The Tuesday Birthday Problem

Writing in 1866, John Venn (of Venn diagram fame) wrote:

To many persons the mention of Probability suggests little else than the notion of a set of rules, very ingenious and profound rules no doubt, with which mathematicians amuse themselves by setting and solving puzzles.

A classic example is the Tuesday Birthday Problem (TBP), which a reader has asked me to comment on. Let’s dive right in by stating the problem:

You meet a man on the street and he says, “I have two children and one is a son born on a Tuesday.” What is the probability that the other child is also a son?

Seems innocent enough. But if you’re first inclination is to say that the information about the Tuesday birthday is irrelevant and the answer is just 1/2, then read on.


Let’s try a simpler problem. Suppose we know that a certain man has two children and we also know that the older one is a boy. In this case we would say that the probability that the other child is a boy is 1/2. After all, the sex of one child is independent of the sex of the other child. That the older child is a boy has no bearing on the sex of the younger child.

Now suppose we know simply that a man has two children and that one of them is a son. This time we would reason that there is no possibility that the person has two girls. It follows that the sexes of his two children, ordered from oldest to youngest, are either BB, BG or GB. Since these cases are equally likely, and since only one of them involves having two boys, we would say the probability that the man has two boys is 1/3.

This is all very cogent, but there are complications we haven’t yet considered. We shall return to this in a moment.

Now let’s consider the TBP. Everyone’s first instinct is to think the information about the Tuesday birthday is simply irrelevant. I mean, seriously, you knew each of the kids was born on some day of the week, so what possible difference could it make that you now know one was born on a Tuesday?

Well, here’s the difference. Let us first assume that it is the older child who was a son born on a Tuesday. In this case the second child could be either of two sexes, and could have been born on any of seven days of the week, for a total of 14 possibilities.

Now let’s suppose it is the younger child who was a son born on a Tuesday. Then the older child could, again, be either of two sexes and could have been born on any of seven days of the week, again providing 14 possibilities. Added to our original 14 that would seem to give 28 possibilities.

But be careful! One possibility got counted twice. Specifically, the one where both children are boys born on Tuesdays. So really there are only 27 possibilities. And since 13 of them involve the second child being a boy, the probability would be 13/27.

Were this the end of the story it would still be a fascinating problem. It is genuinely surprising that the information about the Tuesday birthday could be relevant, but our simple count shows that it is.

Now, we should mention at this point that we are discussing mathematical children who have no existence outside the world of probabilistic brainteasers. So we’re not worrying about the fact that a disproportionate number of children are born on Mondays and Tuesdays, since C-sections aren’t usually scheduled for the weekends. We’re also not worrying about twins or triplets. Or any other sort of “real world” consideration that might occur to you.

But there is a complication we should ponder. When you are thinking about problems in conditional probability, it is not only the new piece of information that is relevant. You also must consider how the information was obtained.

The classic example of this is the Monty Hall problem. (I shall assume you are familiar with this problem. If you are not, I know a good book you should read.) The common fallacy is to ignore what we know about how Monty makes decisions. Thus, when he opens an empty door we tend to think, mistakenly, that we have only learned that that door is empty. In reality we have learned that Monty, who makes his decisions in a rigidly controlled way, chose to take a certain action.

That has relevance to all of these problems about the two children. How we assess the probability that the second child is a boy will depend in part on how we learned that one of the children was a boy in the first place.

Peter Winkler explains the significance of this point:

This puzzle confounds people *legitimately*, however, because most of the ways in which you are likely to find out that X has at least one boy contain an implicit bias which changes the answer. For example, if you happen to meet one of X’s children and it’s a boy, the answer changes to 1/2.

Suppose the puzzle is phrased this way: X says “I have two children and at least one is a boy.” What is the probability that the other is a boy?

