posted
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?
Posts: 538 | Registered: Mar 2004
| IP: Logged |

posted
My first question for the mathmeticians out there is: Does the original problen devolve into 'what is the sex of my second child'? This would be analogous to Mariner's first option of 50:50. My thinking is: the 1:4 (or 1:3) ratio in the second option depends on there being a difference between boy-girl and girl-boy. Would it change the odds if he said 'my oldest child is a boy'?
Posts: 359 | Registered: Nov 2001
| IP: Logged |

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.
Posts: 3742 | Registered: Dec 2003
| IP: Logged |

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

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

But simply because we don't know which one is true (even though we know one of them must be) then the odds drop to 1:3, and thus neither of them is true?
Posts: 359 | Registered: Nov 2001
| IP: Logged |

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

Something's definitely wrong with this reasoning. There's some sneaky mathematical slight of hand going on somewhere. I'm working on figuring out what it is now.

It seems to me that the slight of hand goes like this. You're taking numbers from the general population, and then applying them to a situation that has already been narrowed down to a subset of the general population. The numbers 13, 27, and 196 have no bearing whatsoever on the gender of this guy's other kid, because we've already specified that the guy has one boy.

An equivalent question would be - I flipped two coins. The first one was on Tuesday and came up heads. What are the chances the second one also came up heads? No matter what day of the week it is, or which hand you used, the chances are 50/50 for that second coin. Same with the kids in this sort of a simplified puzzle.
Posts: 7 | Registered: Jun 2006
| IP: Logged |

posted
"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%.
Posts: 4700 | Registered: Oct 2009
| IP: Logged |

posted
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.
Posts: 4387 | Registered: Jul 2006
| IP: Logged |

posted
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.
Posts: 22935 | Registered: Nov 2000
| IP: Logged |

posted
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.
Posts: 255 | Registered: Jan 2008
| IP: Logged |

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

This is not an equivalent statement. It is not known whether it was the first or the second coin (in your example) that came up heads on Tuesday (not that Tuesday matters anyway)
Posts: 10751 | Registered: May 2003
| IP: Logged |

posted
The reason you're having trouble articulating the flaw, Joshua, is because there isn't one, and you're way too smart to accept a fallacy just to make things match your intuition. The reasoning is sound, if a little subtle; this is a much trickier one to explain than the Monty Hall problem, but I'll give it a go.

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

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

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

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

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

quote:You have access to a database of all families in America. You do a filter to get all the families with two children. Now, you ask yourself: what proportion of those families have two boys, GIVEN that the family has one boy who was born on a Tuesday?

Now it might be a bit clearer why the special case is important. Your GIVEN is actually introducing the day of the week that one of the children was born on as an additional filtering factor. The few families in which both children are born on a Tuesday are counted only once in your filtered set of results, even though there are actually two boys in that family who meet your GIVEN, and that's why the probability of the other child being a boy is just a bit less than 50%.

Please tell me this is making a bit of sense?

[ July 23, 2010, 07:35 PM: Message edited by: Jordan ]
Posts: 2147 | Registered: Nov 2004
| IP: Logged |

posted
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.
Posts: 3742 | Registered: Dec 2003
| IP: Logged |

posted
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.
Posts: 3654 | Registered: Sep 2006
| IP: Logged |

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

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

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

[ July 23, 2010, 09:06 PM: Message edited by: Jordan ]
Posts: 2147 | Registered: Nov 2004
| IP: Logged |

posted
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.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

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.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

posted
I think the real problem is more a matter of ambiguous wording. If you phrase the easier question differently, the answer becomes more obvious:

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

Unfortunately, the initial formulation is just ambiguous enough to be misread, which is why so many people go for the 50:50 option by mistake when the actual answer is one-in-three.

Similarly, if you phrase the other question:

quote:Given that I have two children, and that at least one of them is a boy who was born on a Tuesday, what is the probability that I have two boys?

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

[ July 23, 2010, 09:17 PM: Message edited by: Jordan ]
Posts: 2147 | Registered: Nov 2004
| IP: Logged |

posted
(Stupid sodding nuts… Surely you must be able to buy them with those silly little skins removed already?)

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

I'm going to wait till tomorrow to see if you're referring of an á la Monty Hall "the host knows something" situation, some sort of unavoidable initial conditions, or simply assuming that the man might be trying to trick us somehow by randomly selecting the information he wants to tell us. (You should know that I tend to approach problems like this assuming point children and perfectly spherical quiz-show hosts, so the latter possibility just isn't going to cross my mind. )
Posts: 2147 | Registered: Nov 2004
| IP: Logged |

posted
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?

posted
To put it in yet another way: A sexist dad is more likely to have daughters even though he isn't mentioning any.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

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.

