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Posted by

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

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

Posted by

Brian: Yes. If we knew that the

[ July 23, 2010, 01:40 PM: Message edited by: JoshuaD ]

Posted by

quote: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.

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)

Posted by

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

Posted by

quote: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.

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.

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

"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

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

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

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

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

quote: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)

An equivalent question would be - I flipped two coins. The first one was on Tuesday and came up heads.

Posted by

The reason you're having trouble articulating the flaw, Joshua, is because there

The key to understanding this sort of question is spotting the

quote: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.

I have two children, and they're not both girls. What's the chance that they're two boys?

In the second question, there aren't so many special cases. In fact, there's only ONE special case: the one in which

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

quote: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

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?

Please tell me this is making a bit of sense?

[ July 23, 2010, 07:35 PM: Message edited by: Jordan ]

Posted by

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

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

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.

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,

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

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

quote: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.

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.

Posted by

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: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.

Given that I have two children, and at least one of them is a boy, what is the probability that I have two boys?

Similarly, if you phrase the other question:

quote: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.

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?

[ July 23, 2010, 09:17 PM: Message edited by: Jordan ]

Posted by

(Stupid

quote:I'm going to wait till tomorrow to see if you're referring of an

[T]his isn't a pure math problem -- it involves psychological reasoning about the motivation of the guy to give you this information.

Posted by

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

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

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

...which is funny, cause mariner's post explains it.

Teach me to scan and post.

Posted by

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

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

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

quote: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?

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?

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

"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

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:The point is that you don't know whether the result you have comes from the first or second toss.

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?

Posted by

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

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

Posted by

Heck, psycholgical factors could lead a father of two girls to claim that he had a boy born on Tuesday...

Posted by

quote:I don't know what "no psychological factors" means: does it mean the guy randomly selects a kid's gender to reveal?

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.

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: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.

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

quote:Um....As far as I can tell, you're confusing this with an

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%.

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

But in

quote: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.

I don't know what "no psychological factors" means: does it mean the guy randomly selects a kid's gender to reveal?

You

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

quote: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.

As far as I can tell, you're confusing this with an actual Monty Hall problem.

quote: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%.

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: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%.

"He has two kids, one of whom is a boy born on Tuesday. Full stop."

Posted by

quote: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.

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

Posted by

Mariner , thanks for posting this. It's made for a good thinking exercise.

Posted by

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.

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quote:Tom, you're starting from the

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.

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

Of those 27 combinations, there are 13 in which there are two boys.

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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

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

What a lot of people do is

You're not considering every possible combination of boys or girls. You're considering

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

[ July 24, 2010, 03:36 PM: Message edited by: Jordan ]

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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:However, as it was actually expressed, you're left wondering if the guy decided to tell us about a

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?

That's sort of what I thought was up last night, thus my remark about assuming point children. 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:As I said last night, I suspect more people would arrive at the "correct" answer if the question were phrased right.

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?

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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.

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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.

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Of the 27 families with one boy born on a Tuesday, 9 have a second boy. Why do you think otherwise?

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quote:Here is the complete list of possibilities:

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.

- BMon BTue
- BTue BMon
- BTue BTue
- BTue BWed
- BTue BThu
- BTue BFri
- BTue BSat
- BTue BSun
- BWed BTue
- BThu BTue
- BFri BTue
- BSat BTue
- 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.

[ July 24, 2010, 04:16 PM: Message edited by: Jordan ]

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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%)

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Just a little more help, don't forget

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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

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quote:Just checking - you realize this was not part of the original question, right?

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

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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

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quote:Sorry, I should have quoted or linked. I was referring to Brian's modified question in the second post:

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 1/2Just checking - you realize this was not part of the original question, right?

quote:

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

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(Gosh, I don't think I've been this animated since the Monty Hall thread!)

Posted by

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)

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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

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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.

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quote:First, I'm really glad you're looking at the data carefully.

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%.

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

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

Aside from anything else, the fact that this stood out to you means you're well on the way to getting the answer.

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quote: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.

there's only one case in which this happens.

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.

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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.

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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.

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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 ]

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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.

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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

[ July 24, 2010, 07:43 PM: Message edited by: TomDavidson ]

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quote:I'm afraid not. What we're counting are

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.

quote: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.As Aris has noted- what he's talking about matters, not just the basic distribution.

