthful statements possible is not limited to 'my eldest is a boy' or 'my youngest is a boy'.

In the population in question, there is also the possibility of saying 'my eldest is a girl' and 'my youngest is a girl'. This occurs in 1/3 of the cases in the population in question.

So, 50% (boy or girl after identifying a boy) * 2/3 (likelihood of father chosing to identify a boy in this population) still gives 1/3.

Whereas the chance of the 'other' child being a boy when the father identifies a girl (in this population) is 100%. 100% * 1/3 is also 1/3.
 
Posted by vegimo (Member # 6023) on :
 
Brian,

(Another wording of DonaldD's response)
Your issue with the problem leads back to what the question actually asks. We are not trying to figure out whether either child 1 is a boy or child 2 is a boy. We are trying to figure out whether the father belongs to the portion of the population with 2 boys or the portion of the population with a boy and a girl.
 
Posted by PSRT (Member # 6454) on :
 
Well, DonaldD, we don't really know what the population in question is. How was the man chosen? We're told he approaches us. But that doesn't give us enough information about the population he is approaching us from. If he is drawn from the population of all fathers with two children, the answer is 1/2. If he is drawn from the population of fathers with two children, one of whom is male, the answer is 1/3. If he is drawn from the population of fathers who have a son born on Tuesday, the answer is 13/27.

Our assumptions about the English language matter in solving this problem.
 
Posted by Jordan (Member # 2159) on :
 
quote:
JoshuaD:
Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

You are very possibly correct, this could become a situation where the teacher becomes the student. I'll get back to you tomorrow after having a proper think. [Smile]
 
Posted by DonaldD (Member # 1052) on :
 
quote:
Well, DonaldD, we don't really know what the population in question is
PSRT: Actually, we do. Right now, we are working with the simplified population of father with at least one boy.

If we were talking about the population of Father with a Tuesday boy, the math gets more complicated but the methodology would be similar (and yes, we would end up with the 13/27 probability)

The fact that we have used the father's statement to identify the population does not preclude the possibility of different truthful statements being made about the children in question.
 
Posted by Mariner (Member # 1618) on :
 
Cool, I dropped a bomb, ran away, and created lots of interesting discussion.

Thanks Jordan, for articulating clearly the issue that I was gradually understanding as I read the thread. We have a range of probabilities being argued here: 1/3, almost 1/2 (13/27), and 1/2. The reason the second one (Boy born on Tuesday) is almost 1/2 is because, as Jordan said, by adding more information it becomes more and more likely that you're describing a specific child rather than making a general observation.

So with that said, let's pick some scenarios in which we can (mostly) all agree that the probability Joe has 2 boys is 1/2:
- Joe comes up to you and says: "I have two kids. The oldest is a boy."
- Joe comes up to you and says: "I have two kids. The youngest is a boy."
- Joe comes up to you and says: "I have two kids. One of them is named Jake." (barring the highly improbable occurence that he named a girl Jake or that he named both sons Jake)
- Joe comes up to you and says: "I have two kids." A boy then walks up and says "Hi Dad" to Joe.

In each of these cases, Joe is referring to a specific child when the sex of one of them is revealed. When that happens, the question of whether or not Joe has two boys really boils down to the question of whether or not Joe's OTHER child is a boy. And there, we know its 50-50. That's why the "born on Tuesday" bit is relevant under Jordan's interpretation of the question. It increases the odds that the "at least one is a boy..." is referring to a specific boy. That wasn't clear to me at first, but now I understand.

But I don't think Jordan's entirely right that the issue is only due to the ambiguity of the English language. It does come down to when and where the selection is made. Aris is right in that the intention of Joe is important. Or more exactly, I think the key here is less the ambiguity of the English language, and more that you don't know all the rules of the game. I'm going to switch to flipping two pennies here because it's easier to comprehend:

New scenario: Joe flips two pennies. He says "At least one is heads" Jordan responds "I bet you have one heads and one tails" Jordan then provides his reasoning. There are four initial possibilities:
HH - 25% chance
HT - 25% chance
TH - 25% chance
TT - 25% chance
100% total

Joe then specifically eliminated the final possibility. In other words, Joe truthfully saying "at least one is heads" can only happen 75% of the time. Therefore, you eliminate the final possibility, and divide each of the other chances by 75%
HH - 33% chance
HT - 33% chance
TH - 33% chance

Jordan notes that the combination of one head and one tail is twice as likely as the head-head combination, and therefore concludes that its a safe bet.

But is that what actually happens? Let's change the rules of the game a bit. Once again, Joe flips two coins. He then feels absolutely compelled to say "At least one is [heads/tails]" (and he must speak truthfully, obviously). Now, there are four possibilities of what can happen:

HH - Joe is forced to say "At least one is heads" - 25% chance
HT - Joe can choose whether he wants to say "At least one is heads" or "At least one is tails" - 25% chance
TH - Joe can choose whether he wants to say "At least one is heads" or "At least one is tails" - 25% chance
TT - Joe is forced to say "At least one is tails" - 25% chance

Assuming Joe is not biased in terms of picking heads or tails, there are now a grand total of 6 possibilities:
1) HH - "At least one is heads" - 25% chance
2) HT - "At least one is heads" - 12.5% chance (the original 25% multiplied by 50% of choosing to say heads)
3) HT - "At least one is tails" - 12.5% chance
4) TH - "At least one is heads" - 12.5% chance
5) TH - "At least one is tails" - 12.5% chance
6) TT - "At least one is tails" - 25% chance

Joe says "At least one is heads" Looking at the above, you eliminate options 3, 5, and 6. In other words, Joe truthfully saying "At least one is heads" only happens 50% of the time (note that this is the key difference between this and Jordan's logic: Jordan says that he CAN say it 75% of the time, but here I say that he WILL say it 50% of the time). So, once again, we take the available options and divide them by the probability that this happens, and we get:
1) HH - "At least one is heads" - 50% chance
2) HT - "At least one is heads" - 25% chance
4) TH - "At least one is heads" - 25% chance
Now, if Jordan says that he bets that 1 is heads and 1 is tails, he only has a 50-50 shot.

By the way, in a third scenario, Joe flips two coins. Jordan asks "Is at least one heads?" Joe nods. Jordan can then go through his initial reasoning and correctly obtain the 2/3 probability of one head and one tail. This is functionally equivalent to the probability that Joe CAN truthfully say "at least one is heads".

So the question is, is Joe's statement that at least one of his kids is a boy born on Tuesday coupled with the original game (randomly selected two child families) or not? In other words, do we care whether or not Joe CAN say "One is a boy born on a Tuesday" or whether he WILL say "One is a boy born on a Tuesday" if he's going to mention the sex and the date?

To put it into precise terms for methematicians to use, we have two scenarios:

"A family has two children. Given that at least one of them is a boy who was born on a Tuesday, what is the probability that both of the children are boys?" Answer: 13/27

"A family has two children. Given that the father will tell you the sex and gender of one of them, what is the probability of both children being boys if he says that at least one is a boy born on a Tuesday?" Answer: 1/2

And from the original scenario, I don't know if you can with complete certainty choose which scenario is correct.
 
Posted by scifibum (Member # 945) on :
 
Quibble:
quote:
- Joe comes up to you and says: "I have two kids." A boy then walks up and says "Hi Dad" to Joe.
Since you don't know whether this kid is the oldest or the youngest, this does not seem to me to narrow the possibilities beyond the generic "has two kids at least one of whom is a boy".

Edit: Unless knowing the kid's hair color, height, or approximate age has anything to do with it. Which it doesn't feel like they should. But my intuition has been wrong on this from the start, so...

(I'm comforted by the fact that it would be difficult to construct an accurate field of equally-likely variations on these particular attributes. Or would it? *head explodes*)

[ July 26, 2010, 04:53 PM: Message edited by: scifibum ]
 
Posted by JoshuaD (Member # 1420) on :
 
Mariner and I are in agreement on how this question is properly approached. Really cool lesson in probability. I spent years as a successful poker player, and this still sent me through a loop for a few days.
 
Posted by DonaldD (Member # 1052) on :
 
Mariner: in your second scenario, you have made a mistake: the fact that in the HT and TH cases the person chooses to state either heads of tails does not affect the probability of the combination actually occuring.

So even though there may only be a 12.5% likelihood of him saying 'H' in either case, it doesn't change the number of occurences of actual 'HT' or 'TH' combinations.

If, however, you are positing that the guesser knows what is going on inside the first guy's head... well in that case, we could make up just about anything. Essentially, the question comes back to the following: in the puzzle, what is the purpose of the father's initial statement? Is it to limit the population, or to trick you with language?
 
Posted by Mariner (Member # 1618) on :
 
Scifibum, the "oldest" vs "youngest" is just a useful tool to keep them separated, it doesn't actually matter. What matters is that sex determination is independent of one another. When we say that order matters, we don't necessarily mean chronological order. It could be in terms of height, or which one is furthest north at the moment, or which one managed to get through the most posts in this thread without their head exploding. It's just to insure that the two children are independent.

So by having one of the children appear, you are specifically defining one of the children. Because of that, you are specifically dealing with one unknown child when asked if there are two boys.

Imagine if a child was hiding under a cardboard box, and you had to guess whether it was a boy or girl. Imagine then if said child's brother walked in the room. Would that change your guess? No. You are still guessing about the sex of only one child. And if a boy comes up to Joe and says "Hi Dad", then there's only one kid left that you have to guess.

Donald, there is no mistake. Yes, the odds of HT appearing are 25%. HT appears in scenario 2 and 3 in my list above. If you add the probabilities of scenario 2 and 3 together, you get 25%. Exactly what you would expect.

What I did was no different then the mathematical way to solve the Monty Hall problem. Suppose I choose Door #1. There are 4 possibilities:
1) The prize is in #1, and door #2 is opened to show the goat
2) The prize is in #1, and door #3 is opened to show the goat
3) The prize is in #2, and door #3 is opened to show the goat
4) The prize is in #3, and door #2 is opened to show the goat

The odds of each occurring are 1/6, 1/6, 1/3, and 1/3 respectively. Thus, the odds that the prize are in door #1 is still 1/3, since you add the probability of scenarios 1 and 2 together. After a door opens (we'll say it's door #2), you collapse the options and redetermine the probability. In this case, it becomes 1/3 that the prize is in door #1 and 2/3 that it's in door #3. Same thing as I did above.

And you're right that the question comes back to the purpose of the father's initial statement, but your options are misleading. It's actually: is it to describe a specific child (answer is 1/2), or is it to eliminate specific options (answer is 1/3).

