I always much prefer these stated as questions—you stop someone and say “Do you have exactly two children? Is at least one of them a boy born on a Tuesday?” and they say “yes”. Otherwise you get into wondering what the probability they’d say such a strange thing given various family setups might be, which isn’t precisely defined enough...
Very true. The article DanielVarga linked to says:
If you have two children, and one is a boy, then the probability of having two boys is significantly different if you supply the extra information that the boy was born on a Tuesday. Don’t believe me? We’ll get to the answer later.
… which is just wrong: whether it is different depends on how the information was obtained. If it was:
-- On which day of the week was your youngest boy born ?
-- On a Tuesday.
… then there’s zero new information, so the probability stays the same, 1/3rd.
(ETA: actually, to be closer to the original problem, it should be “Select one of your sons at random and tell me the day he was born”, but the result is the same.)
I think the only reasonable interpretation of the text is clear since otherwise other standard problems would be ambiguous as well:
“What is probability that a person’s random coin toss is tails?”
It does not matter whether you get the information from an experimenter by asking “Tell me the result of your flip!” or “Did you get tails?”. You just have to stick to the original text (tails) when you evaluate the answer in either case.
[[EDIT] I think I misinterpreted your comment. I agree that Daniel’s introduction was ambiguous for the reasons you have given.
Still the wording “I have two children, and at least one of them is a boy-born-on-a-Tuesday.” he has given clarifies it (and makes it well defined under the standard assumptions of indifference).
Yesterday I told the problem to a smart non-math-geek friend, and he totally couldn’t relate to this “only reasonable interpretation”. He completely understood the argument leading to 13⁄27, but just couldn’t understand why do we assume that the presenter is a randomly chosen member of the population he claims himself to be a member of. That sounded like a completely baseless assumption to him, that leads to factually incorrect results. He even understood that assuming it is our only choice if we want to get a well-defined math problem, and it is the only way to utilize all the information presented to us in the puzzle. But all this was not enough to convince him that he should assume something so stupid.
I get that assuming that genders and days of the week are equiprobable, of all the people with exactly two children, at least one of whom is a boy born on a Tuesday, 13⁄27 have two boys.
True, but if you go around asking people-with-two-chidren-at-least-one-of-which-is-a-boy “Select one of your sons at random, and tell me the day of the week on which he was born”, among those who answer “Tuesday”, one-third will have two boys.
(for a sufficiently large set of people-with-two-chidren-at-least-one-of-which-is-a-boy who answer your question instead of giving you a weird look)
I’m just saying that the article used an imprecise formulation, that could be interpreted in different ways—especially the bit “if you supply the extra information that the boy was born on a Tuesday”, which is why asking questions the way you did is better.
The Tuesday Boy was loads of fun to think about the first time I came across it—thanks to the parent comment. I worked through the calculations with my 14yo son, on a long metro ride, as a way to test my understanding—he seemed to be following along fine.
The discussion in the comments on this blog on Azimuth has, however, taken the delight to an entirely new level. Just wanted to share.
It seems to me that the standard solutions don’t account for the fact that there are a non-trivial number of families who are more likely to have a 3rd child, if the first two children are of the same sex. Some people have a sex-dependent stopping rule.
P(first two children different sexes | you have exactly two children) > P(first two children different sexes | you have more than two children)
The other issue with this kind of problem is the ambiguity. What was the disclosure algorithm? How did you decide which child to give me information about? Without that knowledge, we are left to speculate.
This issue is also sometimes raised in cultures where male children are much more highly prized by parents.
Most people falsely assume that such a bias, as it stands, changes gender ratios for the society, but its only real effect is that correspondingly larger and rarer families have lots of girls. Such societies typically do have weird gender ratios, but this is mostly due to higher death rates before birth because of selective abortion, or after birth because some parents in such societies feed girls less, teach them less, work them more, and take them to the doctor less.
Suppose the rules for deciding to have a child without selective abortion (and so with basically 50⁄50 odds of either gender) and no unfairness post-birth were: If you have a boy, stop; if you have no boy but have fewer than N children, have another. In a scenario where N > 2, two child families are either a girl and a boy, or two girls during a period when their parents still intend to have a third. Because that window is relatively small relative to the length of time that families exist to be sampled, most two child families (>90%?) would be gender balanced.
Generally, my impression is that parental preferences for one or the other sex (or for gender balance) are generally out of bounds in these kinds of questions because we’re supposed to assume platonicly perfect family generating processes with exact50⁄50 odds, and no parental biases, and so on. My impression is that cultural literacy is supposed to supply the platonic model. If non-platonic assumptions are operating then different answers are expected as different people bring in different evidence (like probabilities of lying and so forth). If real world factors sneak in later with platonic assumptions allowed to stand then its a case of a bad teacher who expects you to guess the password of precisely which evidence they want to be imported, and which excluded.
This issue of signaling which evidence to import is kind of subtle, and people get it wrong a lot when they try to tell a paradox. Having messed them up in the past, I think it’s harder than telling a new joke the first time, and uses similar skills :-)
Actually, a Bayesian and a frequentist can have different answers to this problem. It resides on what distribution you are using to decide to tell me that a boy is born on Tuesday. The standard answer ignores this issue.