Put this way, the puzzle is highly ambiguous. Computer scientists, cryptologists and others who must deal carefully with message-passing know that what counts is not what a person says (even if she is known never to lie) but *under what circumstances would she have said it.*

Here, there is no context and thus no way to know what prompted X to make this statement. Could he instead have said “At least one is a girl”? Could he have said “Both are boys”? Could he have said nothing? If you, the one faced with solving the puzzle, are desperate to disambiguate it, you’d probably have to assume that what really happened was: X (for some reason unconnected with X’s identity) was asked whether it was the case that he had at least one son, and, after being warned–by a judge?–that he had to give a yes-or-no answer, said “yes.” An unlikely scenario, to say the least, but necessary if you want to claim that the solution to the puzzle is 1/3.

This article from Science News helps explain the relevant distinction in these sorts of problems:

Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.

In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male. (Emphasis added.)

This all takes some getting used to. If you’re like me, then it just seems obvious to you that knowing the sex of the older child tells you nothing about the sex of the younger. But the scenario in which you meet one of the children on the street just sounds different. It sounds more like our initial scenario, in which we know simply that a father has two children and one is a boy.

We should be thinking like this: Are you being asked to infer the sex of one person based solely on information about someone else? Or are you being asked to comment on the distribution of sexes among families with multiple children, given information about one of the children?

As the Science News article goes on to note, this has a curious consequence:

The remarkable thing that Foshee’s variation points out is that any piece of information that affects the selection will also affect the probability. If, for example, you selected a family at random among those with two kids, one of whom is a boy who plays the ukulele and wants to become a dancer, the ukulele-playing and dancing ambitions would affect the probabilities about the sex of his sibling.

That still seems weird to me, even though it is precisely the main principle in the Monty Hall problem. Anyway, Tanya Khovanonva provdes some variations on the TBP to illustrate what is going on:

Now let us consider the first scenario. A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: “I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?

The probability that a father of two daughters will make such a statement is zero. The probability that a father of differently-gendered children will produce such a statement is 1/14. Indeed, with a probability of 1/2 the son is chosen over the daughter and with a probability of 1/7 Tuesday is the birthday.

The probability that a father of two sons will make this statement is 1/7. Among the fathers with two children, there are twice as many who have a son and a daughter than fathers who have two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 1/2 for the second child to also be a son.

This fits well with our analysis. In this scenario a child is selected at random, and then information is provided about his sex and birth day. We are thus being asked to infer the sex of one child based solely on information about the sex (and birth day) of the other. Thus, we expect the answer to be 1/2. In other words, knowing the sex of one child tells us nothing about the sex of the other.

Now let us consider the second scenario. A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son’s day of birth. If he has two sons, he has to choose one at random. His statement will be, “I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?

The probability that a father of differently-gendered children will produce such a statement is 1/7. If he has two sons, the probability will likewise be 1/7. Among the fathers with two children, twice as many have a son and a daughter as have two sons. Plugging these numbers into the formula for calculating the conditional probability gives us a probability of 1/3 for the second child to also be a son.

In this scenario our focus has shifted to the distribution of sexes in a family with two children. Thus, the knowledge that there is at least one boy is relevant. But given the procedure we employed in deciding who to ask, the day of the week on which the boy is born is not relevant to the distribution of sexes. Thus, we get precisely the 1/3 answer we discussed earlier.

Now let us consider the third scenario. A father of two children is picked at random. If he doesn’t have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, “I have a son born on Tuesday.” What is the probability that the second child is also a son?

The probability that a father of two daughters will have a son born on Tuesday is zero. The probability that a father of differently-gendered children will have a son who is born on Tuesday is 1/7. The probability that a father of two sons will have a son born on Tuesday is 13/49. Among the fathers with two children, twice as many have a son and a daughter than two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 13/27 for the second child to also be a son.

And here it was built into our selection procedure that we are only surveying families with two children, with the added assumption that one of them is a boy born on Tuesday. If you surveyed all such families, you would find that roughly 13/27 of them have two boys.