Now to confuse myself further. What if Joe had said "I have two kids. But I don't have two girls." Now, it's clear that the purpose of his statement is to eliminate specific options, but it's functionally identical to "at least one is a boy". So now what?
 
Posted by PSRT (Member # 6454) on :
 
quote:
PSRT: Actually, we do.
Not from the question as posed to us. If we wish to specify which set of parents our hypothetical father has been drawn from, we can do that, and then get an exact answer. Without specifying, the question as worded is ambiguous enough that there are multiple correct answers, depending upon interpretation.
 
Posted by scifibum (Member # 945) on :
 
Mariner, that's not satisfying. Back to this scenario:

A man has two kids, at least one of whom is a boy. The probability that he has two boys is 1/3.

The reason that it's 1/3 is that there are three distinct (and equally probable, for argument's sake) ways for the man to get two kids, at least one of whom is a boy. BB, BG, GB. The only thing that distinguishes the latter two scenarios is birth order.

By having a boy who forms one member of one of the three duos above walk into the room, you haven't eliminated ANY of the three scenarios. Whereas knowing the boy is the oldest DOES eliminate the last scenario.

[ July 26, 2010, 09:43 PM: Message edited by: scifibum ]
 
Posted by Badvok (Member # 1085) on :
 
OK, I got a little messed up yesterday in trying to put this straight.

I guess we need to go back to basics:

Example 1:
What is the probability of a child being a boy or a girl?
Answer: Two possibilities (in this simplified mathematical world anyway), therefore 1/2

Example 2:
If there are two children what is the probability of them both being boys?
Answer: Four possibilities (B+B, B+G, G+B, G+G), therefore 1/4

Example 3:
If there are two children what is the probability of one child being older than the other?
Answer: Almost 1, but for the sake of this question lets call it 1 (the chances of finding two children born at exactly the same moment in time is pretty slim and almost impossible if they are from the same mother - sorry girls no matter how horrible it sounds it is not totally impossible).

Example 4:
If there are two children what is the probability of them both being boys if one is a boy?
Answer: OK, this is the main question.
See example 1, we have 2 possibilities for gender of each child.
See example 2, we have 4 possibilities for combinations of two children: B+B, B+G, G+B, G+G.

Now, this is the contentious bit, for the sake of clarity we need to identify the children we are
talking about so we don't get confused, we'll use androgynous names and call them Child1 and Child2.

Child1 is either a boy or a girl, the probability of either is 1/2.
Child2 is either a boy or a girl, the probability of either is 1/2.

If we collapse the first statement to say that Child1 is a boy it has no effect on the second statement and we now have:

Child1 is a boy, probability 1.
Child2 is either a boy or a girl, probability 1/2.

Therefore the probability of both Child1 and Child2 being boys is 1 X 1/2 = 1/2.

Or if we collapse it the other way around:

Child1 is either a boy of a girl, probability 1/2.
Child2 is a boy, probability 1.

Therefore the probability of both Child1 and Child2 being boys is 1/2 X 1 = 1/2.

The classic mistake many are making is to take only part of the probability space from example 2 - only those that contain a boy - doing this is mathematically incorrect - you can't split probability spaces like that!

Example 5:
Does age or birth order affect the result of Example 4?

As we see from Example 3, the probability that one child is older than the other is 1 and therefore it will have no effect on the calculation.

Or to put it another way:

If Child1 is the elder, what is the probability that Child2 is the younger? Answer 1.

And Conversely:

If Child1 is the younger, what is the probability that Child2 is the elder? Answer 1.

If we leave aside real-life biology we can also say that birth order does not affect gender probabilities and therefore it has no relevance to a problem space concerning only gender.

But if we must examine this further to satisfy those who still have doubts then we see that we have the following permutations:

Child1 is eldest and a boy, Child2 is youngest and a boy
Child1 is eldest and a boy, Child2 is youngest and a girl
Child1 is eldest and a girl, Child2 is youngest and a boy
Child1 is eldest and a girl, Child2 is youngest and a girl
Child1 is youngest and a boy, Child2 is eldest and a boy
Child1 is youngest and a boy, Child2 is eldest and a girl
Child1 is youngest and a girl, Child2 is eldest and a boy
Child1 is youngest and a girl, Child2 is eldest and a girl

(Sorry DonaldD, we both missed some yesterday.)

Now because we have introduced age/birth order we have doubled the size of the problem space but we haven't actually changed anything. The probability of both children being boys is now 2/8 = 1/4.

And if we collapse the problem space by saying Child1 is a boy we get:

Child1 is eldest and a boy, Child2 is youngest and a boy
Child1 is eldest and a boy, Child2 is youngest and a girl
Child1 is youngest and a boy, Child2 is eldest and a boy
Child1 is youngest and a boy, Child2 is eldest and a girl

So the probability of both children being boys is 2/4 = 1/2.

The key thing to remember is that there are two children and they are distinct and separate entities. One child's gender is not linked in any way to their sibling's gender (in this mathematical problem space anyway, even if in real-life there are quite significant links).

Q.E.D.

[ July 27, 2010, 05:41 AM: Message edited by: Badvok ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Originally posted by JoshuaD:
[QB]
Now, same scenario. A room full of 1,000,000 dads who each have two children and at least one is a son. Instead I first ask, because I'm trying to kill time, "What day of the week was your son born on?" and receive an answer.

Really? What the hell of an answer will people with two sons give you? This question is meaningless and unanswerable for people with two sons.

quote:
I then continue to ask "Is your other child a boy?" How many will answer yes?
Other of what? If someone with two sons says "I have at least one son", then asking if your other child is a boy is meaningless question because there's no "other". He wasn't referring to any particular boy.

quote:
Scenario 5: I have a room of 1,000,000 fathers who have 2 children where one child is a boy. I stand them in a line and ask them each "What day was your son born on?" If answer any day other than Tuesday, I ask them to leave.
Again a meaningless question for fathers with two sons.

quote:
the specific boy mentioned is older, then the odds are 50:50. If the specific boy mentioned is younger, then the odds are 50:50.
If the father is referring to a specific child, then the odds are 50:50.
If the father isn't referring to a specific child, then the odds are 1/3.

That was what my whole contribution to the thread was about.

quote:
If the only two choices lead to a 50:50 chance, why would the intermediate odds be different just because we don't know which one it is yet?
Because when a father says "atleast one of my sons" is a boy, the possibilities don't collapse to
a- I chose to mention younger son.
b- I chose to mention older son.

they collapse to this:
a- I fathered one boy then one girl.
b- I fathered one girl then one boy.
c- I fathered two boys.

I urge you to look at my own earlier contribution to this thread where I argue that the different reasons for the father mentioning the info he mentioned, affects the probability.

-----

This thread has derailed back into an earlier stage, with people forgetting the fact we've mentioned several times already: "atleast one son" doesn't pinpoint to one particular child. You can't ask a father the day his "atleast one son" was born, because they may be more than one.

Read the thread before you post, people.

[ July 27, 2010, 06:34 AM: Message edited by: Aris Katsaris ]
 
Posted by Aris Katsaris (Member # 888) on :
 
Look at conditional probability please.

Probability(A given B) = (Probability of A AND B) / Probability (B)

Probability(two boys, given atleast one boy) = Probability(two boys) / probability(atleast one boy) = (1/4) / (3/4) = 33%
 
Posted by Aris Katsaris (Member # 888) on :
 
The thing that Badvok is arguing (correctly) is that the probability for gender of the second child is independent of the gender of the first child.

What he's arguing wrongly is that he thinks this means the probability of two boys is independent of the probability of one boy.

You can't merge probabilities the way Badvok is doing.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
Look at conditional probability please.

Probability(A given B) = (Probability of A AND B) / Probability (B)

Probability(two boys, given atleast one boy) = Probability(two boys) / probability(atleast one boy) = (1/4) / (3/4) = 33%

Not relevant:
quote:
When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B.
In this case the events are not linked and therefore conditional probability does not apply.
Unless you are categorically stating that having one boy child affects the likelihood of the other being a boy (though true biologically this is not really the case here).
 
Posted by DonaldD (Member # 1052) on :
 
quote:
And if we collapse the problem space by saying Child1 is a boy we get:
Badvok, you went through all that effort, and you just repeated the semantic issue in the end. By limiting the statement to a very specific child, yes you reach the 1/2 value. But your statement above is exactly equivalent to saying "And if we collapse the problem space by saying my son John is a boy we get". Do you see how this is different from the interpretation "at least one of my children is a boy"?

Now take that same interpretation, which in your case is exactly equivalent to 'either child1, child2 or both are boys' and apply it to your 8 combinations and see what you get.

As Aris has so testily observed, been there, done that.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
What he's arguing wrongly is that he thinks this means the probability of two boys is independent of the probability of one boy.

The probability of one boy is 1 - this is stated in the question.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Now take that same interpretation, which in your case is exactly equivalent to 'either child1, child2 or both are boys' and apply it to your 8 combinations and see what you get.

But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
 
Posted by DonaldD (Member # 1052) on :
 
quote:
In this case the events are not linked and therefore conditional probability does not apply.
This conditional probability calculation requires the events to be independent (ie, random, not linked) so it very much could apply.
 
Posted by DonaldD (Member # 1052) on :
 
quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".
It is "Child1 is a boy". John has declared the gender of one of his children - this child can be labelled as Child1. The question is then: What is the probability the other child (labelled as Child2 so as to mark it as a separate and distinct entity to Child1) is also a boy?

[ July 27, 2010, 07:38 AM: Message edited by: Badvok ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Not relevant:
Not relevant?? That's what conditional probability is all about. That's pretty much the ONLY relevant thing.

quote:
What is the probability the other child is a boy if one is a boy?
No, the question is what is the probability that *BOTH* are boys if "at least one" is a boy.

You are misquote the quote to indicate that one particular child was mentioned to be a boy. NO. No particular child was declared a boy. The only thing declared was that ATLEAST ONE OF THEM was a boy.

Not "the other" - "both"
Not "if one" - "if atleast one"

quote:
The probability of one boy is 1 - this is stated in the question.
That's what Probability of (A given B) means. That you assume the B happens and it. To calculate (A given B) with the formula i gave, you calculate the probability that B would have to happen (WITHOUT you knowing it did happen).

You don't understand conditional probability.
You take B by itself. The you take the possibilities of A and B together. Then to calculate (A given B) you divide the probabilities of (A and B) by probability of (B).


If we know one *particular* child is a boy, then the probabilities are:
(Both boys, given one particular child a boy) = (Both boys)/(one particular child a boy)
= (1/4) / (1/2) = 50%.

If we don't know that one particular child is a boy, then:
(Both boys, given atleast one boy)= (Both boys)/(atleast one boy) = (1/4)/(3/4) = 33%

Math proves you wrong.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
Math proves you wrong.