I don’t know much about the philosophy of statistical inference. But I am dead sure that if the Bayesian and the frequentist really do ask the same question, then they will get the same answer. There is a nice spoiler post where the possible interpretations of the puzzle are clearly spelled out. Do you suggest that some of these interpretations are preferred by either a frequentist or a Bayesian?
Well, essentially, focusing on that coin flip is a very Bayesian thing to do. A frequentist approach to this problem won’t imagine the prior coin flip often. See Eliezer’s post about this here. I agree however that a careful frequentist should get the same results as a Bayesian if they are careful in this situation. What results one gets depends in part on what exactly one means by a frequentist here.
Just so it’s clear, since it didn’t seem super clear to me from the other comments, the solution to the Tuesday Boy problem given in that article is a really clever way to get the answer wrong.
The problem is the way they use the Tuesday information to confuse themselves. For some reason not stated in the problem anywhere, they assume that both boys cannot be born on Tuesday. I see no justification for this, as there is no natural justification for this, not even if they were born on the exact same day and not just the same day of the week! Twins exist! Using their same bizarre reasoning but adding the extra day they took out I get the correct answer of 50% (14/28), instead of the close but incorrect answer of 48% (13/27).
Using proper Bayesian updating from the prior probabilities of two children (25% boys, 50% one each, 25% girls) given the information that you have one boy, regardless of when he was born, gets you a 50% chance they’re both boys. Since knowing only one of the sexes doesn’t give any extra information regarding the probability of having one child of each sex, all of the probability for both being girls gets shifted to both being boys.
No, that’s not right. They don’t assume that both boys can’t be born on Tuesday. Instead, what they are doing is pointing out that although there is a scenario where both boys are born on Tuesday, they can’t count it twice—of the situations with a boy born on Tuesday, there are 6 non-Tuesday/Tuesday, 6 Tuesday/non-Tuesday, and only 1, not 2, Tuesday/Tuesday.
Actually, “one of my children is a boy born on Tuesday” is ambiguous. If it means “I picked the day Tuesday at random, and it so happens that one of my children is a boy born on the day I picked”, then the stated solution is correct. If it means “I picked one of my children at random, and it so happens that child is a boy, and it also so happens that child was born on Tuesday”, the stated solution is not correct and the day of the week has no effect on the probability
No, read it again. It’s confusing as all getout, which is why they make the mistake, but EACH child can be born on ANY day of the week. The boy on Tuesday is a red herring, he doesn’t factor into the probability for what day the second child can be born on at all. The two boys are not the same boys, they are individuals and their probabilities are individual. Re-label them Boy1 and Boy2 to make it clearer:
Here is the breakdown for the Boy1Tu/Boy2Any option:
Seven options for both. For some reason they claim either BTu/Tuesday isn’t an option, or Tuesday/BTu isn’t an option, but I see no reason for this. Each boy is an individual, and each boy has a 1⁄7 probability of being born on a given day. In attempting to avoid counting evidence twice you’ve skipped counting a piece of evidence at all! In the original statement, they never said one and ONLY one boy was born on Tuesday, just that one was born on Tuesday. That’s where they screwed up—they’ve denied the second boy the option of being born on Tuesday for no good reason.
A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you’ll approach 50% probability. That is true because the actual probability is 50%.
Here is the breakdown for the Boy1Tu/Boy2Any option:
Boy1Tu/Boy2Tuesda
Then the BAny/Boy1Tu option:
Boy2Tuesday/Boy1Tu
You’re double-counting the case where both boys are born on Tuesday, just like they said.
A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you’ll approach 50% probability.
If you find a trait rarer than being born on Tuesday, the double-counting is a smaller percentage of the scenarios, so being closer to 50% is expected.
In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.
That’s why the “born on tuesday” is a red herring, and doesn’t add any information. How could it?
This sounds like you are trying to divide “two boys born on Tuesday” into “two boys born on Tuesday and the person is talking about the first boy” and “two boys born on Tuesday and the person is talking about the second boy”.
That doesn’t work because you are now no longer dealing with cases of equal probability. “Boy 1 Monday/Boy 2 Tuesday”, “Boy 1 Tuesday/Boy 2 Tuesday”, and “Boy 1 Tuesday/Boy 1 Monday” all have equal probability. If you’re creating separate cases depending on which of the boys is being referred to, the first and third of those don’t divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.
doesn’t add any information. How could it?
As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by “one is a boy born on Tuesday”. If you picked “boy” and “Tuesday” at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said “one is a girl born on a Monday”, you are correct that no information is provided.
The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.
Rephrase it this way:
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
Now, to see why Tuesday is irrelevant, I’ll re-state it thusly:
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1⁄7 chance that child 1 was born on each day of the week, and there is a 1⁄7 chance that child 2 was born on each day of the week. There is a 1⁄49 chance that both children will be born on any given day (1/7*1/7), for a 7⁄49 or 1⁄7 chance that both children will be born on the same day. That’s your missing 1⁄7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
1⁄3 (you either got hh, heads/tails,or tails/heads). You didn’t tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.