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:The subtlety in Aris' interpretation is that we aren't told

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?

[ July 24, 2010, 07:43 PM: Message edited by: Jordan ]

Posted by

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.

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quote:Oh,

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 haveanydistinguishing characteristic beyond sex, this breaks down the same way.

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.

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quote: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

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.

Consider the question using the "at least one child" phrasing instead of asking how he "selected" one particular child might help.

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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

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.

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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 ]

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I can't wrap my brain around boy Tuesday.

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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)."

quote:You're dangerously close. If the numbers game is getting in the way, you can get quite close to understanding theI 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.

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

—

To start with, you don't know

—

You ask:

**{Heads Tails}**, and he was talking about the first one;**{Tails Heads}**, and he was talking about the last one; or**{Heads Heads}**, and he could have been talking about either of them.

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

—

You ask:

**{Heads Tails}**; or**{Heads Heads}**.

In other words, there's a ½ probability that the toss came up

—

You ask:

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:

**{SilverHeads CopperTails}**;**{SilverHeads SilverTails}**;**{SilverHeads CopperHeads}**;**{CopperTails SilverHeads}**;**{SilverTails SilverHeads}**;**{CopperHeads SilverHeads}**; and**{SilverHeads SilverHeads}**.

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

—

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:

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 ]

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quote:No, don't pollute my beautiful maths with tricksy semantics!

JoshuaD:I suspect the following two cases are distinct. Do you agree?

The first one is very ambiguous. My first reading of it is that the man has to actually

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

So my first reading is that they're distinct; but if you had asked, "tell me the day that

Posted by

LOL@beboaed.

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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?

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I see. No, it wouldn't matter. If you replied, "Tell me a day of the week on which

It doesn't matter

[ July 25, 2010, 01:38 PM: Message edited by: Jordan ]

Posted by

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.

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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

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?

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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).

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quote:Hee. Badvok, Google the problem we discuss at the beginning of this scenario, the Monty Hall problem.

The gender of the unknown child is not dependent on the gender of the known child (excepting any biological factors).

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quote:Elder Boy

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?

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

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quote:The Monty Hall problem is totally different because there is a direct causal link.

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.

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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

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?

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?

Now, the final scenario. I have a room of 1,000,000 dads who each have two children, one of which is a boy

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

- Scenario 1: I have a room of 1,000,000 fathers with 2 children. I stand them in a line and ask them one at a time "Are your children the same sex?" Approximately how many of them will answer yes?
**Approximately 500,000**.

- - Scenario 2: I have a room of 1,000,000 fathers with 2 children. I ask them one at a time "Do you have at least one son?" If they answer no, they are asked to leave and their answer is not tallied. If they answer yes, I continue: "Are your children the same sex?" Approximately how many of them will answer yes?
**Approximately 750,000 will have been asked the 2nd question, and approximately 250,000 will have answered yes to the second question. Approximately 250,000 will have been asked to leave without being tallied. One third of those asked the second question will have answered yes. One quarter of the entire group will have answered yes.**

- - Scenario 3: I have a room of 1,000,000 fathers who have 2 children where one of the children is a boy. I stand them in a line and ask them each: "Are both your children boys?" How many of them will answer yes?
**Approximately 333,333.**

- - Scenario 4: 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?" and receive an answer. I then continue to ask "Are both your children boys?" How many of of them will answer yes?
.*Approximately 333,333*

- - 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. If they answer Tuesday I then continue to ask "Are both your children boys?" How many will answer yes to this question?
.*Approximately 68783. (1,000,000 * 1/7 * 13/27)*

[ July 26, 2010, 10:50 AM: Message edited by: JoshuaD ]

Posted by

quote: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.

The Monty Hall problem is totally different because there is a direct causal link.

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

quote: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).

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.

quote: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.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).

[ July 26, 2010, 11:16 AM: Message edited by: Badvok ]

Posted by

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

quote:What question? John

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.

Posted by

quote:And how exactly does the goat transform into a car? Higher math?

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 anactualMonty 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 werenotrequired 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.

Posted by

quote:Read my long post above.

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? Johnstatesthat he has one son! So the probability space only concerns the other child.

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

quote: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.