LOL, I give up, I'll let you all head off to kindergarten now while I get on with some real work.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by DonaldD:
[qb]
quote:
But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".
Yes, sort of: if "one of them is a boy" not "child1 is a boy".

It is "Child1 is a boy". John has declared the gender of one of his children - this child can be labelled as Child1.
If he did pinpoint to a particular child, you could label it.

But he didn't, so you can't.
 
Posted by Aris Katsaris (Member # 888) on :
 
These are simple highschool math, Badvok. I don't know what country you're from, but it needs an improvement of its educational system.
 
Posted by Aris Katsaris (Member # 888) on :
 
And not just in math, since you don't even seem to understand what the word "given" means.
 
Posted by TomDavidson (Member # 99) on :
 
Aris, could you try to refrain from just straight-up insulting people?
 
Posted by Aris Katsaris (Member # 888) on :
 
For the sake of anyone who's less of an ass than Badvok:

(Both boys, given atleast one boy born on Sunday) =
(Both Boys AND a boy born on Tuesday) / (one boy born on Tuesday) = (13/196) / (27/196) = 13/27

--

Since nobody bothers reading the explanations any more, I'll stick with the so-called kindergarten math.
 
Posted by Aris Katsaris (Member # 888) on :
 
Tom, I told him that math proves him wrong, and in response he told all of us to go to kindergarten.

So, **** him. He's a moron and an *******.
 
Posted by TomDavidson (Member # 99) on :
 
I think what Badvok's saying is that the mere fact that one of the boys can now be distinguished as "the boy you were told about" changes the matrix, even if you don't know which boy you were told about, as follows:

Boy I told you about / Girl
Girl / Boy I told you about
Boy I told you about / Boy
Boy / Boy I told you about

The reason this isn't true is that in the latter case, there is no way to distinguish between the two boys. Ergo, the real space is:

Boy I told you about / Girl
Girl / Boy I told you about
Boys, one of which I told you about
 
Posted by DonaldD (Member # 1052) on :
 
I think Badvok just got a little frustrated, Aris. Patience.

Basically, it is not a math issue with Badvok, but a semantic one. It comes down to why the 'child I told you about who I will name Boy 1' is not the same as 'one of the two children' ("I have two kids. One of them is a boy"). Badvok, what you need to get your head around is that the father might also be talking about child2 instead of child1. There is no way for us to know (see Tom's statement above). And this ambiguity does not increase the probability of 2 boys being born out of two.

Again, Badvok, look at the possible combinations from a first born/second born perspective:
1st. 2nd
Girl Girl
Girl Boy
Boy Girl
Boy Boy

That is it. By naming the boys in the last case (labelling them if your prefer) you don't make that case more probable.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
I think Badvok just got a little frustrated, Aris. Patience.

Yep, sorry.

quote:
Again, Badvok, look at the possible combinations from a first born/second born perspective:
1st. 2nd
Girl Girl
Girl Boy
Boy Girl
Boy Boy

That is it. By naming the boys in the last case (labelling them if your prefer) you don't make that case more probable.

Nope it doesn't but you have now labelled them 1st and 2nd instead [Smile] I thought labelling would help understanding but it obviously didn't.

If 1st is a boy then 2nd can be either a boy or a girl.
If 2nd is a boy then 1st can be either a boy or a girl.

The probability of both being boys is 1/2.

The error people are making is in using the known to collapse the space [GG,GB,BG,BB] to [GB,BG,BB] which is incorrect. Because one half of the pair has become known the space becomes [kG,kB,kG,kB] or [Gk,Gk,Bk,Bk] where k is the known. In this space the probability that the unknown is B is 2/4 or 1/2.
 
Posted by Ciasiab (Member # 6390) on :
 
First let me say that I am not a mathematician / statistician, so I am not conversant in statistical spaces. That being said, as an engineer, I like to check my answer to make sure it is in the right ballpark.


Here is the problem statement:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?

Since John Doe basically volunteered the information without being probed, to me this would be the same as the following hypothetical situation (ignoring day of the week for now):

I have a room full of 1,000,000 dads. Each is the dad of two children. I ask each dad to tell me the gender of one of their children. Based on my simple statistical training, I would predict that ~500,000 dads would tell me they have a boy and ~500,000 dads would tell me they have a girl. I would also expect ~500,000 dads to have mixed gender families.

Based on the logic by Donald, Aris, and others, if the Dad said that he had at least one boy, then there is a 66% chance he has a boy and a girl. So that gives me 333,333 mixed gender families. Using the same logic on the dad's who said they had at least one girl gives another 333,333 mixed gender families. Totaling the mixed gender families gives 666,666 (impossible!).

The only way the problem works out to 2/3 is if we restate the problem as follows:

John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

In this case the probability of two boys is 66%.
 
Posted by threads (Member # 5091) on :
 
quote:
Originally posted by Badvok:
[GG,GB,BG,BB] to [snip] [kG,kB,kG,kB].

You transformed GG to kG where k is the known BOY.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by threads:
quote:
Originally posted by Badvok:
[GG,GB,BG,BB] to [snip] [kG,kB,kG,kB].

You transformed GG to kG where k is the known BOY.
It doesn't matter whether the known is a boy or a girl, the known is simply a known and becomes a constant that no longer affects the probability.
If the known is a boy the probability of both being boys is equal to the probability the unknown is a boy.

[ July 27, 2010, 09:28 AM: Message edited by: Badvok ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
I have a room full of 1,000,000 dads. Each is the dad of two children. I ask each dad to tell me the gender of one of their children.
No, no, NO! A million times NO, you don't ask that!

You don't ask each dad to tell the gender of one of their children. YOU ASK THEM WHETHER THEY HAVE atleast one boy.

75% of dads will have atleast one boy, and say "YES!"

This splits into 25% of dads with GB, 25% of dads with BG, and 25% of dads with BB.

So, 1 out of 3 dads that replied YES will have a second boy, and 2 out of 3 dads (out of those that replied YES) will NOT have a second boy.


quote:
Based on the logic by Donald, Aris, and others, if the Dad said that he had at least one boy, then there is a 66% chance he has a boy and a girl.
I'm getting close to tears here. We've all repeated a hundred times that asking dad to name the gender of one of his children is DIFFERENT to asking him about whether he has atleast one boy.

You know why? Because 75% of dads have atleast one boy, but only 50% of dads would select a boy if you ask them to pick randomly a gender of one of their kids.

Do you understand now, please?
 
Posted by Pete at Home (Member # 429) on :
 
Badvok, think of this:

Situation 1: Man A tells you he has 2 kids. What's the possibility that his oldest kid is a boy?

Situation 2: Man B tells you that he has two kids, and that at least one of them is a boy. What's the possibility that the oldest kid is a boy?

But I still don't get boy Tuesday.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
The only way the problem works out to 2/3 is if we restate the problem as follows:

John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

In this case the probability of two boys is 66%.

The probability of two boys here is unambiguously 33%.

Thanks for the rephrasing, btw, it's the exact thing that I suggested needed be done, for the statistics to be clear.
 
Posted by DonaldD (Member # 1052) on :
 
Well, this is entertaining.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Pete at Home:
Badvok, think of this:

Situation 1: Man A tells you he has 2 kids. What's the possibility that his oldest kid is a boy?

1/2 = the same as whether any individual kid is a boy.
quote:

Situation 2: Man B tells you that he has two kids, and that at least one of them is a boy. What's the possibility that the oldest kid is a boy?

Now that is a totally different question and is more like the Monty Hall problem because you have made a selection (the oldest) that directly affects the probability. You have now narrowed the possibilities to 3 (eByB,eByG,eGyB) and hence the probability that the oldest is a boy is 2/3.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
But I still don't get boy Tuesday.
I don't blame people that don't get boy Tuesday, he's counter to intuition. The 7x7x2x2=196 possibilities are btw too much to keep in mind, so it gets even more confusing.

So instead of days of the week, let's say Democrat/Republican. Let's consider even distributions. The 4 possibilities:
BB, BG, GB, GG
now become 16 possibilities:
BD/BD, BD/BR, BR/BD, BR/BR
BD/GD, BD/GR, BR/GD, BR/GR
GD/BD, GD/BR, GR/BD, GR/BR
GD/GD, GD/GR, GR/GD, GR/GR

John Doe: I have two children
Me: Is at least one a republican boy?
John Doe: Yes.

Out of those 16 possibilities the following 7 remain:
BD/BR, BR/BD, BR/BR,
BR/GD, BR/GR,
GD/BR, GR/BR

We know for sure dad is one of these possibilities.
Out of these 7 possibilities however only 3 of them contain a second boy.

So there's a 3/7 chance for a second boy.

--

Again math reaches the same conclusion:
(Probability of two boys, given probability of a republican boy) = (Probability of two boys AND a republican boy)/(Probability of a republican boy) = (3/16) / (7/16) = 3/7
 
Posted by DonaldD (Member # 1052) on :
 
quote:
If the known is a boy the probability of both being boys is equal to the probability the unknown is a boy.
Badvok, you still don't know which child is 'known'. The father is a black box. And since you don't 'know' that, the child is not 'known' in fact.

"I have two kids. One of them is a boy". Do you agree that this statement is effectively the same as "I have two kids. They are not both girls"?
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
You don't ask each dad to tell the gender of one of their children. YOU ASK THEM WHETHER THEY HAVE atleast one boy.
I guess I can't see how you can possibly get this from the original problem statement of John Doe walking up and declaring he has a boy.

The only way I see your logic working is if the problem is restated with probing questions being asked, instead of information declared. If I declare (unasked) information about one of my children, it does not change the probability of my other child being a particular gender. That just doesn't work as shown above.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
"The only way I see your logic working is if the problem is restated with probing questions being asked, instead of information declared."
Which is pretty much what I argued at the whole first page of this thread, so we're in agreement here.

Probing questions is the best way to create unambiguous situations, because declared information has the problem of selection bias on behalf of the father.

IF the father is randomly selecting the gender of one his kids to reveal, there's 50% chance that he has two boys. But IF the father only wants to reveal the existence of a boy, there's 33% chance that he has two boys.

In short, everything I said in lots and lots of detail in the 1st page of this thread.
 
Posted by Badvok (Member # 1085) on :
 
I think it is a lost cause trying to explain this yet again but here goes:

There are two distinct children.
Their genders are not interdependent.
Their genders are not determined by birth order.
Their gender is not dependent on the day of the week they were born on.

We know one is a boy.

Therefore there is only one single sole distinct individual solitary unknown gender value and that can have only one of two possible values.