This is a pretty well known stats problem, a variant of Gardern’s boy/girl paradox. You’ll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).
If you have two boys (probability 1⁄4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1⁄7 for the boy hitting on Tuesday).
No one is suggesting one flip informs the other, rather that when you say “one coin came up heads” you are giving some information about both coins.
I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?
This is 1⁄2, because there are two scenarios, hh, ht. But its different information then the other question.
If you say “one coin is heads,” you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).
No, it’s the exact same question, only the labels are different.
The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options—HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.
It’s exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn’t the monty hall problem. Knowing half the problem (that there’s at least one boy) doesn’t change the probability of the other boy, it just changes what our possibilities are.
No, it’s the exact same question, only the labels are different.
No, it isn’t. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.
Now, consider “I flip two coins” the possible outcomes are
hh,ht,th,tt
I hope we can agree on that much.
Now, I give you more information and I say “one of the coins is heads,” so we Bayesian update by crossing out any scenario where one coin isn’t heads. There is only 1 (tt)
hh,ht,th
So it should be pretty clear the probability I flipped two heads is 1⁄3.
Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information “the first coin is heads,” so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1⁄2.
I know it’s not the be all end all, but it’s generally reliable on these types of questions, and it gives P = 1⁄2, so I’m not the one disagreeing with the standard result here.
Do the math yourself, it’s pretty clear.
Edit: Reading closer, I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions. However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
Did you actually read it? It does not agree with you. Look under the heading “second question.”
Do the math yourself, it’s pretty clear.
I did the math in the post above, enumerating the possibilities for you to try to help you find your mistake.
Edit, in response to the edit:
I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions.
Which is exactly analogous to what Jiro was saying about the Tuesday question. So we all agree now? Tuesday can raise your probability slightly above 50%, as was said all along.
However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
And you are immediately making the exact same mistake again. You are told ONE child is a boy, you are NOT told the FIRST child is a boy. You do understand that these are different?
The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a bit sticky. Just how do we know that “at least” one is a boy? One description of the problem states that we look into a window, see only one child and it is a boy. This sounds like the same assumption. However, this one is equivalent to “sampling” the distribution (i.e. removing one child from the urn, ascertaining that it is a boy, then replacing). Let’s call the statement “the sample is a boy” proposition “b”. Now we have:
P(BB|b) = P(b|BB) P(BB) / P(b) = 1 1⁄4 / 1⁄2 = 1⁄2.
The difference here is the P(b), which is just the probability of drawing a boy from all possible cases (i.e. without the “at least”), which is clearly 0.5.
The Bayesian analysis generalizes easily to the case in which we relax the 50⁄50 population assumption. If we have no information about the populations then we assume a “flat prior”, i.e. P(GG) = P(BB) = P(G.B) = 1⁄3. In this case the “at least” assumption produces the result P(BB|B) = 1⁄2, and the sampling assumption produces P(BB|b) = 2⁄3, a result also derivable from the Rule of Succession.
We have no general population information here. We have one man with at least one boy.
I’m not at all sure you understand that quote. Lets stick with the coin flips:
Do you understand why these two questions are different:
I tell you- “I flipped two coins, at least one of them came out heads, what is the probability that I flipped two heads?” A:1/3
AND
“I flipped two coins, you choose one at random and look at it, its heads.What is the probability I flipped two heads” A: 1⁄2
For the record, I’m sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I’m right about which question we are answering.
Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn’t click until the last post or two.
Like I said, I still think I’m right, but not because my prior analysis was any good. The 1⁄3 case was a major hole in my reasoning. I’m happily waiting to see if you’re going to destroy my latest analysis, but I think it is pretty solid.
Yes, and we are dealing with the second question here.
Is that not what I said before?
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is 1⁄2, so the probability that both are boys is 1⁄2.
The possible options for the “Boy born on Tuesday” are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.
The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).
There are two Boy/Boy combinations, not one. You don’t have enough information to throw one of them out.
As long as you realize there is a difference between those two questions, fine. We can disagree about what assumptions the wording should lead us to, thats irrelevant to the actual statistics and can be an agree-to-disagree situation. Its just important to realize that what the question means/how you get the information is important.
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information.
If we have one family with two children, of which one is a boy, they are (by definition) a member of the set “all families that have at least one boy.” So it matters how we got the information.
If we got that information by grabbing a kid at random and looking at it (so we have information about one specific child), that is sampling, and it leads to the 1⁄2 probability.
If we got that information by having someone check both kids, and tell us “at least one is a boy” we have different information (its information about the set of kids the parents have, not information about one specific kid).
This is NOT a case of sampling.
If it IS sampling (if I grab a kid at random and say “whats your Birthday?” and it happens to be Tuesday), then the probability is 1⁄2. (we have information about the specific kid’s birthday).
If instead, I ask the parents to tell me the birthday of one of their children, and the parent says ‘I have at least one boy born on Tuesday’, then we get, instead, information about their set of kids, and the probability is the larger number.