If you read the Science News article I linked to, you will find that this problem was raised at the Ninth Gathering for Gardner, a conference held biennially in honor of Martin Gardner, in Atlanta. As it happens, I was at that conference, as was Tanya. We discussed this problem at length. My initial reaction was that it should be irrelevant that one child was born on a Tuesday. But then, thinking that the problem would not have been raised if the obvious answer were correct, I managed to enumerate the sample space and came up with 13/27. I was pleased with myself, but Tanya gave me an earful about unwarranted assumptions. She was right, of course.

On the other hand, as I noted in the BMHB (that’s the big Monty Hall book), and as I suggested to Tanya at the conference, there really does come a time when the hair-splitting must stop, unless we are specifically trying to drive people insane with these problems!

Comments

  1. #1 heddle
    November 8, 2011

    I wrote on this puzzle once, just in case you are interested.

    http://helives.blogspot.com/2010/07/tuesday-child-puzzle.html

  2. #2 miller
    November 8, 2011

    And what if the same father undergoes all three scenarios sequentially, and, by coincidence, produces the same statement each time? :)

  3. #3 Russell
    November 8, 2011

    Let’s change this a little bit:

    Now let us consider the third scenario. A father of two children is picked at random. If he doesn’t have a son, he is sent home and another is picked at random until one who has a son is found. He is instructed to tell you, “I have a son born on whatever-day-of-the-week one of his sons was born.”

    It seems to me impossible to distinguish these scenarios, in how the problem is first posed. The assumption to the counting argument is that the right set to consider is all men with two children including a son born on Tuesday, rather than all men including a son, born on some day of the week.

  4. #4 Another Matt
    November 8, 2011

    There’s another complication in the wording, where in normal speech the question could be understood as:

    ‘You meet a man on the street and he says, “I have two children and [only] one is a son born on a Tuesday.” What is the probability that the other child is also a son?’

    Or ‘I have two children and one is a son born on a Tuesday [and the other is not].’

    Understood this way, it would eliminate completely the two boys born on Tuesday scenario, which would make it even less likely that the other was a son (12/26)

    Imagine I said, “I have two dogs. One of them is a poodle.” In everyday conversation this means that I have one poodle and one of some other breed because I’m assumed to be trying to convey information clearly and efficiently, and I would just say “both are poodles” rather than “one is a poodle and the other is a poodle” if indeed I had two poodles.

    The problem with the TBP is that a question is posed after the informative statement. If I said instead, “I have two dogs. One of them is a poodle. What do you think the other is?” The way you go about answering that question has a lot to do with whether or not you have reason to believe I’m trying to trick you.

    A lot of punks on the playground exploit ambiguities in language to win bets.

  5. #5 John Jacob Lyons
    November 9, 2011

    An intriguing problem Jason but, as far as I know, it is only of academic interest. No jury has ever been fooled into a ‘guilty’ verdict because of the fallacy, for example. This is not so for the ‘Prosecutor’s Fallacy’. This has certainly caused several appalling miscarriages of justice in UK courts. On these occasions, innocent mothers have been found guilty of murder after a so-called ‘expert witness’ (falsely) told the court that the probability of two cot-deaths in one family was about 73 million to 1. The case of Sally Clark several years ago was a well-known, and tragic, example.

    The fallacy arises when the (irrelevant) ‘conditional probability of the occurrence of the event(s) assuming innocence’ is confused with the (relevant) ‘conditional probability of innocence given the happening of the event(s)’.

    Lack of knowledge of probability theory can cause very rough justice in courts of law.

  6. #6 Wow
    November 9, 2011

    There isn’t anything saying that ONLY one is a Son born on Tuesday, though.

    Like the old gambler joke: I have two pair: two pairs of Aces.

    This is somewhat different than the Monty Haul proposition (IIRC the name), where you’re allowed to change your mind after selecting which of three doors you’re going choose a prize behind when after your choice, one of the dud doors is revealed as empty of a prize.

    Should you change?

    Answer: yes.

  7. #7 Another Matt
    November 9, 2011

    “There isn’t anything saying that ONLY one is a Son born on Tuesday, though.”