All this collapsing of probability sets/spaces incorrectly just confuses the issue and you are bound to get weird and wacky numbers.
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
IF the father is randomly selecting the gender of one his kids to reveal, there's 50% chance that he has two boys. But IF the father only wants to reveal the existence of a boy, there's 33% chance that he has two boys.
It sounds like we don't disagree. From the problem statement, there is no way I would assume that the father only wants to reveal the existence of a boy. All we can tell from the problem statement is that the father has chosen one of his kids, and revealed that this particular kid was born on a Tuesday and happens to be a boy.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
"We know one is a boy."
We know the existence of a boy. And the existence of a boy IS interdependent with the probability that a specific child is a boy.

quote:
"Therefore there is only one single sole distinct individual solitary unknown gender value and that can have only one of two possible values."
The combination "one girl and one boy" is twice more likely to occur in a population than the combination "two boys".

Again, man: out of 100% dads with two boys: 75% will have at least one boy. But only 25% of the dads will have two boys.

Please, can you respond whether you agree or not with the above fact?
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
It sounds like we don't disagree. From the problem statement, there is no way I would assume that the father only wants to reveal the existence of a boy. All we can tell from the problem statement is that the father has chosen one of his kids, and revealed that this particular kid was born on a Tuesday and happens to be a boy.
Yes, that's what I argued in the first page of this thread.
 
Posted by Ciasiab (Member # 6390) on :
 
Badvok

If the problem were stated as follows:
John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

Would you agree the probability of two boys changes from 1/2 to 1/3?
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Ciasiab:
Badvok

If the problem were stated as follows:
John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes.

Would you agree the probability of two boys changes from 1/2 to 1/3?

Nope, sorry. It is still 1/2. No matter how you re-phrase it. There is still only one unknown that can have one of only two values.
 
Posted by DonaldD (Member # 1052) on :
 
Badvok:

"I have two kids. One of them is a boy".

Do you agree that this statement is effectively the same as:

"I have two kids. They are not both girls"?
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
The combination "one girl and one boy" is twice more likely to occur in a population than the combination "two boys".

Yep. But that is the wrong question.
The answer to the original question is 'The combination "one girl and one boy" is just as like to occur in a population (of fathers with at least one son) as the combination "two boys"'.

quote:
Again, man: out of 100% dads with two boys: 75% will have at least one boy. But only 25% of the dads will have two boys.
No I think out of 100% dads with two boys 100% will have at least one [Smile]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Badvok:

"I have two kids. One of them is a boy".

Do you agree that this statement is effectively the same as:

"I have two kids. They are not both girls"?

Well yes, but turning the question around doesn't change the number of unknowns.
 
Posted by DonaldD (Member # 1052) on :
 
quote:
Well yes, but turning the question around doesn't change the number of unknowns.
Correct. Bear with me.

In the case of a 2-child family, do you agree the the following four cases are the only possibilities?
first child born is Girl, second born is Girl
first born is Girl, second born is Boy
first born is Boy, second born is Girl
first born is Boy, second born is Boy
 
Posted by Badvok (Member # 1085) on :
 
Yep, I think I said that earlier.
 
Posted by DonaldD (Member # 1052) on :
 
I assume you agree that each case has an equal chance of occuring. I also assume you agree that the structure of the puzzle implies that all these births predate the father's statements and as such the father's statements can have no effect on the probability of each combination.

Now, given that you are presented with a father of two who claims that he has no daughters, what are the remaining possible family structures available?
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
'The combination "one girl and one boy" is just as like to occur in a population (of fathers with at least one son) as the combination "two boys"'.
Except, that it's really really not.

Think about it like this. A population of fathers with 7 children, and at least 6 daughters. What's the chance they'll have 6 daughters and 1 son?

The possibilities are the following:
GGGGGGG
GGGGGGB
GGGGGBG
GGGGBGG
GGGBGGG
GGBGGGG
GBGGGGG
BGGGGGG

So 6 daughters and 1 son are SEVEN TIMES MORE LIKELY than 7 daughters, if it is given that the collective number of daughters is at least 6.

If however the six FIRST children are daughters, then the possible populations are only these:
GGGGGGG
GGGGGGB

So, we're back at 50% chance for a boy existing.
--
In short, not knowing *which* children the father is referring to, increases the number of possibilities for the non-matching children.

If the daughters' locations are perfectly pinned down, there's only possible position for the boy, which reduces his chance to a mere 50%.

[ July 27, 2010, 11:30 AM: Message edited by: Aris Katsaris ]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
I assume you agree that each case has an equal chance of occuring. I also assume you agree that the structure of the puzzle implies that all these births predate the father's statements and as such the father's statements can have no effect on the probability of each combination.

Yes.

quote:
Now, given that you are presented with a father of two who claims that he has no daughters, what are the remaining possible family structures available?
Do you want to rephrase that - I don't think it is asking what you wanted to ask. If he has no daughters then he can have only two sons.
 
Posted by DonaldD (Member # 1052) on :
 
oops - "who claims he does not have two daughters."
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Now, given that you are presented with a father of two who claims he does not have two daughters, what are the remaining possible family structures available?

Either two boys or a boy and a girl.

[ July 27, 2010, 11:37 AM: Message edited by: Badvok ]
 
Posted by DonaldD (Member # 1052) on :
 
Uh uh. Go by family structure. You did accept that there were initially 4 possibilities, and each had an equal chance of occuring already...

Now that you know the father's family is not a two-daughter family, what is left?
 
Posted by Badvok (Member # 1085) on :
 
Yeah, I guessed where you were going [Smile]

The statement "does not have two daughters" gives the following mutually exclusive possibilities:

first born is Boy or Girl, second born is Boy
or
first born is Boy, second born is Boy or Girl

Only the gender of one child can be indeterminate/unknown and that gender can only be one of two values.
 
Posted by DonaldD (Member # 1052) on :
 
I think you are trying to obfuscate now. You do see how you repeated yourself above, correct? [Smile]

Let me answer the question, and you tel me how this is wrong. Of the 4 possible family structures:

first child born is Girl, second born is Girl (25%)
first born is Girl, second born is Boy (25%)
first born is Boy, second born is Girl (25%)
first born is Boy, second born is Boy (25%)

The first family structure is excluded since the father does not have two daughters. That leaves you with:
first born is Girl, second born is Boy (25%)
first born is Boy, second born is Girl (25%)
first born is Boy, second born is Boy (25%)

Do you disagree?
 
Posted by Pete at Home (Member # 429) on :
 
c, don't you mean changes from 1/4 to 1/3?
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
Originally posted by Badvok:
Yep. But that is the wrong question.
The answer to the original question is 'The combination "one girl and one boy" is just as like to occur in a population (of fathers with at least one son) as the combination "two boys"'.



But he wasn't picked out of the population of fathers with one son. He was picked out of the population of all fathers with two children.

And even in the population of fathers with two kids, one of who is a boy, twice as many have a boy and a girl than do two boys.


quote:
quote:
Again, man: out of 100% dads with two boys: 75% will have at least one boy. But only 25% of the dads will have two boys.
No I think out of 100% dads with two boys 100% will have at least one [Smile]
Let's get even simpler here since you missed that the 25% was a subset of the original 75%

You have 100 fathers with two kids.
How many have at least one boy?
How many have only one boy?
How many have two boys?
 
Posted by Ciasiab (Member # 6390) on :
 
Pete:

No. I was making the distinction between the case where we are asking the father questions (i.e. do you have at least one son?) to the case where the Dad declares information about one of his two children. In the first case, we ask the dad if he has at least one son. If he answers in the affirmative, then the odds are 1/3 that he has two sons.

In the second case (this is how I think the problem is stated), the father chooses one of his kids and starts giving you information about that child (i.e. he is a boy, he was born on Tuesday, he likes the color green and votes Republican). In this case the odds that the other child is a boy is 1/2. Since the father arbitrarily chooses one child to give random information about, it does not affect the probability of the other child's gender.
 
Posted by Pyrtolin (Member # 2638) on :
 
Ciasiab:
That doesn't change the fact that _you_ don't know which child he's giving information about, until he narrows it down for you. You can only assess the situation from the information that you have, not what he's got up his sleeve, until he lets you know it. You don't know for sure that he's chosen a specific one to think about until he goes on to give you specifics (And then you don't know for sure that those specifics don't apply to both of them and that he's still talking in the general sense)
 
Posted by Ciasiab (Member # 6390) on :
 
Pyrtolin,

It doesn't matter that I don't know which child he is giving information about. The key is that he does. The way the problem is stated, the man walks up and tells you information about one of his kids.
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday."
John is clearly giving information about one of his two kids. Regardless of how much information John gives you about this particular kid, the odds of the other one being a boy or a girl is 50-50.

Now if the problem was stated as an interview, then things change. That's when we don't know for sure he's chosen a specific one to think about. If we ask do you have a boy that was born on a Tuesday, and he answers yes, the odds are 13/27 that the other one is a boy.
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
John is clearly giving information about one of his two kids. Regardless of how much information John gives you about this particular kid, the odds of the other one being a boy or a girl is 50-50.
Unless both of his kids were born on a Tuesday. From your perspective you cannot be clear that he's talking about a specific one until he tells you that he's talking about a specific one. Intuition and common sense don't apply here; if the information isn't explicitly given, you can't imagine that you have it.

Even if he's volunteering the information, he still came out of the full distribution, where it is twice as likely that a parent will have a boy and a girl than two boys.

He may know which one he's talking about, but you don't have the information to say whether he's talking about one or the other until he provides it.
 
Posted by Ciasiab (Member # 6390) on :
 
Pyrtolin,

It absolutely does matter how we come across the information. Say we have 100 dads with two kids. 25 have two girls, 25 have two boys and 50 have one of each. Now we ask each father in the room to tell us information about at least one of their kids. 50 dads will likely choose to tell us about a daughter. 50 dads will likely choose to tell us about a son. If I understand your logic correctly you predict that of the 50 dads of girls, 2/3 will have a boy. Of the 50 dads of boys 2/3 will have a girl. This just doesn't add up. Where am I going wrong?

[ July 27, 2010, 02:01 PM: Message edited by: Ciasiab ]
 
Posted by Ciasiab (Member # 6390) on :
 
In the problem statement John Doe volunteers that he has at least one boy. We know he comes from the general population, and 2/3 of the population with at least one boy also have one girl. 1/3 of the population would have 2 boys given that at least one is a boy. (We agree to this point).