Sampling is what leads to the answer you are supporting.
The answer I’m supporting is based on flat priors, not sampling. I’m saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not 1⁄3.
Sampling to the “Boy on Tuesday” problem gives roughly 48% (as per the original article), not 50%.
We are simply told that the man has a boy who was born on tuesday. We aren’t told how he chose that boy, whether he’s older or younger, etc. Therefore we have four possibilites, like I outlined above.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don’t see it.
BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
No. For any day of the week EXCEPT Tuesday, boy and girl are equivalent. For the case of both children born on Tuesday you have for girls: Boy(tu)/Girl(tu),Girl(tu)/Boy(tu), and for boys: boy(tu)/boy(tu).
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
I agree with this.
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
I agree with this if they said something along the lines of “One and only one of them was born on Tuesday”. If not, I don’t see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it’s twice as likely as the other two configurations that that is the configuration they are talking about when they say “One was born on Tuesday”.
Here’s my breakdown with 1000 families, to try to make it clear what I mean:
1000 Families with two children, 750 have boys.
Of the 750, 500 have one boy and one girl. Of these 500, 1⁄7, or roughly 71 have a boy born on Tuesday.
Of the 750, 250 have two boys. Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
71 = 71, so it’s equally likely that there are two boys as there are a boy and a girl.
Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.
And I don’t think I’m confused about the sampling, because I didn’t use the sampling reasoning to get my result*, but I’m not super confident about that so if I am just keep giving me numbers and hopefully it will click.
*I mean in the previous post, not specifically this post.
Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
This is wrong. With two boys each with a probability of 1⁄7 to be born on Tuesday, the probability of at least one on a Tuesday isn’t 2⁄7, its 1-(6/7)^2
How can that be? There is a 1⁄7 chance that one of the two is born on Tuesday, and there is a 1⁄7 chance that the other is born on Tuesday. 1⁄7 + 1⁄7 is 2⁄7.
There is also a 1⁄49 chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn’t change the probability that either of them are born on Tuesday, and both of those probabilities add.
You overcount, the both on Tuesday is overcounted there. Think of it this way- if I have 8 kids do I have a better than 100% probability of having a kid born on Tuesday?
There is a 1/7x6/7 chance the first is born on Tuesday and the second is born on another day. There is a 1/7x6/7 chance the second is born on Tuesday and the first is born on another day. And there is a 1⁄49 chance that both are born on Tuesday.
All together thats 13⁄49. Alternatively, there is a (6/7)^2 chance that both are born not-on-Tuesday, so 1-(6/7)^2 tells you the complementary probability.
I’ve seen that same explanation at least five times and it didn’t click until just now. You can’t distinguish between the two on tuesday, so you can only count it once for the pair.
Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.
Thanks for your patience!
ETA: I also think I see where I’m going wrong with the terminology—sampling vs not sampling, but I’m not 100% there yet.
“The first coin comes up heads” (in this version) is not the same thing as “one of the coins comes up heads” (as in the original version). This version is 50%, the other is not.
How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?
If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.
The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn’t matter that you don’t know which is which, one flip doesn’t have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50⁄50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don’t get both of them!).
Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It’s a fair coin, it stays 50%.
Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).
Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).
By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you’ve actually learned (“would this person have equally likely said ‘one of my children is a girl’ if they had both a boy and girl?”).
Yeah, probably the biggest thing I don’t like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.
. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl.
No there’s not. The cases where the second child is a boy and the second child is a girl are not equal probability.
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
If you picked “heads” before flipping the coins, then the probability is 1⁄3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
If you picked “heads” and “Tuesday” before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren’t any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn’t affect the result.
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born.
If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say “there aren’t any of the sex I picked” and the sexes in the cases where you would continue are not equally distributed.
Note before you start calculating this: There’s a distinction between the “first” and the “second” child made in the article. To avoid the risk of having to calculate all over again, take this into account if you want to compare your results to theirs.
I calculated the probability without knowing this, so that I just counted BG and GB as one scenario, where there’s one girl and one boy. That means that without the Tuesday fact the probability of another boy is 1⁄2, not 1⁄3.
(I ended up at a posterior probability of 2⁄3, by the way.)
I’m so happy: I’ve just got this one right, before looking at the answer. It’s damn beautiful.
Same here. It was a perfect test, as I’ve never seen the Tuesday Boy problem before. Took a little wrangling to get it all to come out in sane fractions, and I was staring at the final result going, “that can’t be right”, but sure enough, it was exactly right.
(Funny thing: my original intuition about the problem wasn’t that far off. I was simply ignoring the part about Tuesday, and focusing on the prior probability that the other child was a boy. It gives a close, but not-quite-right answer.)
Wonderful. Are you aware of the Tuesday Boy problem? I think it could have been a more impressive second example.
(The intended interpretation is that I have two children, and at least one of them is a boy-born-on-a-Tuesday.)
I found it here: Magic numbers: A meeting of mathemagical tricksters
I always much prefer these stated as questions—you stop someone and say “Do you have exactly two children? Is at least one of them a boy born on a Tuesday?” and they say “yes”. Otherwise you get into wondering what the probability they’d say such a strange thing given various family setups might be, which isn’t precisely defined enough...