What question? John states that he has one son! So the probability space only concerns the other child

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

Apologies to all - I seem to have gone down the wrong track here, I thought this long meandering thread was talking about:

quote: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.

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?

Posted by

Err, no,it's not

Posted by

quote:Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.

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]

Posted by

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

Badvok: No. If birth order cancels out, then the odds of me having two children of the same sex is 2/3: [B 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

OK, to clarify how birth order is irrelevant:

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

You forgot about:

Youngest+Boy:Eldest+Girl

Eldest+Boy:Youngest+Girl

= 6 combinations

Posted by

quote:Yes, and?

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].

quote:I'll throw the same back at you!

Originally posted by JoshuaD:Please try to refrain from knee-jerk responses.

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

quote:Eh? I think I listed those.

Originally posted by DonaldD:You forgot about:

Youngest+Boy:Eldest+Girl

Eldest+Boy:Youngest+Girl

= 6 combinations

Edit: Ooops sorry, no I didn't - need to go home now!

[ July 26, 2010, 12:19 PM: Message edited by: Badvok ]

Posted by

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

quote:Nope, you listed their inverse cases only.

Eh? I think I listed those

Posted by

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

quote: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:

Birth order cancels itself out. You are only counting boy-boy once - it should be twice Boy1-Boy2 and Boy2-Boy1.

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

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

Tom:

quote: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?

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.

If so, then those are the only two possibilities. That specific boy is either younger or older. Unless his last name is Schroedinger.

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

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

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

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

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

quote: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.

JoshuaD:Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

Posted by

quote:PSRT: Actually, we do. Right now, we are working with the simplified population of father with at least one boy.

Well, DonaldD, we don't really know what the population in question is

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

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

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

Quibble:

quote: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".

- Joe comes up to you and says: "I have two kids." A boy then walks up and says "Hi Dad" to Joe.

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

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

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

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

quote: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.

PSRT: Actually, we do.

Posted by

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

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

quote: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.

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.

quote: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.

I then continue to ask "Is your other child a boy?" How many will answer yes?

quote:Again a meaningless question for fathers with two sons.

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.

quote:If the father is referring to a specific child, then the odds are 50:50.

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 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:Because when a father says "atleast one of my sons" is a boy, the possibilities don't collapse to

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?

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

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

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

quote:Not relevant:

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%

quote:In this case the events are not linked and therefore conditional probability does not apply.

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.

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

quote: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"?

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

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

quote:The probability of one boy is 1 - this is stated in the question.

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.

Posted by

quote:But that is not the question! The question is "What is the probability the

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.

Posted by

quote:This conditional probability calculation requires the events to be independent (ie, random, not linked) so it very much could apply.

In this case the events are not linked and therefore conditional probability does not apply.

Posted by

quote:Yes, sort of: if "one

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

quote: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?

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 "oneof themis a boy" not "child1 is a boy".

[ July 27, 2010, 07:38 AM: Message edited by: Badvok ]

Posted by

quote:Not relevant?? That's what conditional probability is all about. That's pretty much the ONLY relevant thing.

Not relevant:

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

What is the probability the other child is a boy if 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: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).

The probability of one boy is 1 - this is stated in the question.

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.

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quote:LOL, I give up, I'll let you all head off to kindergarten now while I get on with some real work.

Originally posted by Aris Katsaris:Math proves you wrong.

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quote:If he did pinpoint to a particular child, you could label it.

Originally posted by Badvok:It is "Child1 is a boy". John has declared the gender of one of his children - this child can be labelled as Child1.quote:

Originally posted by DonaldD:

[qb]quote:Yes, sort of: if "one

But that is not the question! The question is "What is the probability the other child is a boy if one is a boy?".of themis a boy" not "child1 is a boy".

But he didn't, so you can't.

Posted by

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

And not just in math, since you don't even seem to understand what the word "given" means.

Posted by

Aris, could you try to refrain from just straight-up insulting people?

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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

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

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

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

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

quote:Yep, sorry.

Originally posted by DonaldD:I think Badvok just got a little frustrated, Aris. Patience.

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

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

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: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):

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?

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

quote:You transformed GG to kG where k is the known BOY.

Originally posted by Badvok:[GG,GB,BG,BB] to [snip] [kG,kB,kG,kB].

Posted by

quote: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.

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.

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

quote:No, no, NO! A million times NO, you don't ask that!