Since he is volunteering information, if he has one girl and one boy, there is only a 50% chance he would tell us about a boy. If he has two boys there is a 100% chance he would tell us about a boy. Hence the final odds of a boy and a girl can be calculated as follows: ((2/3)*(1/2))/(2/3)=1/2
 
Posted by Jordan (Member # 2159) on :
 
quote:
JoshuaD:
Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

After some more careful parsing, I think you're right—I was wrong. But I think you may be partially wrong about why I was wrong. [Wink]

Aris identified where I flubbed: I asked a meaningless question. There isn't any way that you can ask that question without forcing the father to select one of his sons. In fact, no matter how I think of it, the closest to a meaningful means of asking the question is: "Go and stand over there if you have at least one son who was born on a Tuesday." The fathers who go where you indicated have a 13/27 probability of having two sons.

Incidentally, Aris has shown possibly the clearest understanding of the problem throughout, though his patience obviously got a bit frayed towards the end! [Smile]
 
Posted by Ciasiab (Member # 6390) on :
 
Jordan,

How about: "Do you have at least one son who was born on Tuesday?"
 
Posted by Brian (Member # 588) on :
 
Pyrtolin:
quote:
Intuition and common sense don't apply here; if the information isn't explicitly given, you can't imagine that you have it.
Where the heck were you 3 pages ago?!?

I'm almost certain I asked if deduction is the same as information.

This is why I suck at some forms of math. I don't "understand" it, I intuitively know how it works. When intuition doesn't work (I'm looking at you, quantum mechanics!) then the only result I get is a pounding headache.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Ciasiab:
How about: "Do you have at least one son who was born on Tuesday?"

Basically what I was doing, but addressed to Joshua's hypothetical million fathers instead of just one. It's kind of hard to know who's saying "yes"! [Wink]
 
Posted by Wayward Son (Member # 210) on :
 
quote:
When intuition doesn't work (I'm looking at you, quantum mechanics!) then the only result I get is a pounding headache.
Intuition works with quantum mechanics.

Just not human intuition. [Smile]
 
Posted by Badvok (Member # 1085) on :
 
I think both Mariner and Aris have attempted to complicate this and confuse the issue to a very high degree but I do not know whether their motives for doing so are malicious or mischievous.

Unfortunately some have fallen into the trap cunningly presented to them.

If we look at this from a pure mathematical perspective and not from a motivational psycho-analysis perspective then:

The original proposition contained three statements of fact made by one person (i.e. not a person selected from a population or a randomly selected person being interviewed):

I have two kids.
(We now have two unknown genders.)
One of them is a boy.
(We now have only one unknown gender.)
He was born on a Tuesday.
(We still have only one unknown gender.)

The second statement does not remove one possibility from the two unknown gender probability space. It alters the probability space to be that of only a single unknown gender.

I don't know if you have ever seen this one:

"One day 3 women each obtained £10 from work making £30 in total. They located a TV priced £30 and decided to buy it. They went in the shop and gave the manager the £30 and started to carry the TV home. The manager realized that the TV was on sale giving £5 off. The manager told his assistant to take £5 from the till and return it to the women. However the assistant decided to keep £2 for him self leaving only £3 to give back to the women. So each lady paid £9 pounds each instead of £10 each. So adding the 3 lots of £9 gives £27 and including the £2 the assistant kept gives £29. So what happened to the missing pound?"

But the answer is the same - if you do the maths wrong you get the wrong answer!

[ July 28, 2010, 06:54 AM: Message edited by: Badvok ]
 
Posted by TomDavidson (Member # 99) on :
 
quote:
The second statement does not remove one possibility from the two unknown gender probability space. It alters the probability space to be that of only a single unknown gender.
No, it does not. You still do not know which of his children is a boy. Saying, "I have two children, at least one of whom is a boy," leaves you with a 33% chance of having two boys.

Seriously. I have explained to you several times why this is the case. But I'll do it again:

quote:

"I have two children."
Result set:
Boy/Boy
Girl/Girl
Boy/Girl
Girl/Boy

quote:

"At least one of them is a boy."
Result set:
Boy/Boy
Boy/Girl
Girl/Boy

quote:

"The boy I mentioned was born on a Tuesday."
Result set:
Boy Tuesday/Boy Sunday
Boy Tuesday/Boy Monday
Boy Tuesday/Boy Tuesday *collapsed
Boy Tuesday/Boy Wednesday
Boy Tuesday/Boy Thursday
Boy Tuesday/Boy Friday
Boy Tuesday/Boy Saturday
Boy Tuesday/Girl Sunday
Boy Tuesday/Girl Monday
Boy Tuesday/Girl Tuesday
Boy Tuesday/Girl Wednesday
Boy Tuesday/Girl Thursday
Boy Tuesday/Girl Friday
Boy Tuesday/Girl Saturday



[ July 28, 2010, 09:10 AM: Message edited by: TomDavidson ]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by TomDavidson:
You still do not know which of his children is a boy.
Seriously. I have explained to you several times why this is the case.

And I have been trying (and obviously failing) to point out that it doesn't matter which one is a boy just that one is a boy and therefore the other is either a boy or a girl. Two options only = 50:50 chance of each.
It wouldn't make the slightest difference if he said he had 50 kids and that at least 49 of them were boys. The probability that they are all boys still comes down to only one single gender question.

[ July 28, 2010, 09:12 AM: Message edited by: Badvok ]
 
Posted by Badvok (Member # 1085) on :
 
Try thinking of it a different way:

I put a child in a room and ask you to guess the gender. What are your chances of being right?

I now say that I'm going to put a boy in the room too. Now what are your chances of being right about the genders?

[ July 28, 2010, 09:19 AM: Message edited by: Badvok ]
 
Posted by TomDavidson (Member # 99) on :
 
quote:
And I have been trying (and obviously failing) to point out that it doesn't matter which one is a boy just that one is a boy and therefore the other is either a boy or a girl.
Except that you're wrong. Seriously. You really are.

The question is not "what is the probability that my other child is a boy." It is "what is the probability that both my children are boys."

I know it seems ridiculous. It seems like these should be the same answer, because on the face of it these appear to be the same question. But they're not. And this is because you have no mechanism that would permit you to "fix" in place one of the children and say "this particular child, of two, is a boy, and therefore we only need to solve for the other child." As the question is framed, there is no way to tell which child is a boy.

If it would help to imagine this problem with four children, go ahead. You have four children. Three of them are boys. What are the odds that they are all boys? (Hint: it is not 50%.)

[ July 28, 2010, 09:23 AM: Message edited by: TomDavidson ]
 
Posted by Aris Katsaris (Member # 888) on :
 
We have given you the numbers, Badvok.

100 fathers with 2 children: With random even distribution 75 of these have atleast one son, but only 25 of them have two boys.

This means that only 1/3 of fathers with atleast one son have two sons.

Can you pinpoint to us which one of these numbers you don't accept?
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
We have given you the numbers, Badvok.

100 fathers with 2 children: With random even distribution 75 of these have atleast one son, but only 25 of them have two boys.

This means that only 1/3 of fathers with atleast one son have two sons.

Can you pinpoint to us which one of these numbers you don't accept?

Your numbers are not wrong in that sense.
1/3 of the 75 fathers (from a population of 100) who have at least one son have two sons.
But you are just solving the wrong problem and hence using the wrong numbers for the problem. We have ONE father, ONE son, and ONE other child who's gender we do not know.

[ July 28, 2010, 09:41 AM: Message edited by: Badvok ]
 
Posted by DonaldD (Member # 1052) on :
 
I told you all it was semantics [Smile]

Badvok, why did you quit on me above. You were soooo close [Wink]

Here is the original statement of fact, Badvok: "I have two kids. One of them is a boy"

Which of the following is/are equivalent to the above?

"I have two kids. The first born is a boy"
"I have two kids. Either the first or the second born (or both) is a boy"
"I have two kids. My child John is a boy"
"I have two kids. At least one of them is a boy"
"I have two kids. Exactly one of them is a boy"
"I have two kids. They are not both girls"
"I have two kids. The child I am currently thinking of is a boy"

Choose as many as you think are correct, as there is some overlap.

[ July 28, 2010, 10:21 AM: Message edited by: DonaldD ]
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
But you are just solving the wrong problem and hence using the wrong numbers for the problem. We have ONE father, ONE son, and ONE other child who's gender we do not know.
Right- and given the overall distribution of choices, there are two chances that that unknown child will be a girl and one that it will be a boy, because boy-girl families are twice as common as boy-boy families.

We have one father and one boy, sure. but they were picked from the full set of families, some of which had no boys.

quote:
I put a child in a room and ask you to guess the gender. What are your chances of being right?

I now say that I'm going to put a boy in the room too. Now what are your chances of being right about the genders?

That's not a comparable situation. When you put two kids into the room, the possibility existed that you put two girls in as well. It's not till _after_ you randomly picked them that you checked one at random and found out that it was a boy. The process of selecting both kids was completely random.

If I put my hand into a bag with an equal number of red and blue marbles and grab two at the same time, what's the chance that I pull out two blue ones together?

Does this change if, after I've pulled the two out, I randomly pick one to look at and see that it's blue? It's twice as likely that, in the initial grab, I got one of each than it is that I got either specific case of having two of the same. Knowing what one of them is after the random selection doesn't change that initial likelihood.

This is the point your missing here. We're not locking the parent in as having a boy then asking what the next/other kid will be. We're assigning two kids randomly up front, then discovering that one of the assignments happened to be a boy.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Badvok, why did you quit on me above. You were soooo close [Wink]

LOL, I had to head off to catch my commuter train home.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Pyrtolin:
We have one father and one boy, sure. but they were picked from the full set of families, some of which had no boys.

Nope that is not what was stated!
quote:

When you put two kids into the room, the possibility existed that you put two girls in as well. It's not till _after_ you randomly picked them that you checked one at random and found out that it was a boy. The process of selecting both kids was completely random.

Nope, I put one child in the room then added a boy. I didn't select two random children.

quote:
If I put my hand into a bag with an equal number of red and blue marbles and grab two at the same time, what's the chance that I pull out two blue ones together?

If you want to use balls then it is one blue ball and one ball that might be either blue or red in the bag. What is the probability there are two blue balls in the bag?

Or how about:

I put a stuffed toy and a child in a room and ask you to guess the child's gender. What are your chances?

I magically change the stuffed toy into a boy and tell you that is what I have done. What are your chances now?
 
Posted by DonaldD (Member # 1052) on :
 
Badvok, you are chasing your tail.

You are arguing that the puzzle involves only a single unknown, and then trying to show us the math.

Here is the thing: everybody on this thread, if they were to agree that there is only a single unknown and that unknown has two possibilities of equal probability, would agree that the chance is 50/50. But nobody is arguing against that math.