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

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

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.
Posts: 7 | Registered: Jun 2006
| IP: Logged |

quote:If I tell you I tossed a coin twice and it came up heads the first time. Wouldn't you agree that there's a 50/50 chance that the second toss was also heads?

Yes. But that's not the question. The question, rather, is: I tossed a coin twice, and it came up heads at least once. What are the chances that both tosses came up heads?

This is a very important distinction.

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

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

posted
"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.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

Yeah, I missed it too. It's a probability problem in a riddle-ish form. The math is simple--it's just worded in the same riddle type as the "I have two coins with a summed value of 35 cents. One of them is not a quarter..."

quote:In a single run of two coin tosses, how can knowledge of the result of the first toss possibly affect the probability of the second toss?

The point is that you don't know whether the result you have comes from the first or second toss.
Posts: 3654 | Registered: Sep 2006
| IP: Logged |

posted
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.
Posts: 10751 | Registered: May 2003
| IP: Logged |

posted
Aris is making the question more complicated than it needs to be. Yes, the odds of a boy birth vs. a girl birth (as hobsen noted) are not actually 50% in any given instance. Yes, unless the problem explicitly rules out psychological factors, psychological factors will probably affect the outcome.

But I think it's pretty safe to say that the "puzzle" here fairly means us to assume both an equal chance of boy/girl births and no psychological factors. To quibble over these, on this particular question, is to miss the forest for the trees.
Posts: 22935 | Registered: Nov 2000
| IP: Logged |

posted
Heck, psycholgical factors could lead a father of two girls to claim that he had a boy born on Tuesday...
Posts: 10751 | Registered: May 2003
| IP: Logged |

quote:But I think it's pretty safe to say that the "puzzle" here fairly means us to assume both an equal chance of boy/girl births and no psychological factors.

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

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

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

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

quote:Aris, what you are missing is that the fact of the sexes of the children is independent of any choice being made by the person making the statement. Unless you are positing that some misogynist fathers would exclude themselves by being unable not to say they have two boys when they do, but in that case, you would be better off factoring in female birth rates and twin factors into the equation.

This selective reporting can be done with coin-flips if you don't want to cloud the issue with gender politics. I can easily construct a program simulating coin-flips and using any of the 3 logics I listed, I could completely confuse your attempts to figure out the probability.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

quote:If he randomly selected a kid whose gender he wanted to reveal, then the fact he mentioned it was a boy, leaves the other one's chance at 50%, not 33%.

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

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

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

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

No. It means that we are invited to assume the guy's motivations and methods are irrelevant. He has two kids, one of whom is a boy born on Tuesday. Full stop.

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

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

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

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

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

Yes. Assuming he was being random with his choice of kid whose birth/gender information he revealed, the odds he has two boys are 50%. All you guys saying it's 33% are wrong. My math coincides with common sense in this: No matter how many further info he gives, (hey, it's a Capricorn redhaired boy born on Tuesday, on a moonless night, while the wolves howled and the omens looked favourably down upon him) it won't budge this from 50%.

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

Okay. 50% then. The kids were born before he gave *any* information about them, so ANY information he gives about the gender, day of birth, zodiac signs whatever, won't budge this percentage from 50%.
Posts: 3318 | Registered: Feb 2003
| IP: Logged |

quote: A clever man would then say it is actually 1/3, since there are 4 possibilities of a two child family: GG, GB, BG, and BB. Since GG is eliminated, there's only 1/3 chance

We're looking at combinations, not permutations. There are only three combinations of two children: GG, BG, BB. We've eliminated one. There are only two combinations left.
Posts: 2152 | Registered: Apr 2009
| IP: Logged |

posted
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 ˝ 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 ˝ 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.
Posts: 255 | Registered: Jan 2008
| IP: Logged |

quote:TomDavidson: Why is it mathematically sound that 13 of the 27 families with a boy born on Tuesday will have two boys? As 1/3 of all two-child families with one boy have two boys, there are thus 9 families in that result set.

9/27 = 1/3.

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

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

There are two sexes, each with probability ½.

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

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

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

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

Of those 27 combinations, there are 13 in which there are two boys.
Posts: 2147 | Registered: Nov 2004
| IP: Logged |

posted
OK, I'm going to take my best shot at explaining why the day of the week is NOT irrelevant.

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

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

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

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

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

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

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

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

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

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

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

[ July 24, 2010, 03:36 PM: Message edited by: Jordan ]
Posts: 2147 | Registered: Nov 2004
| IP: Logged |