    Not in the words themselves, but in the information the words are assumed to imply in everyday conversation. If I see someone on the street that I haven’t seen in a decade, and I say “how the hell are ya?!” and she says “I’m great! I have four kids now, and one of them is a girl!” — I just have to assume that the other three are boys because nobody ever says “one of them is a girl, and one of the remaining three is a girl, and one of the remaining two is a girl, and the yet unenumerated child is a boy.”

    It’s why we have the phrase “at least.” If the original problem were phrased “I have two children. At least one of them is a son born on a Tuesday.” then there would be no ambiguity of meaning.

    It’s really the same problem with inclusive and exclusive OR, isn’t it? — it’s all about context. If I’m at an amusement park and I tell my child, “You can have ice cream or a funnel cake.” and he says “ICE CREEEEAAAM,” when he’s done he doesn’t get to say, “I shall have a funnel cake as well, please, for I understood your use of ‘or’ to be in the inclusive sense, dad, and my behavior will not be contrary to the permission you gave if I have ice cream AND a funnel cake. Next time be sure to include ‘but not both,’ dad, if that’s what you really mean.”

  8. #8 Wow
    November 9, 2011

    “but in the information the words are assumed to imply in everyday conversation”

    But in everyday conversation, one child is a son doesn’t say that the second one isn’t a son.

    It is more information to say “at least”, but in a society and language where “Hi” is enough to greet someone, where “‘sup” means “What’s up” and all those other places where contraction is used to reduce information, missing “at least” isn’t providing any information.

    And you’re right: it does depend on context. The context being here “lets trick you into giving the ‘wrong’ answer!”.

    The context missing means that we have the words as said. And they don’t say “only one of which is a Son”.

  9. #9 eric
    November 9, 2011

    Wow: But in everyday conversation, one child is a son doesn’t say that the second one isn’t a son.

    I think Another Matt is right, and “one child…” does imply the second child isn’t a son. Because the speaker has a choice of picking to use the word “one” or “two.” And if the speaker chooses to use the word “one,” in common speech you can assume she didn’t mean “two.”

    In your argument, Wow, every time we hear “one” in such a sentence, we should interpret that to mean “greater than or equal to one.” But I do not see why “greater than or equal to” should be presumed. What is the logic behind presuming that is what the person meant?

    It seems a lot more verbally sensible to assume that the word “one” typically refers to the number 1 and not the mathematical ‘>=1.

  10. #10 MKR
    November 9, 2011

    Another Matt makes a good point. Getting the correct solution to a mathematical “story problem”–that is, the solution which would be deemed correct by those who devised the problem–very often depends on a strict interpretation of the words used, which in some cases is at odds with the pragmatics of ordinary conversation. Those who are adept at such problems (call them the skillful) may think that persons who interpret “one of them is a boy” to imply that the other is not a boy (call those who assume this interpretation the unskillful) are misinterpreting the words. But I would say that what is going on is that the skillful are wise to the pragmatic rules that govern story problems while the unskillful simply follow the rules of ordinary conversation.

  11. #11 Wow
    November 9, 2011

    “Those who are adept at such problems (call them the skillful)”

    Or weird.

    I’d prefer to call those who pose such problems “Silly”.

    After all, 1/2 is right if you take one meaning of the question, therefore you’re right saying 1/2. If you take another meaning of the question, you get a different number. They are all correct, even though they may not be the one that the one posing the question thought was correct.

    Which is why it’s a silly question: there’s several right answers. Not just one.

  12. #12 Wow
    November 9, 2011

    “I think Another Matt is right, and “one child…” does imply the second child isn’t a son.”

    If both are sons, then is one child not a son?

    No.

    “Only one child is a Son” means specifically only one. “At least one” means one or two are sons. “One may be a son” is that there may or may not be at least one son.

    If you’re going to put words in there to make the meaning clear, then the question is no longer the one asked, it’s the one you’re answering.

    Which means that unless you get that wrong, you’ll be right, even if the one asking thinks you’re wrong.

  13. #13 Another Matt
    November 9, 2011

    But in everyday conversation, one child is a son doesn’t say that the second one isn’t a son.