But that is not how the vast majority of people here interpret the puzzle in question. Go back to the puzzle statement, the words themselves, and we can discuss what they mean first.

You could start with my post from 10:18 as a shortcut...
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
You are arguing that the puzzle involves only a single unknown

Yep! The only math I have tried to correct is the erroneous elimination of an option from a probability space. This is, I think, the key cause of most people here misinterpreting the problem.
Like the nonsense math problem I posted earlier it is mangled maths that leads to the 1/3 probability.
The use of children seems to cloud the issue with the introduction of birth order and birth day.
If I take the original statement:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
and rephrase it without using children:
quote:
I have a box that contains two chess pieces. One of them is white. It is also scratched. What are the odds that they are both white?
Does this help at all?

[ July 28, 2010, 11:02 AM: Message edited by: Badvok ]
 
Posted by DonaldD (Member # 1052) on :
 
OK, so you are now working with the original puzzle (less the Tuesday information, but OK for now).

But before going off on your tangent, let's finish with mine, alright?

Which of the following is/are equivalent to puzzle statement above?

"I have two kids. The first born is a boy"
"I have two kids. Either the first or the second born (or both) is a boy"
"I have two kids. My child John is a boy"
"I have two kids. At least one of them is a boy"
"I have two kids. Exactly one of them is a boy"
"I have two kids. They are not both girls"
"I have two kids. The child I am currently thinking of is a boy"

Choose as many as you think are correct, as there is some overlap.
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by Pyrtolin:
We have one father and one boy, sure. but they were picked from the full set of families, some of which had no boys.

Nope that is not what was stated!


It very explicitly is. This isn't a world where everyone starts with one boy and then has children after that, it's a world that includes the initial possibility that the man could have had two girls before he explicitly eliminated that possibility.

quote:
quote:

When you put two kids into the room, the possibility existed that you put two girls in as well. It's not till _after_ you randomly picked them that you checked one at random and found out that it was a boy. The process of selecting both kids was completely random.

Nope, I put one child in the room then added a boy. I didn't select two random children.


And that's not a parallel situation. The parent in question had both children randomly before he offered you any information. The fact that there was a boy to reveal in the first place was part of the random generation of the possibilities. Letting you know after the fact doesn't change the initial field from which the possibilities were drawn.

quote:
quote:
If I put my hand into a bag with an equal number of red and blue marbles and grab two at the same time, what's the chance that I pull out two blue ones together?

If you want to use balls then it is one blue ball and one ball that might be either blue or red in the bag. What is the probability there are two blue balls in the bag?

I put a stuffed toy and a child in a room and ask you to guess the child's gender. What are your chances?

I magically change the stuffed toy into a boy and tell you that is what I have done. What are your chances now?

[/qb][/quote]

Those aren't comparable to the initial statement. He didn't say "I have a boy, what will my next child be" He said "I have to randomly selected kids. One of them turned out to be a boy." Both were picked at random before the information was given. Revealing the information after the fact doesn't alter the initial odds; until he said something, there was still a 25% chance that he had two girls.

Go back to the room with 100 people. 75 of them will have one boy. If you pick on of those 75 at random, 66% of them will have a girl as well and only 33% will have two boys.

The parent in the initial question was one from that pool of 100, not one from a more limited pool of only ones whose first child was a boy. So revealing that they have a boy only eliminates the possibility that they were one 25 who have no boys, none of the rest. You still have 75 possibilities, only 25 of which have two boys.

The information was only fixed after both selection were made randomly.
 
Posted by Badvok (Member # 1085) on :
 
Please! Someone please tell me that my English is not that bad that no one can understand me!
Why do people like Pyrtolin keep quoting me but then not understanding what I actually said?
There is no field from which a selection was made in the original statement:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
Where is the field? I can't see it! All I see is one father, one son and one other child.
Please open my eyes! Or are we talking about an assumed field? As we all know Assume makes an ...
 
Posted by Jordan (Member # 2159) on :
 
quote:
Badvok:
Unfortunately some have fallen into the trap cunningly presented to them.

I can assure you it's not a trap. The question is badly worded for the conclusion that you're meant to draw from it, but it is absolutely not a trap.

Let's see exactly how you're approaching this. I'm going to try and reword the question a little more clearly:

quote:
In front of you are two screens, standing side by side. You are told that there is one person behind both of them. You are further told that at least one of the two people behind the screeens is male. What is the probability that both of the people behind the screens are male?
How would you go about answering this?

[ July 28, 2010, 11:27 AM: Message edited by: Jordan ]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Jordan:
How would you go about answering this?

You're having a laugh aren't you?
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
Originally posted by Badvok:
Please! Someone please tell me that my English is not that bad that no one can understand me!
Why do people like Pyrtolin keep quoting me but then not understanding what I actually said?
There is no field from which a selection was made in the original statement:
quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
Where is the field? I can't see it! All I see is one father, one son and one other child.
Please open my eyes! Or are we talking about an assumed field? As we all know Assume makes an ...

The field is that of all possible fathers. You're not in a magical world that only has this father or only fathers whose first child was a boy. Until he opened his mouth he could have had 100 daughters, for all you knew. When he does talk, he reveals that he's from the specific set of fathers with two children, and then narrows that to the set of fathers with two children, one of which is a boy. But all of those conditions were established randomly before he said a word- the only thing that has changed is the information that you have to work with.

He didn't say We had a boy, and then another child. So you have to count in the situations where he had a girl first and then a boy as well.

So if you really must have a situation where you put a known sex into the room, you have to count both ways: Put a boy in the room, and then have a child of unknown sex, you'll get a boy 50% of the time in that case. Then put girl in the room and count the cases where the unknown, from that perspective, happened to be a boy, as the statement about having a boy applies to that situation as well.

In the first situation, you've got a 50% chance of getting two boys, but in the second, your chances are 0%, but only half of the case in the second scenario are valid. So out of 3 possible valid situations, only one results in two boys.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Pyrtolin:
The field is that of all possible fathers. You're not in a magical world that only has this father or only fathers whose first child was a boy.

Where did I ever say that the first child was a boy? Where does order come into it?
Where did I say he was the only father in the world? He is simply the only father in the problem space.
 
Posted by DonaldD (Member # 1052) on :
 
Now, as to your tangent.

I'm going to simplify it a bit, because your 'scratch' bit is not equivalent to 'Tuesday', but we're not anywhere near discussing Tuesday yet.

quote:
I have a box that contains two chess pieces.
Full stop. What are the chances that the box contains two white pieces? Two black? A mix?

The answer: 25%/25%/50%. Agreed?

Now, someone looks at one of the pieces. He does not touch them. He does not say anything - yet.

By looking, has he changed the above probabilities?

Think about it this way - there are (yes) 100 boxes, 100 men, and an infinite number of chess pieces in a barrel (with an equal number of white and black pieces.)

The 100 men each grab 2 pieces and put them in a box. How many boxes would you expect to contain 2 black pieces? 2 white pieces? A mix?

The answer (on average) would be 25/25/50, right? In fact, you look into each of the 100 boxes and verify this fact. you then leave the room and the men trade around their boxes.

Now, one of these men looks into his box at the pieces. By looking, he has not changed the pieces. He tells you that one of the pieces is white. What can you immediately, unequivocally state about the actual contents of the box? That there are not 2 black pieces. Agreed? Anything else?

If the only additional fact you have now is that there are not two black pieces, you can logically exclude the possibility that your box is one of the twenty five boxes with both black, but nothing else.

In which case, you know that your box is one of the 75. You also know the contents of these 75 boxes, and that only 25 of them have two white pieces.

Now, seriously, answer my previous post [Smile]
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by DonaldD:
Now, one of these men looks into his box at the pieces. By looking, he has not changed the pieces. He tells you that one of the pieces is white.

Noooo! You are changing the problem space again! The set of possibilities is only those where one of the pieces is white. There are no boxes where there is not at least one white piece because this is a stated fact! (Plus there is actually only one box.)
quote:
Now, seriously, answer my previous post [Smile]
I don't think there is much point do you?

[ July 28, 2010, 12:02 PM: Message edited by: Badvok ]
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by Pyrtolin:
The field is that of all possible fathers. You're not in a magical world that only has this father or only fathers whose first child was a boy.

Where did I ever say that the first child was a boy? Where does order come into it?
Order comes into it because the order of the children introduces additional possibilities. The ambiguity of the order is exactly what's at key here and shifts the overall probability. The problem in all of your attempts to model the situation is explicitly that you don't account for the possibility of different orders or that he may have had two girls until he revealed order ambiguous information.

quote:
Where did I say he was the only father in the world? He is simply the only father in the problem space.
No- all fathers are in the problem space. He just happens to be the one that stepped forward and asked to be analyzed.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Pyrtolin:
He just happens to be the one that stepped forward and asked to be analyzed.

Now that is a very big assumption! What did I say about assume?

[ July 28, 2010, 12:07 PM: Message edited by: Badvok ]
 
Posted by TomDavidson (Member # 99) on :
 
I'd also like to present my previous example.

You meet a man who says he has four children. He then tells you "at least three of them are boys."

What are the odds that he has four boys? (As I said before, it is not 50%.)

----------

The problem you're having is that you're looking at the remaining child in question and saying "there is a 50% chance that this child is a boy." That's correct.

But you don't know which child in the set is the remaining child. Let's say he has four boys. Which one of the four boys was the one he didn't mention? Now let's say that he has three boys and one girl. Of the four possible combinations of three boys and a girl, which combination is actually the correct one? Based on the information provided, we don't know. Ergo, we see one possible valid combination of four boys and four possible valid combinations of three boys and a girl.
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by TomDavidson:
Ergo, we see one possible valid combination of four boys and four possible valid combinations of three boys and a girl.

If you want to take order into account then you need to treat each child as a distinct entity and then there is a lot more than one combination for four boys - I'm too tired to work it all out now (time for me to head home) but I'm sure you can.
 
Posted by Pyrtolin (Member # 2638) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by Pyrtolin:
He just happens to be the one that stepped forward and asked to be analyzed.

Now that is a very big assumption! What did I say about assume?
That's not an assumption. That's axiomatic. Using anything less than the general population when such a limit isn't explicitly stated in the problem is making an assumption.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Originally posted by Badvok:
quote:
Originally posted by Pyrtolin:
He just happens to be the one that stepped forward and asked to be analyzed.

Now that is a very big assumption! What did I say about assume?
A big assumption? It is indeed a big ASSUMPTION as I detailed in the first page of this thread, which you don't seem to have read.

However, you've repeatedly refused to acknowledge that this assumption affects the probabilities in any way. The quiz was rephrased to you as following:

"John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes."