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.

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: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.

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.

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

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

quote:The probability of two boys here is unambiguously 33%.

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%.

Thanks for the rephrasing, btw, it's the exact thing that I suggested needed be done, for the statistics to be clear.

Posted by

Well, this is entertaining.

Posted by

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

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?

quote: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.

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?

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quote: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.

But I still don't get boy Tuesday.

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

quote: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.

If the known is a boy the probability of both being boys is equal to the probability the unknown is a boy.

"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

quote: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.

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

quote:Which is pretty much what I argued at the whole first page of this thread, so we're in agreement here.

"The only way I see your logic working is if the problem is restated with probing questions being asked, instead of information declared."

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

I think it is a lost cause trying to explain this yet again but here goes:

There are

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

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.

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.

Posted by

quote: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.

"We know one is a boy."

quote:The combination "one girl and one boy" is twice more likely to occur in a population than the combination "two boys".

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

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

quote:Yes, that's what I argued in the first page of this thread.

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

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

quote: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.

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?

Posted by

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

quote:Yep. But that is the wrong question.

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".

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:No I think out of 100% dads with two boys 100% will have at least oneAgain, 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.

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quote:Well yes, but turning the question around doesn't change the number of unknowns.

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"?

Posted by

quote:Correct. Bear with me.

Well yes, but turning the question around doesn't change the number of unknowns.

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

Yep, I think I said that earlier.

Posted by

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

quote:Except, that it's really really not.

'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"'.

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

quote:Yes.

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.

quote: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.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

oops - "who claims he does not have two daughters."

Posted by

quote:Either two boys or a boy and a girl.

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?

[ July 27, 2010, 11:37 AM: Message edited by: Badvok ]

Posted by

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

Yeah, I guessed where you were going

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

I think you are trying to obfuscate now. You do see how you repeated yourself above, correct?

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

c, don't you mean changes from 1/4 to 1/3?

Posted by

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:Let's get even simpler here since you missed that the 25% was a subset of the original 75%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

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

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

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

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

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.

John Doe comes up to you and says: "I have two kids. One of them is a boy. He was born on a Tuesday."

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

quote: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.

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.

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

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

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

quote: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.

JoshuaD:Jordan: I was up all night thinking about this, and I'm now nearly certain that you're partially wrong. :-)

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

Incidentally, Aris has shown possibly the clearest understanding of the problem throughout, though his patience obviously got a bit frayed towards the end!

Posted by

Jordan,

How about: "Do you have at least one son who was born on Tuesday?"

Posted by

Pyrtolin:

quote:Where the heck were you 3 pages ago?!?

Intuition and common sense don't apply here; if the information isn't explicitly given, you can't imagine that you have it.

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

quote: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"!

Ciasiab:How about: "Do you have at least one son who was born on Tuesday?"

Posted by

quote:Intuition works with quantum mechanics.

When intuition doesn't work (I'm looking at you, quantum mechanics!) then the only result I get is a pounding headache.

Just not human intuition.

Posted by

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

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

quote: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.

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.

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

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. Two options only = 50:50 chance of each.

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.

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

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

quote:Except that you're wrong. Seriously. You really are.

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.

The question is not "what is the probability that my

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

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

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

quote:Your numbers are not wrong in that sense.

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?

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

I told you all it was semantics

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

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

quote: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.

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.

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: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.

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?

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

quote:LOL, I had to head off to catch my commuter train home.

Originally posted by DonaldD:Badvok, why did you quit on me above. You were soooo close

Posted by

quote:Nope that is not what was stated!

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.

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

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.

quote: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?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?

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

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:

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

quote: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.

Originally posted by DonaldD:You are arguing that the puzzle involves only a single unknown

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:and rephrase it without using children:

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?

quote:Does this help at all?

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?

[ July 28, 2010, 11:02 AM: Message edited by: Badvok ]

Posted by

OK, so you are now working with the original puzzle (less the Tuesday information, but OK for now).

But before going off on

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

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:[/qb][/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?

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

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:

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?

Please open my eyes! Or are we talking about an assumed field? As we all know Assume makes an ...

Posted by

quote: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.

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

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

quote:How would you go about answering this?

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 thatat least oneof the two people behind the screeens is male. What is the probability that both of the people behind the screens are male?