    I have three bridges for sale. One is a bridge connecting Manhattan and Brooklyn.

  14. #14 eric
    November 9, 2011

    Wow:

    If you’re going to put words in there to make the meaning clear, then the question is no longer the one asked.

    We must mentally put words in there because, without clarification, the sentence can have multiple meanings. The question is what the speaker was trying to communicate. Did the speaker intend to communicate 1.0 son or ‘>=1 son?

    What I’m saying is absent any contextual clues, the first makes a lot more sense to assume. Because if she’d wanted to communicate to you that she’d had two sons, she would’ve said ‘two’ instead of ‘one.’

    Now, I’ll concede that in some contexts – if a speaker is giving a riddle, playing a verbal joke, or just being an ass – then your interpretation could be warranted. If you phrase Matt’s scenario as “One logician jokingly said to the other ‘I now have two kids, and one is a son!’” then yeah, you could be right in interpreting 1 to mean ‘>=1.” But absent those contextual clues, I don’t think your position holds much water.

  15. #15 Wow
    November 9, 2011

    “We must mentally put words in there because, without clarification, the sentence can have multiple meanings”

    And then you no longer have the question asked.

    You have the question you decided to answer.

    If it doesn’t say that only one son is there, then there is nothing saying there cannot be two sons.

    But, like I said several times before, the question is silly because you “have to” add words to the question to get a question.

    But the question asked said nothing about there being only one son.

  16. #16 Wow
    November 9, 2011

    “I have three bridges for sale. One is a bridge connecting Manhattan and Brooklyn.”

    Or, maybe, I have the London Bridge to sell you. :-)

    Assuming words not in evidence is not a good move, in general, if someone is trying to trick you.

    There are two bridges crossing the Severn. Just saying “there’s one bridge crossing it, what does the other bridge cross?” isn’t going to mean the second bridge can’t be crossing the Severn too. If I were to tell you there was only one bridge crossing the Severn, then you’d be right (though only because I lied)

  17. #17 eric
    November 9, 2011

    If it doesn’t say that only one son is there, then there is nothing saying there cannot be two sons.

    Yes, there is: the speaker’s choice of the word “one” instead of “two.” You keep ignoring this information as if it doesn’t exist.

    Which is, now that I think about it, exactly the sort of error Jason was talking about in his post.

  18. #18 Wow
    November 9, 2011

    “Yes, there is: the speaker’s choice of the word “one” instead of “two.””

    No there isn’t: the lack of “only”.

    You need “only one” for there to be only one.

    I keep ignoring this information because it DOESN’T exist.

    There is no information there saying only one son.

    If you wish to point out where it does say “only one”, I shall investigate and maybe recant.

  19. #19 Wow
    November 9, 2011

    I think the problem might be what my original claim stated has morphed by virtue of misapprehension of what I meant and poor wording allowing it.

    I’ll see if I can work out another wording that makes my point clearer.

    It may be some time…

  20. #20 Another Matt
    November 9, 2011

    If you wish to point out where it does say “only one”, I shall investigate and maybe recant.

    I think it’s only in the context, and you’ve been making this point yourself by saying things like “Assuming words not in evidence is not a good move, in general, if someone is trying to trick you.” But assuming words not in evidence has to happen all the time for efficient communication in everyday language. There are two contexts in the original problem – one is the one asserted in the problem (viz. that you meet a man on the street and he says something to you – the “everyday language” context) and the other is the statement of the problem itself (the “uh oh a logician might be trying to trick me” context). It’s just an ambiguity that is usually resolved one way normally but possibly another way if there’s a scheme afoot.

    Here’s another example. What are the subtle differences between the following?
    1) “I flipped a coin 20 times. One flip came up heads.”
    2) “I flipped a coin 20 times. One of those times it came up heads.”
    3) “I flipped a coin 20 times. It came up heads one of those times.
    4) “I flipped a coin 20 times. It came up heads once.”
    5) “I flipped a coin 20 times. Of those 20 flips, one was heads.”
    6) “I flipped a coin 20 times. One of those times it came up heads. ONE!!”

    etc.