And you KEPT insisting that the probability for two sons is 1/2 even with this rephrasing.

So yes, with the initial phrasing it's very ambiguous, but you were given the unambiguous phrasing in which the dad volunteers no other information except "two children" -- and you still believed it was 1/2.

The new rephrasing with the probing question is unambiguous: and therefore you're unambiguously wrong in calculating the probabilities as 50%. They're absolutely 33.3%.
 
Posted by TomDavidson (Member # 99) on :
 
quote:
If you want to take order into account then you need to treat each child as a distinct entity and then there is a lot more than one combination for four boys...
This may be the source of your confusion. We don't particularly want to take order into account, but we have to take order into account. However, while there are a lot of possible permutations of four boys, we're only interested in the single combination of four boys; those permutations all collapse into one combination, because without any additional detail we can't tell any of the boys apart. They're all boys. The possible permuations, as written, are BBBB, BBBB, BBBB, and BBBB (etc.); they're all the same. The situation is different with one girl in the mix, since we can distinguish the permutation of BGBB from BBGB, BBBG, and GBBB.

[ July 28, 2010, 12:25 PM: Message edited by: TomDavidson ]
 
Posted by threads (Member # 5091) on :
 
Badvok:

There are 100 two-child families in the room. The world is naturally perfect so 25 of the families have 2 boys, 25 of the families have 2 girls, and 50 of the families have a boy and a girl. A father comes up to you and says "at least one of my children is a boy". What is the probability that he has two boys?

[ July 28, 2010, 12:26 PM: Message edited by: threads ]
 
Posted by DonaldD (Member # 1052) on :
 
quote:
I don't think there is much point do you?
Actually, yes. Your problem is one of semantics, and clarifying the initial statement is the only way to agree on what the puzzle actually means. You keep saying things like "Noooo! You are changing the problem space again!" but you refuse to clarify what you think the puzzle meant by "I have two kids. One of them is a boy".

So once again:

"I have two kids. The first born is a boy"
"I have two kids. Either the first or the second born (or both) is a boy"
"I have two kids. My child John is a boy"
"I have two kids. At least one of them is a boy"
"I have two kids. Exactly one of them is a boy"
"I have two kids. They are not both girls"
"I have two kids. The child I am currently thinking of is a boy"

Choose as many as you think are correct, as there is some overlap.

Tom - Aris actually gave the same example on the previous page, but with 7 children I think.

Badvok, if you play Tom and Aris' game, it will become quite clear to you.
 
Posted by TomDavidson (Member # 99) on :
 
It's important to note that this is why the Tuesday detail matters, because it becomes possible to distinguish a boy born on a Tuesday from another boy except if the second boy was also born on Tuesday.
 
Posted by Ciasiab (Member # 6390) on :
 
If the information is volunteered the odds of the second kid being a boy is 50/50.


quote:
Let's see exactly how you're approaching this. I'm going to try and reword the question a little more clearly:


quote:
In front of you are two screens, standing side by side. You are told that there is one person behind both of them. You are further told that at least one of the two people behind the screeens is male. What is the probability that both of the people behind the screens are male?
How would you go about answering this?

You are assuming that if at least one of the people is a male the announcer will tell you that there is a male. Without further information this is a bad assumption. If there is one male and one female, 50% of the time the announcer will choose male and 50% of the time the announcer will choose female.

So we have two people behind the screens, and have been told that one is male. It is tempting to say that there is a 2 in 3 chance of the other person being female, but this is not the case since if the other person is female, the odds of being told about a male have been reduced by half.

If your logic is right, once we know the gender of one contestant, we can conclude that 2/3 of the time the gender of the other contestant is the opposite. This cannot be true.
 
Posted by Aris Katsaris (Member # 888) on :
 
Badvok, because each child (of a given father) has an independent 50% chance of being born girl or a boy, that's why we have to take "order" into account. We don't have to take the order of their births: we can take alphabetical order, or order of prettiness, or order of weight, or any other order we seek.

But the point is treating them as each having a distinct 50% chance of having been born boy or girl.

The problem is the following:
Father: I have two children.
Me: Is atleast one of them a boy?
Father: Yes.

Given the above, the probability of two boys is 33%. UNAMBIGUOUSLY.

FFS, we're not being either 'malicious' or 'mischievous', we're trying to make you see how logic works.
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
Originally posted by threads:
Badvok:

There are 100 two-child families in the room. The world is naturally perfect so 25 of the families have 2 boys, 25 of the families have 2 girls, and 50 of the families have a boy and a girl. A father comes up to you and says "at least one of my children is a boy". What is the probability that he has two boys?

Of the 50 families with a boy and a girl, what are the odds of a father choosing to tell you he has a boy?
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
So yes, with the initial phrasing it's very ambiguous, but you were given the unambiguous phrasing in which the dad volunteers no other information except "two children" -- and you still believed it was 1/2.

The intial problem is not ambiguous. The answer is 1/2. In order for the answer 1/3 (or 13/27) we have to assume the father would only tell us the gender if he had a boy and only tell us the day of the week if it happened to be Tuesday.

[ July 28, 2010, 12:36 PM: Message edited by: Ciasiab ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
If the information is volunteered the odds of the second kid being a boy is 50/50
Sorta. It depends on the guy's reason for volunteering the information. He may have volunteered it for reasons that resolve to the same 33% possibility (e.g. asking army recruitment information, so all the fathers with atleast one boy would have to go there, while fathers with only girls wouldn't).

The following phrasing:
"Father: I have two children.
Me: Is atleast one of them a boy?
Father: Yes."

is unambiguous, though, and I wish more people from both sides used it, instead of insisting that the phrasing doesn't matter. It does matter, very much.
 
Posted by threads (Member # 5091) on :
 
quote:
Originally posted by Ciasiab:
quote:
Originally posted by threads:
Badvok:

There are 100 two-child families in the room. The world is naturally perfect so 25 of the families have 2 boys, 25 of the families have 2 girls, and 50 of the families have a boy and a girl. A father comes up to you and says "at least one of my children is a boy". What is the probability that he has two boys?

Of the 50 families with a boy and a girl, what are the odds of a father choosing to tell you he has a boy?
You should read Aris' discussion of this point. Relevant post:
quote:
Originally posted by Aris Katsaris:
Let me put it in another way.
Possibilities for two children:
G/G 1/4
G/B 1/4
B/G 1/4
B/B 1/4
--
If guy decides to reveal one of these genders randomly, possibilities now become
G/G (reveals G) 1/4
G/B (reveals 1st: G) 1/8
G/B (reveals 2nd: B) 1/8
B/G (reveals 1st: B) 1/8
B/G (reveals 2nd: G) 1/8
B/B (reveals B) 1/4

Now if we know "B" was revealed, this corresponds to 1/8 + 1/8 + 1/4 = 50%. Out of these, it's even odds that the other one was a boy, or that the other one was a girl.

So if the guy randomly selected the kid whose gender he'd reveal, it's even odds that the other kid is either gender. 50% says common sense, and 50% it indeed is.
--
HOWEVER if the guy thinks: I will NOT mention there's a girl, but I will only reveal if there exists a boy. The possibilities become:
G/G (mentions no information) 1/4
G/B (mentions there's a B) 1/4
B/G (mentions there's a B) 1/4
B/B (mentions there's a B) 1/4

Now, knowing he revealed it was a boy, there only 33.3% chances that the other kid is a boy too, and 66.7% chances that the other kid is a girl.
--
AND if the guy thinks: I will mention ALL my boys, but none of my girls. The possibilities become:
G/G (mentions no information) 1/4
G/B (mentions there's a B) 1/4
B/G (mentions there's a B) 1/4
B/B (mentions there's two boys) 1/4

Now, with the knowledge he mentioned only *one* B for certain, we can be 100% sure that the other kid is a girl -- because he'd have mentioned two boys if B/B was the reality.
--
That's what I mean when I say motivation matters. WHY did he reveal the particular gender? Was he randomly picking a kid, or was he choosing that gender for some reason?



[ July 28, 2010, 12:38 PM: Message edited by: threads ]
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
Sorta. It depends on the guy's reason for volunteering the information. He may have volunteered it for reasons that resolve to the same 33% possibility (e.g. asking army recruitment information, so all the fathers with atleast one boy would have to go there, while fathers with only girls wouldn't).

You have to project additional reasons to get there. You have to add an assumption about army recruitment or a sexist dad.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
"The initial problem is not ambiguous. The answer is 1/2. In order for the answer 1/3 (or 13/32) we have to assume the father would only tell us the gender if he had a boy and only tell us the day of the week if it happened to be Tuesday. "
The original phrasing is ambiguous exactly because we don't know if these assumptions are right or wrong. A Spartan soldier volunteering to be among the 300 would want at least one son to carry the family line. The existence or non-existence of girls wouldn't matter to him.
 
Posted by threads (Member # 5091) on :
 
Ciasiab, you are projecting just as much by assuming that the father is randomly choosing which kid to talk about. That's not stated anywhere in the initial problem. Just agree that it's ambiguous and let it be.
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
You should read Aris' discussion of this point. Relevant post:

Aris' post proves my point. Unless you assume the father is not providing random infomation (i.e. he's at an army recruiting station), the problem as stated resolves to 1/2.
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
"You have to project additional reasons to get there."
Sure, but "I've randomly picked a kid of mine whose gender I'm telling you" is also an assumption, as you don't know that.

That's why I say the phrasing is ambiguous.
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
Originally posted by threads:
Ciasiab, you are projecting just as much by assuming that the father is randomly choosing which kid to talk about. That's not stated anywhere in the initial problem. Just agree that it's ambiguous and let it be.

Most problems can be ambiguous. You can read it as random, or you can add assumptions in your head. The only way to solve the problem is to assume it is random. If you add another assumption you are adding to the problem (like saying you happen to be at a military recruitment center). If you assume (another assumption I know) that all the information is contained in the problem statement, then I don't see how not to conclude he is choosing a kid at random.
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
Originally posted by Ciasiab:
quote:
Originally posted by threads:
Ciasiab, you are projecting just as much by assuming that the father is randomly choosing which kid to talk about. That's not stated anywhere in the initial problem. Just agree that it's ambiguous and let it be.

Most problems can be ambiguous. You can read it as random, or you can add assumptions in your head. The only way to solve the problem is to assume it is random. If you add another assumption you are adding to the problem (like saying you happen to be at a military recruitment center). If you assume (another assumption I know) that all the information is contained in the problem statement, then I don't see how not to conclude he is choosing a kid at random.
Edit to add: How about we call it ambiguous and leave it at that. [Embarrassed]
 
Posted by threads (Member # 5091) on :
 
Agreed!
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
You can read it as random, or you can add assumptions in your head. The only way to solve the problem is to assume it is random.
Eh. What if the randomness is that he randomly chooses a gender, and then mentions its existence or non-existence?