Very true. The article DanielVarga linked to says:
… which is just wrong: whether it is different depends on how the information was obtained. If it was:
… then there’s zero new information, so the probability stays the same, 1/3rd.
(ETA: actually, to be closer to the original problem, it should be “Select one of your sons at random and tell me the day he was born”, but the result is the same.)
I think the only reasonable interpretation of the text is clear since otherwise other standard problems would be ambiguous as well:
“What is probability that a person’s random coin toss is tails?”
It does not matter whether you get the information from an experimenter by asking “Tell me the result of your flip!” or “Did you get tails?”. You just have to stick to the original text (tails) when you evaluate the answer in either case.
[[EDIT] I think I misinterpreted your comment. I agree that Daniel’s introduction was ambiguous for the reasons you have given.
Still the wording “I have two children, and at least one of them is a boy-born-on-a-Tuesday.” he has given clarifies it (and makes it well defined under the standard assumptions of indifference).
Yesterday I told the problem to a smart non-math-geek friend, and he totally couldn’t relate to this “only reasonable interpretation”. He completely understood the argument leading to 13⁄27, but just couldn’t understand why do we assume that the presenter is a randomly chosen member of the population he claims himself to be a member of. That sounded like a completely baseless assumption to him, that leads to factually incorrect results. He even understood that assuming it is our only choice if we want to get a well-defined math problem, and it is the only way to utilize all the information presented to us in the puzzle. But all this was not enough to convince him that he should assume something so stupid.
For me, the eye opener was this outstanding paper by E.T. Jaynes:
http://bayes.wustl.edu/etj/articles/well.pdf
IMO this describes the essence of the difference between the Bayesian and frequentist philosophy way better than any amount of colorful polygons. ;)
I get that assuming that genders and days of the week are equiprobable, of all the people with exactly two children, at least one of whom is a boy born on a Tuesday, 13⁄27 have two boys.
True, but if you go around asking people-with-two-chidren-at-least-one-of-which-is-a-boy “Select one of your sons at random, and tell me the day of the week on which he was born”, among those who answer “Tuesday”, one-third will have two boys.
(for a sufficiently large set of people-with-two-chidren-at-least-one-of-which-is-a-boy who answer your question instead of giving you a weird look)
I’m just saying that the article used an imprecise formulation, that could be interpreted in different ways—especially the bit “if you supply the extra information that the boy was born on a Tuesday”, which is why asking questions the way you did is better.
The Tuesday Boy was loads of fun to think about the first time I came across it—thanks to the parent comment. I worked through the calculations with my 14yo son, on a long metro ride, as a way to test my understanding—he seemed to be following along fine.
The discussion in the comments on this blog on Azimuth has, however, taken the delight to an entirely new level. Just wanted to share.
It seems to me that the standard solutions don’t account for the fact that there are a non-trivial number of families who are more likely to have a 3rd child, if the first two children are of the same sex. Some people have a sex-dependent stopping rule.
P(first two children different sexes | you have exactly two children) > P(first two children different sexes | you have more than two children)
The other issue with this kind of problem is the ambiguity. What was the disclosure algorithm? How did you decide which child to give me information about? Without that knowledge, we are left to speculate.
This issue is also sometimes raised in cultures where male children are much more highly prized by parents.
Most people falsely assume that such a bias, as it stands, changes gender ratios for the society, but its only real effect is that correspondingly larger and rarer families have lots of girls. Such societies typically do have weird gender ratios, but this is mostly due to higher death rates before birth because of selective abortion, or after birth because some parents in such societies feed girls less, teach them less, work them more, and take them to the doctor less.
Suppose the rules for deciding to have a child without selective abortion (and so with basically 50⁄50 odds of either gender) and no unfairness post-birth were: If you have a boy, stop; if you have no boy but have fewer than N children, have another. In a scenario where N > 2, two child families are either a girl and a boy, or two girls during a period when their parents still intend to have a third. Because that window is relatively small relative to the length of time that families exist to be sampled, most two child families (>90%?) would be gender balanced.
Generally, my impression is that parental preferences for one or the other sex (or for gender balance) are generally out of bounds in these kinds of questions because we’re supposed to assume platonicly perfect family generating processes with exact 50⁄50 odds, and no parental biases, and so on. My impression is that cultural literacy is supposed to supply the platonic model. If non-platonic assumptions are operating then different answers are expected as different people bring in different evidence (like probabilities of lying and so forth). If real world factors sneak in later with platonic assumptions allowed to stand then its a case of a bad teacher who expects you to guess the password of precisely which evidence they want to be imported, and which excluded.
This issue of signaling which evidence to import is kind of subtle, and people get it wrong a lot when they try to tell a paradox. Having messed them up in the past, I think it’s harder than telling a new joke the first time, and uses similar skills :-)
Actually, a Bayesian and a frequentist can have different answers to this problem. It resides on what distribution you are using to decide to tell me that a boy is born on Tuesday. The standard answer ignores this issue.