[ July 28, 2010, 11:27 AM: Message edited by: Jordan ]

Posted by

quote:You're having a laugh aren't you?

Originally posted by Jordan:How would you go about answering this?

Posted by

quote: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.

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:

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 ...

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

quote:Where did I ever say that the first child was a boy? Where does order come into it?

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 say he was the only father in the world? He is simply the only father in the problem space.

Posted by

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:Full stop. What are the chances that the box contains two white pieces? Two black? A mix?

I have a box that contains two chess pieces.

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

Posted by

quote:Noooo! You are changing the problem space again! The set of possibilities is only those where one of the pieces

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.

quote:I don't think there is much point do you?Now, seriously, answer my previous post

[ July 28, 2010, 12:02 PM: Message edited by: Badvok ]

Posted by

quote: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.

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?

quote:No- all fathers are in the problem space. He just happens to be the one that stepped forward and asked to be analyzed.Where did I say he was the only father in the world? He is simply the only father in the problem space.

Posted by

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

Originally posted by Pyrtolin:He just happens to be the one that stepped forward and asked to be analyzed.

[ July 28, 2010, 12:07 PM: Message edited by: Badvok ]

Posted by

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

Posted by

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 - I'm too tired to work it all out now (time for me to head home) but I'm sure you can.

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.

Posted by

quote: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.

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?

Posted by

quote: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.

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?

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

quote:This may be the source of your confusion. We don't particularly

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...

[ July 28, 2010, 12:25 PM: Message edited by: TomDavidson ]

Posted by

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

quote: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".

I don't think there is much point do you?

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

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

Posted by

If the information is volunteered the odds of the second kid being a boy is 50/50.

quote: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.

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

quote:How would you go about answering this?

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?

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

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

quote: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?

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?

Posted by

quote: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.

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.

[ July 28, 2010, 12:36 PM: Message edited by: Ciasiab ]

Posted by

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).

If the information is volunteered the odds of the second kid being a boy is 50/50

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

quote:You should read Aris' discussion of this point. Relevant post:

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?

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

quote:You have to project additional reasons to get there. You have to add an assumption about army recruitment or a sexist dad.

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).

Posted by

quote: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.

"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. "

Posted by

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

quote: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.

You should read Aris' discussion of this point. Relevant post:

Posted by

quote: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.

"You have to project additional reasons to get there."

That's why I say the phrasing is ambiguous.

Posted by

quote: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.

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.

Posted by

quote:Edit to add: How about we call it ambiguous and leave it at that.

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.

Posted by

Agreed!

Posted by

quote:Eh. What if the randomness is that he randomly chooses a gender, and then mentions its existence or non-existence?

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.

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

quote:You make a very excellent point.

Eh. What if the randomness is that he randomly chooses a gender, and then mentions its existence or non-existence?

Posted by

quote: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.

Badvok:You're having a laugh aren't you?

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

Badvok, can I ask another question? Here's the following narrative, please examine it:

(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.

(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

quote: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?

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?

Posted by

quote: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.

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.

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

quote:Even still, his children come in the following permutations:

To me he is one man, he has one son and there is one child we do not know the gender of.

Boy / Girl

Girl / Boy

Boy / Boy

Posted by

quote:No drama, Badvok. 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!

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

The main thing that people are confused about is, simply, if you

(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

[ July 29, 2010, 08:09 AM: Message edited by: Jordan ]

Posted by

quote:*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.

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

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

quote: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.

"To me he is one man, he has one son and there is one child we do not know the gender of."

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

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

Coolness. I must have missed or forgotten that post.

Posted by

I saw that post, but when you followed with:

quote: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?

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.

Posted by

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:

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?

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

quote: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.

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!

So if I gather correctly, you accept that it's 1/3 if you don't know

Though there's still time to discuss how knowing that

Posted by

quote:Completely agreed with this. As the pronoun specifically identifies the child, the odds are at 50%.

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 the puzzle was however phrased like this:

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.

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

Jordan...

quote:BTW, Badvok

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.

quote: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.

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

Sorry Donald—I can see you're drawing my attention to what Badvok said, but I'm not sure why yet! Have I missed something?

Isn't English fun?

[ July 30, 2010, 08:33 AM: Message edited by: Jordan ]

Posted by

quote: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.

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.

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