In that case from an original set of 200 dads:
GG 50
GB 50
BG 50
BB 50
we go to
GG 25 (mentions absence of B)
GG 25 (mentions existence of G)
GB 25 (mentions existence of B)
GB 25 (mentions existence of G)
BG 25 (mentions existence of B)
BG 25 (mentions existence of G)
BB 25 (mentions existence of B)
BB 25 (mentions absence of G)

So you have 75 dads that randomly chose to mention existence of B. Out of them only 25 have a second B. It reverts to 33%

The original phrasing is just ambiguous. Did he randomly choose a child whose gender he revealed (50% chance of two boys), or did he randomly choose a gender whose absense/presence he revealed (33% chance of two boys)?

[ July 28, 2010, 12:54 PM: Message edited by: Aris Katsaris ]
 
Posted by Ciasiab (Member # 6390) on :
 
quote:
Eh. What if the randomness is that he randomly chooses a gender, and then mentions its existence or non-existence?

You make a very excellent point.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Badvok:
You're having a laugh aren't you?

No, Badvok; I'm trying to make an opening—and something of a break from the prior discussion—with what I hope is a somewhat cleaner (and clearer) example. [Smile]

So that you know where I'm going with this, I'm asking you to start by explaining how you would approach it with the intention of asking you some more questions to follow up.
 
Posted by Aris Katsaris (Member # 888) on :
 
Badvok, can I ask another question? Here's the following narrative, please examine it:

Father: I have exactly two children.

(Probabilities now stand at:
25% chance of two girls.
25% chance of two boys.
50% chance of one boy and one girl)

DO YOU AGREE WITH ME SO FAR? Please answer this question first: If you disagree there's no point in continuing further, but if you agree with me, please continue reading.

Me: Are they both girls?
Father: No.

(The probabilities as I calculate them now stand at:
0% chance of two girls.
33% chance of two boys.
66% chance of one boy and one girl)

----------------

When you argue that the chance of two boys is now at 50% you're essentially saying that by excluding the probability of two daughters, we are ONLY increasing the probability of two sons, but we're not increasing the probability of one son and one daughter.

Does that seem logical to you? Isn't it common sense that all other possibilities must increase in probability (at least slightly) by the exclusion of *one* previously possible scenario?
 
Posted by edgmatt (Member # 6449) on :
 
quote:
There are 100 two-child families in the room. The world is naturally perfect so 25 of the families have 2 boys, 25 of the families have 2 girls, and 50 of the families have a boy and a girl. A father comes up to you and says "at least one of my children is a boy". What is the probability that he has two boys?
Badvok - This post by Threads seems to be the best way to phrase the question so as to be understood. Does it make more sense when it is put this way?
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Aris Katsaris:
"John Doe: I have two children.
Me: Is at least one a boy?
John Doe: Yes."

And you KEPT insisting that the probability for two sons is 1/2 even with this rephrasing.

LOL, no I don't think I KEPT insisting that phrasing meant the probability was 1/2. I admit that I did mistakenly respond that way once though.

I see that I really don't get what this thread is all about (as I said in my first post to the thread). I don't understand why the assumption that John Doe comes from a population of random families was made. To me he is one man, he has one son and there is one child we do not know the gender of.

You are obviously not talking about the original statement and I am just confusing things by not accepting your assumptions about it.

I'll bow out in disgrace now.
 
Posted by TomDavidson (Member # 99) on :
 
quote:
To me he is one man, he has one son and there is one child we do not know the gender of.

Even still, his children come in the following permutations:

Boy / Girl
Girl / Boy
Boy / Boy
 
Posted by Jordan (Member # 2159) on :
 
quote:
Badvok:
You are obviously not talking about the original statement and I am just confusing things by not accepting your assumptions about it.

No drama, Badvok. [Smile] I think everyone here accepts that there are two very different interpretations of the question floating about—whether or not they accept that either one of them is a more natural or correct way of reading it!

The main thing that people are confused about is, simply, if you a) understand the logic behind the 1/3 answer, and b) agree that if the scenario is constructed carefully enough, 1/3 will definitely be the correct answer.

(By the way, I did exactly what I didn't want to do and read back through all your posts so far—and even after that, I honestly can't work out what your position is on b)! [Big Grin] )

[ July 29, 2010, 08:09 AM: Message edited by: Jordan ]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
You are obviously not talking about the original statement and I am just confusing things by not accepting your assumptions about it.
*I*, for one, wasn't making assumptions, I rephrased the question so that no assumptions needed be made, and I detailed in the first page of this thread how different assumptions (that other people were making) affect the probability.

I'm disappointed that you didn't respond in my last post. That you bowed out and said this isn't the original situations, doesn't mean you couldn't have taken half a min to say "yes" or "no" to my question. It's frustrating that even with the different rephrasing you can't bring yourself to actually state unambigously "Yes, in this case it's 33%".
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
"To me he is one man, he has one son and there is one child we do not know the gender of."
No, he is one man, and he has two children whose gender we don't know of -- with the knowledge that atleast one of them is a son affecting the corresponding probabilities for each child being a son.

With this bit of knowledge, any given child has 66% chance of being a son, and the situation for both being sons either way reverts to:
(if first child is a boy, second child has 50% possibility) (2/3)*(1/2) = 1/3
(if first child is a girl, second child has 100% chance of being a boy) (1/3)*(1) = 1/3
 
Posted by Badvok (Member # 1085) on :
 
quote:
Originally posted by Badvok:
Your numbers are not wrong in that sense.
1/3 of the 75 fathers (from a population of 100) who have at least one son have two sons.


 
Posted by Aris Katsaris (Member # 888) on :
 
Coolness. I must have missed or forgotten that post.
 
Posted by Jordan (Member # 2159) on :
 
I saw that post, but when you followed with:

quote:
Badvok:
But you are just solving the wrong problem and hence using the wrong numbers for the problem. We have ONE father, ONE son, and ONE other child who's gender we do not know.

I interpreted that, combined with your earlier comparison of bad arithmetic, as meaning that you didn't believe the same logic would apply in the case of one father about whom we know only that he has at least one son. Was I wrong? [Smile]
 
Posted by Badvok (Member # 1085) on :
 
Oh my, I am such an idiot!

I just couldn't let this go, my brain kept on working on it and so I've reviewed what I've said again. I never once explained why there is only a single unknown - stupid!

Let's review the original statement again:

quote:
John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday." What are the odds that he has two boys?
All I see is one father, one son and one other child.

Why is there only one other child and hence only a single unknown? The statement 'HE was born on a Tuesday." is not totally irrelevant but identifies one of the two children because of the pronoun used!

If he had said "one is a boy born on Tuesday" that would be different because there is no pronoun and hence no identification of an individual.

The "Tuesday" value itself is irrelevant because the question is only about gender. My understanding in this case can be shown by replacing this bit with "He has light brown hair." - we are now unable to enumerate all the possibilities but does this change the answer to the question of whether both are boys?

Of course this is all about interpreting English language and hence it is always possible to interpret it another way. However that was my interpretation which I failed to previously explain.

Thanks to DonaldD for questioning my inclusion of "It was scratched." in my alternate version - it is not irrelevant it is key to reducing the problem to a single unknown.
 
Posted by Jordan (Member # 2159) on :
 
quote:
Badvok:
Why is there only one other child and hence only a single unknown? The statement 'HE was born on a Tuesday." is not totally irrelevant but identifies one of the two children because of the pronoun used!

I'm unimpressed enough by the original phrasing of the question that by now I'm willing to grant anyone's particular interpretation rather than bothering to argue. As I said on the first page it's pretty ambiguous.

So if I gather correctly, you accept that it's 1/3 if you don't know which child is being referred to, but you consider the phrasing of the question to preclude that condition. I suspect you've tamed a few savage breasts with that! [Smile]

Though there's still time to discuss how knowing that at least one is born on a Tuesday affects the probabilities, if you're up for it! [Big Grin]
 
Posted by Aris Katsaris (Member # 888) on :
 
quote:
Why is there only one other child and hence only a single unknown? The statement 'HE was born on a Tuesday." is not totally irrelevant but identifies one of the two children because of the pronoun used!
Completely agreed with this. As the pronoun specifically identifies the child, the odds are at 50%.

If the puzzle was however phrased like this:
Father: I have two children.
Me: Was at least one of them a boy born on a Tuesday?
Father: Yes.
The possibility for two sons would now be 13/27 as elsewhere detailed.

The more specificity about the child in our probing question the more the probability goes from 33% to 50%. For example.
Father: I have two children.
Me: Was at least one of them a boy that is the current Governor of Alabama?
Father: Yes.

This is 50% again since the boy is specifically identified by the fact only one governor of Alabama can exist at a time.
 
Posted by DonaldD (Member # 1052) on :
 
Jordan...
quote:
If he had said "one is a boy born on Tuesday" that would be different because there is no pronoun and hence no identification of an individual.

The "Tuesday" value itself is irrelevant because the question is only about gender.

BTW, Badvok
quote:
Thanks to DonaldD for questioning my inclusion of "It was scratched." in my alternate version - it is not irrelevant it is key to reducing the problem to a single unknown.
I don't think I agree with your interpretation of my statement, here. I suggested something being scratched as not equivalent to a child being born on Tuesday not because it identifies the item (it does not necessarily, any more than does a Tuesday birth) but rather because the probability of a scratch is indeterminate.
 
Posted by Jordan (Member # 2159) on :
 
Sorry Donald—I can see you're drawing my attention to what Badvok said, but I'm not sure why yet! [Smile] Have I missed something?

Edit: in case it's not clear, I get the impression now that Badvok understands and accepts that there is a 1/3 probability of there being two boys when the question is carefully phrased, but I went with interpreting the second sentence that you bolded as implying that he doesn't yet accept the 13/27 answer to the case of at least one boy born on a Tuesday.

Isn't English fun? [Wink]

[ July 30, 2010, 08:33 AM: Message edited by: Jordan ]
 
Posted by Jordan (Member # 2159) on :
 
quote:
Aris Katsaris:
The more specificity about the child in our probing question the more the probability goes from 33% to 50%. For example.
Father: I have two children.
Me: Was at least one of them a boy that is the current Governor of Alabama?
Father: Yes.

This is 50% again since the boy is specifically identified by the fact only one governor of Alabama can exist at a time.

Aside from being the only time I recall seeing the word "specificity" used to refer to anything other than CSS selectors, that's an admirably clear summary of the principle at work, Aris.
 


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