I don’t know much about the philosophy of statistical inference. But I am dead sure that if the Bayesian and the frequentist really do ask the same question, then they will get the same answer. There is a nice spoiler post where the possible interpretations of the puzzle are clearly spelled out. Do you suggest that some of these interpretations are preferred by either a frequentist or a Bayesian?
Well, essentially, focusing on that coin flip is a very Bayesian thing to do. A frequentist approach to this problem won’t imagine the prior coin flip often. See Eliezer’s post about this here. I agree however that a careful frequentist should get the same results as a Bayesian if they are careful in this situation. What results one gets depends in part on what exactly one means by a frequentist here.
Just so it’s clear, since it didn’t seem super clear to me from the other comments, the solution to the Tuesday Boy problem given in that article is a really clever way to get the answer wrong.
The problem is the way they use the Tuesday information to confuse themselves. For some reason not stated in the problem anywhere, they assume that both boys cannot be born on Tuesday. I see no justification for this, as there is no natural justification for this, not even if they were born on the exact same day and not just the same day of the week! Twins exist! Using their same bizarre reasoning but adding the extra day they took out I get the correct answer of 50% (14/28), instead of the close but incorrect answer of 48% (13/27).
Using proper Bayesian updating from the prior probabilities of two children (25% boys, 50% one each, 25% girls) given the information that you have one boy, regardless of when he was born, gets you a 50% chance they’re both boys. Since knowing only one of the sexes doesn’t give any extra information regarding the probability of having one child of each sex, all of the probability for both being girls gets shifted to both being boys.
No, that’s not right. They don’t assume that both boys can’t be born on Tuesday. Instead, what they are doing is pointing out that although there is a scenario where both boys are born on Tuesday, they can’t count it twice—of the situations with a boy born on Tuesday, there are 6 non-Tuesday/Tuesday, 6 Tuesday/non-Tuesday, and only 1, not 2, Tuesday/Tuesday.
Actually, “one of my children is a boy born on Tuesday” is ambiguous. If it means “I picked the day Tuesday at random, and it so happens that one of my children is a boy born on the day I picked”, then the stated solution is correct. If it means “I picked one of my children at random, and it so happens that child is a boy, and it also so happens that child was born on Tuesday”, the stated solution is not correct and the day of the week has no effect on the probability
No, read it again. It’s confusing as all getout, which is why they make the mistake, but EACH child can be born on ANY day of the week. The boy on Tuesday is a red herring, he doesn’t factor into the probability for what day the second child can be born on at all. The two boys are not the same boys, they are individuals and their probabilities are individual. Re-label them Boy1 and Boy2 to make it clearer:
Here is the breakdown for the Boy1Tu/Boy2Any option:
Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday
Then the BAny/Boy1Tu option:
Boy2Monday/Boy1Tu Boy2Tuesday/Boy1Tu Boy2Wednesday/Boy1Tu Boy2Thursday/Boy1Tu Boy2Friday/Boy1Tu Boy2Saturday/Boy1Tu Boy2Sunday/Boy1Tu
Seven options for both. For some reason they claim either BTu/Tuesday isn’t an option, or Tuesday/BTu isn’t an option, but I see no reason for this. Each boy is an individual, and each boy has a 1⁄7 probability of being born on a given day. In attempting to avoid counting evidence twice you’ve skipped counting a piece of evidence at all! In the original statement, they never said one and ONLY one boy was born on Tuesday, just that one was born on Tuesday. That’s where they screwed up—they’ve denied the second boy the option of being born on Tuesday for no good reason.
A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you’ll approach 50% probability. That is true because the actual probability is 50%.
You’re double-counting the case where both boys are born on Tuesday, just like they said.
If you find a trait rarer than being born on Tuesday, the double-counting is a smaller percentage of the scenarios, so being closer to 50% is expected.
I see my mistake, here’s an updated breakdown:
Boy1Tu/Boy2Any
Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday
Then the Boy1Any/Boy2Tu option:
Boy1Monday/Boy2Tu Boy1Tuesday/Boy2Tu Boy1Wednesday/Boy2Tu Boy1Thursday/Boy2Tu Boy1Friday/Boy2Tu Boy1Saturday/Boy2Tu Boy1Sunday/Boy2Tu
See 7 days for each set? They aren’t interchangeable even though the label “boy” makes it seem like they are.
Do the Bayesian probabilities instead to verify, it comes out to 50% even.
What’s the difference between
and
?
In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.
That’s why the “born on tuesday” is a red herring, and doesn’t add any information. How could it?
This sounds like you are trying to divide “two boys born on Tuesday” into “two boys born on Tuesday and the person is talking about the first boy” and “two boys born on Tuesday and the person is talking about the second boy”.
That doesn’t work because you are now no longer dealing with cases of equal probability. “Boy 1 Monday/Boy 2 Tuesday”, “Boy 1 Tuesday/Boy 2 Tuesday”, and “Boy 1 Tuesday/Boy 1 Monday” all have equal probability. If you’re creating separate cases depending on which of the boys is being referred to, the first and third of those don’t divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.