    There are other contexts where it’s pretty clear that “one” does not necessarily mean “only one” – especially cases where one is learning or teaching.

    “I have two coins that together equal 10 cents. One of them is a nickel. What is the other?” This is a deduction problem for kindergarten, and it’s clear that the other one could be a nickel as well; the image is more like “I have a coin in my right hand and one in my left. They equal 10 cents. Pick a hand and I’ll open it, and then you tell me what’s in the other hand. Oooh! My right hand has a nickel; so we know that one of them is a nickel, and remember that together they equal 10 cents. What do you think the other one is?”

    This is actually pretty close in context to the original problem – it’s just the “man on the street” scenario that bugs me, and especially if extra words can be added to disambiguate the point at hand.

  21. #21 cwfong
    November 9, 2011

    Have you considered the reputation of the speaker? If he’s a notorious trickster and liar, then it’s no longer a math problem – assuming that it ever was.

  22. #22 CRM-114
    November 9, 2011

    You meet a man on the street and he says, “I have two children and one is a son born on a Tuesday.” What is the probability that you will be correct in your assumptions about why he picked you to tell, and why he chose to tell you this, and why he chose that particular wording?

    A true mathematician will convince himself that all his assumptions will all be perfectly correct.

  23. #23 Another Matt
    November 9, 2011

    A true mathematician will convince himself that all his assumptions will all be perfectly correct.

    This is the “no true mathematician” fallacy.

  24. #24 BKsea
    November 10, 2011

    epic fail! (I think)

    When you were enumerating the possibilities, you also have to enumerate the choice of days that the father has when both children are sons. If they are born on different days, he has 2 choices and there is only a 50% chance he will pick Tuesday. If they are both born on Tuesday, there is a 100% chance he picks Tuesday. If you correct for this factor, the information about Tuesday becomes irrelevant and order is restored to the universe.

  25. #25 Wow
    November 10, 2011

    I don’t think it’s an *epic* fail.

    However, it does illustrate that if you put words in the question that doesn’t exist in the question to make sense of it, you’re changing the question.

    The sons could be twins. He has one son born on the same day as the other.

    We leave it there, we don’t add “and the other son is born on the same day as the other other one”.

    Two sons. Two tuesdays. One son IS born on Tuesday.

  26. #26 Markk
    November 10, 2011

    All of these are just ways that show how we map English (or other natural languages) to sets of possible outcomes. You could say things like this are optical illusions for our mind because they are doing similar things – making our heuristic or natural interpretation come in conflict with a not obvious bit in information. Drwing lines around possible outcomes and drawing the possible outcomes often makes things real clear that the ground – the set of possible outcomes being considered, is shifting under you’re feet.

  27. #27 John Jacob Lyons
    November 10, 2011

    I suggest that a question can only be answered unambiguously if the question itself is unambiguous. It is the responsibility of the person posing the question to ensure that this is the case. Despite this, if the person attempting to answer the question spots an ambiguity he/she can either (a) seek clarification from the poser or (b) provide answers for each of the possible meanings he/she can identify.

    With ambiguity, the only exception to the above would be when the answer is the same whichever meaning is intended.

  28. #28 CherryBombSim
    November 11, 2011

    Thanks. I had never heard of the TBP before. It made me think of this scenario:

    A man tells you he has a son who is left-handed, the probability that he has two sons would be close to 1/2. If he tells you he has a son who is right-handed, the probability of two sons is closer to 1/3. Cute.

  29. #29 jahigginbotham
    November 11, 2011

    #4,

    True for poodles, but not other necessarily other animals:

    A farmer wanted to control snakes, so he decided to buy a some predators. The only problem was that he didn’t know the plural of “mongoose.”
    He started the letter: “To whom it may concern, I need two mongeese.”
    But that didn’t sound right so he tried again: “To whom it may concern, I need two mongooses.”
    “Is that right?” he thought to himself.
    Finally, he got an idea: “To whom it may concern, I need one mongoose, and while you’re at it, please send me another one.”