As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by “one is a boy born on Tuesday”. If you picked “boy” and “Tuesday” at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said “one is a girl born on a Monday”, you are correct that no information is provided.
The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.
Rephrase it this way:
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
Now, to see why Tuesday is irrelevant, I’ll re-state it thusly:
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1⁄7 chance that child 1 was born on each day of the week, and there is a 1⁄7 chance that child 2 was born on each day of the week. There is a 1⁄49 chance that both children will be born on any given day (1/7*1/7), for a 7⁄49 or 1⁄7 chance that both children will be born on the same day. That’s your missing 1⁄7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.
1⁄3 (you either got hh, heads/tails,or tails/heads). You didn’t tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.
This is a pretty well known stats problem, a variant of Gardern’s boy/girl paradox. You’ll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).
If you have two boys (probability 1⁄4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1⁄7 for the boy hitting on Tuesday).
Lets add a time delay to hopefully finally illustrate the point that one coin toss does not inform the other coin toss.
I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?
No one is suggesting one flip informs the other, rather that when you say “one coin came up heads” you are giving some information about both coins.
This is 1⁄2, because there are two scenarios, hh, ht. But its different information then the other question.
If you say “one coin is heads,” you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).
No, it’s the exact same question, only the labels are different.
The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options—HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.
Here’s the probability in bayesian:
P(BoyBoy) = 0.25 P(Boy) = 0.5 P(Boy|BoyBoy) = 1
P(BoyBoy|Boy) = P(Boy|BoyBoy)*P(BoyBoy)/P(Boy)
P(BoyBoy|Boy)= (1*0.25) / 0.5 = 0.25 / 0.5 = 0.5
P(BoyBoy|Boy) = 0.5
It’s exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn’t the monty hall problem. Knowing half the problem (that there’s at least one boy) doesn’t change the probability of the other boy, it just changes what our possibilities are.
No, it isn’t. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.
Now, consider “I flip two coins” the possible outcomes are hh,ht,th,tt
I hope we can agree on that much.
Now, I give you more information and I say “one of the coins is heads,” so we Bayesian update by crossing out any scenario where one coin isn’t heads. There is only 1 (tt)
hh,ht,th
So it should be pretty clear the probability I flipped two heads is 1⁄3.
Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information “the first coin is heads,” so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1⁄2.
I don’t know how to make this any more simple.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
I know it’s not the be all end all, but it’s generally reliable on these types of questions, and it gives P = 1⁄2, so I’m not the one disagreeing with the standard result here.
Do the math yourself, it’s pretty clear.
Edit: Reading closer, I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions. However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
Did you actually read it? It does not agree with you. Look under the heading “second question.”
I did the math in the post above, enumerating the possibilities for you to try to help you find your mistake.
Edit, in response to the edit:
Which is exactly analogous to what Jiro was saying about the Tuesday question. So we all agree now? Tuesday can raise your probability slightly above 50%, as was said all along.
And you are immediately making the exact same mistake again. You are told ONE child is a boy, you are NOT told the FIRST child is a boy. You do understand that these are different?
Re-read it.
The relevant quote from the Wiki:
We have no general population information here. We have one man with at least one boy.
I’m not at all sure you understand that quote. Lets stick with the coin flips:
Do you understand why these two questions are different: I tell you- “I flipped two coins, at least one of them came out heads, what is the probability that I flipped two heads?” A:1/3 AND “I flipped two coins, you choose one at random and look at it, its heads.What is the probability I flipped two heads” A: 1⁄2
For the record, I’m sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I’m right about which question we are answering.
Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn’t click until the last post or two.
Like I said, I still think I’m right, but not because my prior analysis was any good. The 1⁄3 case was a major hole in my reasoning. I’m happily waiting to see if you’re going to destroy my latest analysis, but I think it is pretty solid.
Yes, and we are dealing with the second question here.
Is that not what I said before?
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is 1⁄2, so the probability that both are boys is 1⁄2.
The possible options for the “Boy born on Tuesday” are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.
The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).
There are two Boy/Boy combinations, not one. You don’t have enough information to throw one of them out.
This is NOT a case of sampling.
As long as you realize there is a difference between those two questions, fine. We can disagree about what assumptions the wording should lead us to, thats irrelevant to the actual statistics and can be an agree-to-disagree situation. Its just important to realize that what the question means/how you get the information is important.
If we have one family with two children, of which one is a boy, they are (by definition) a member of the set “all families that have at least one boy.” So it matters how we got the information.
If we got that information by grabbing a kid at random and looking at it (so we have information about one specific child), that is sampling, and it leads to the 1⁄2 probability.
If we got that information by having someone check both kids, and tell us “at least one is a boy” we have different information (its information about the set of kids the parents have, not information about one specific kid).
If it IS sampling (if I grab a kid at random and say “whats your Birthday?” and it happens to be Tuesday), then the probability is 1⁄2. (we have information about the specific kid’s birthday).
If instead, I ask the parents to tell me the birthday of one of their children, and the parent says ‘I have at least one boy born on Tuesday’, then we get, instead, information about their set of kids, and the probability is the larger number.
Sampling is what leads to the answer you are supporting.
The answer I’m supporting is based on flat priors, not sampling. I’m saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not 1⁄3.
Sampling to the “Boy on Tuesday” problem gives roughly 48% (as per the original article), not 50%.
We are simply told that the man has a boy who was born on tuesday. We aren’t told how he chose that boy, whether he’s older or younger, etc. Therefore we have four possibilites, like I outlined above.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don’t see it.
BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
No. For any day of the week EXCEPT Tuesday, boy and girl are equivalent. For the case of both children born on Tuesday you have for girls: Boy(tu)/Girl(tu),Girl(tu)/Boy(tu), and for boys: boy(tu)/boy(tu).
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
I agree with this.
I agree with this if they said something along the lines of “One and only one of them was born on Tuesday”. If not, I don’t see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it’s twice as likely as the other two configurations that that is the configuration they are talking about when they say “One was born on Tuesday”.
Here’s my breakdown with 1000 families, to try to make it clear what I mean:
1000 Families with two children, 750 have boys.
Of the 750, 500 have one boy and one girl. Of these 500, 1⁄7, or roughly 71 have a boy born on Tuesday.
Of the 750, 250 have two boys. Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
71 = 71, so it’s equally likely that there are two boys as there are a boy and a girl.
Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.
And I don’t think I’m confused about the sampling, because I didn’t use the sampling reasoning to get my result*, but I’m not super confident about that so if I am just keep giving me numbers and hopefully it will click.
*I mean in the previous post, not specifically this post.
This is wrong. With two boys each with a probability of 1⁄7 to be born on Tuesday, the probability of at least one on a Tuesday isn’t 2⁄7, its 1-(6/7)^2
How can that be? There is a 1⁄7 chance that one of the two is born on Tuesday, and there is a 1⁄7 chance that the other is born on Tuesday. 1⁄7 + 1⁄7 is 2⁄7.
There is also a 1⁄49 chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn’t change the probability that either of them are born on Tuesday, and both of those probabilities add.
The problem is that you’re counting that 1/49th chance twice. Once for the first brother and once for the second.
I see that now, it took a LOT for me to get it for some reason.
You overcount, the both on Tuesday is overcounted there. Think of it this way- if I have 8 kids do I have a better than 100% probability of having a kid born on Tuesday?
There is a 1/7x6/7 chance the first is born on Tuesday and the second is born on another day. There is a 1/7x6/7 chance the second is born on Tuesday and the first is born on another day. And there is a 1⁄49 chance that both are born on Tuesday.
All together thats 13⁄49. Alternatively, there is a (6/7)^2 chance that both are born not-on-Tuesday, so 1-(6/7)^2 tells you the complementary probability.
Wow.
I’ve seen that same explanation at least five times and it didn’t click until just now. You can’t distinguish between the two on tuesday, so you can only count it once for the pair.
Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.
Thanks for your patience!
ETA: I also think I see where I’m going wrong with the terminology—sampling vs not sampling, but I’m not 100% there yet.
“The first coin comes up heads” (in this version) is not the same thing as “one of the coins comes up heads” (as in the original version). This version is 50%, the other is not.
How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?
If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.
The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn’t matter that you don’t know which is which, one flip doesn’t have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50⁄50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don’t get both of them!).
Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It’s a fair coin, it stays 50%.
Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).
Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).
By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you’ve actually learned (“would this person have equally likely said ‘one of my children is a girl’ if they had both a boy and girl?”).
Yeah, probably the biggest thing I don’t like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.
No there’s not. The cases where the second child is a boy and the second child is a girl are not equal probability.
If you picked “heads” before flipping the coins, then the probability is 1⁄3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.
If you picked “heads” and “Tuesday” before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren’t any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn’t affect the result.
If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say “there aren’t any of the sex I picked” and the sexes in the cases where you would continue are not equally distributed.
Which boy did I count twice?
Edit:
BAny/Boy1Tu in the above quote should be Boy2Any/Boy1Tu.
You could re-label boy1 and boy2 to be cat and dog and it won’t change the probabilities—that would be CatTu/DogAny.
Note before you start calculating this: There’s a distinction between the “first” and the “second” child made in the article. To avoid the risk of having to calculate all over again, take this into account if you want to compare your results to theirs.
I calculated the probability without knowing this, so that I just counted BG and GB as one scenario, where there’s one girl and one boy. That means that without the Tuesday fact the probability of another boy is 1⁄2, not 1⁄3.
(I ended up at a posterior probability of 2⁄3, by the way.)
I’m so happy: I’ve just got this one right, before looking at the answer. It’s damn beautiful.
Thanks for sharing.
Same here. It was a perfect test, as I’ve never seen the Tuesday Boy problem before. Took a little wrangling to get it all to come out in sane fractions, and I was staring at the final result going, “that can’t be right”, but sure enough, it was exactly right.
(Funny thing: my original intuition about the problem wasn’t that far off. I was simply ignoring the part about Tuesday, and focusing on the prior probability that the other child was a boy. It gives a close, but not-quite-right answer.)