  30. #30 Antaeus Feldspar
    November 12, 2011

    Now, we should mention at this point that we are discussing mathematical children who have no existence outside the world of probabilistic brainteasers.

    Well, this is fine, but…

    Here, there is no context and thus no way to know what prompted X to make this statement. Could he instead have said “At least one is a girl”? Could he have said “Both are boys”? Could he have said nothing? If you, the one faced with solving the puzzle, are desperate to disambiguate it, you’d probably have to assume that what really happened was: X (for some reason unconnected with X’s identity) was asked whether it was the case that he had at least one son, and, after being warned–by a judge?–that he had to give a yes-or-no answer, said “yes.”

    Why specify that we’re talking about mathematical children and then express such disbelief that the people talking about those mathematical children would be talking mathematically? Sure, in a more real-world conversation, people are unlikely to say that a descriptor such as “male” or “born on a Tuesday” applies to one of their children when it actually applies to both… but then for consistency you’d have to consider that C-sections are rarely scheduled for weekends, et cetera.

    Anyways, my real problem is this:

    That still seems weird to me, even though it is precisely the main principle in the Monty Hall problem.

    Ummm… no, it really isn’t, except in the most general possible terms. The Two Children Problem, and Tuesday Birthday Problem, both deal with two random events; the Monty Hall problem has only one random event. In addition, the TCP and TBP both deal with certain combinations that are possible outcomes of the random events but which are prohibited by the problem description (e.g. the two children both being girls); many people believe that this is also true of the MHP but that’s only the case in the misunderstood version of the problem where it’s believed that Monty only opens a door that’s empty by chance, not because he knows which doors are empty and deliberately opens an empty door. The principle really isn’t the same.

  31. #31 Another Matt
    November 13, 2011

    …the TCP and TBP both deal with certain combinations that are possible outcomes of the random events but which are prohibited by the problem description (e.g. the two children both being girls); many people believe that this is also true of the MHP but that’s only the case in the misunderstood version of the problem where it’s believed that Monty only opens a door that’s empty by chance, not because he knows which doors are empty and deliberately opens an empty door.

    Except that what you’ve just described is where the dispute about sampling with the TBP comes in – how the information about the father was obtained (or equivalently, how the sample pool of fathers was selected, or which population you take the father as representing – the population of all fathers with two children, the population of all fathers with two children and at least one son, or the population of all fathers with two children and at least one son born on tuesday) leads to one of three different probabilities for a two-son father outcome. Here’s a more simple example that will illustrate this more easily:

    Say I’ve put each of two six-sided dice in its own separate cup. I turn the cups over, and then I call you on the phone and say “I can tell you that at least one of the dice came up ’1′. Would you rather bet that I’ve rolled snake-eyes, or that I’ve rolled a ’7′?”

    Your answer to this depend a little on what you think led me to say “at least one of the dice has come up ’1′.” In the case that I have lifted only one of the cups and it happened to be a 1, then the probability for snake-eyes and for ’7′ is still just 1/6. But if I’ve looked at both of them, then the possible rolls that will have elicited “at least one has come up 1″ is:

    1 1
    1 2
    1 3
    1 4
    1 5
    1 6
    2 1
    3 1
    4 1
    5 1
    6 1

    Which leaves a 1/11 probability for snake-eyes and 2/11 for ’7′. So clearly in that scenario I would do better to bet on 7 than snake-eyes. Since I don’t know how you’ve gotten your information I should bet on 7 each time.

    The first scenario – choosing one cup randomly and finding that its die has happened to come up ’1′ is somewhat analogous to the MHP where Monty really does choose a door at random that happens to be empty. The second scenario where I’ve looked at both dice is analogous to the MHP where Monty knows which doors are empty and always chooses the empty one. In both cases – the MHP and the dice game – you don’t lose anything by assuming Monty has more information.

    BTW you could recast the TBP as a dice game where I have four cups, two red and two blue, and I put one “7-sided die” in one red cup and one in a blue cup, and a coin in the other red and blue cup each, and flipped them all over. If I tell you that “at least one of the colors had a coin that came up heads, and at least one heads was associated with a dice-roll of ’3′ under a cup with the same color,” your answer to the probability that both coins were heads would be determined by how many of the cups (and in what order) I’ve lifted to give you that information. If I lifted two cups of the same color and found that it happened to be “heads, 3″ then the probability the coin under the other cup is heads would be 1/2. If I lifted the two coin cups and happened to find one or two heads, and then lifted one of the dice cups with the same color as a heads and happened to find a 3, then the probability they were both heads is 1/3. If I lifted all four cups and found that at least one pairing was “heads, 3″ then the probability the other is heads would be the 13/27 from the original problem — all of these are the same as the analysis in Jason’s post.

  32. #32 Wow
    November 14, 2011

    “people are unlikely to say that a descriptor such as “male” or “born on a Tuesday” applies to one of their children when it actually applies to both”

    But then again, people are even less likely to ask the question posed.

    And why then did they ask what sex their other child was if they’ve said one is male when “male” applies to both?

  33. #33 Antaeus Feldspar
    November 14, 2011

    The first scenario – choosing one cup randomly and finding that its die has happened to come up ’1′ is somewhat analogous to the MHP where Monty really does choose a door at random that happens to be empty. The second scenario where I’ve looked at both dice is analogous to the MHP where Monty knows which doors are empty and always chooses the empty one. In both cases – the MHP and the dice game – you don’t lose anything by assuming Monty has more information.

    Only the latter of those is the actual Monty Hall problem, however. I will stipulate that yes, sometimes people state the problem ambiguously, or even incorrectly, and produce the impression that Monty opening an empty door is simply what happens by chance on this particular occasion – but that isn’t the Monty Hall problem.

    Ironically, the door-is-empty-by-chance variant of the MHP (let’s call it the EBCV) has more in common with TBP than MHP does. MHP has only one meaningful random event, and that is whether the prize winds up after the player’s initial pick a) in the one-member set of “doors that the player picked” or b) in the two-member set of “doors that the player didn’t pick.” In EBCV, there is the random event of whether the prize wound up behind the door that the player picked, and then the random event of whether the prize wound up behind the door Monty picked out of the two remaining. Except that since the problem description states that Monty opens an empty door, we discount the two cases where he picks the prize door. TBP and EBCV have that basic similarity in common, but I really can’t see how one can say that TBP and MHP are working on the same principle.

  34. #34 Another Matt
    November 16, 2011

    I think we’re disagreeing on the level of generality or abstraction. Here’s the principle from the original post:

    …any piece of information that affects the selection will also affect the probability.

    It’s pretty clear that this principle is at play in the difference in resulting probabilities between the MHP and your EBCV, and also that the various flavors of TBP with their resulting probability outcomes differ because of this principle. I don’t think it’s meant to extend beyond this level of generality.

  35. #35 Antaeus Feldspar
    November 17, 2011
    …any piece of information that affects the selection will also affect the probability.

    It’s pretty clear that this principle is at play in the difference in resulting probabilities between the MHP and your EBCV, and also that the various flavors of TBP with their resulting probability outcomes differ because of this principle. I don’t think it’s meant to extend beyond this level of generality.

    Well, if that’s what was meant, I certainly agree that that principle applies to both problems. I just didn’t think that could be what was meant, because it’s certainly not ‘precise’ (as that adjective is distinguished from ‘accurate’.)

  36. #36 Wow
    November 17, 2011

    Although in the given TBP there is no apparent choice in telling you the one and only tuesday’s child. You have to put one in there.

    Since they’re already admitting that despite having said one is a son, there is a probability that they have TWO sons, why is it not that, despite having said that one is born on Tuesday, there is no possibility that they have TWO children born on Tuesday?

    With the doors, Monty KNOWS which doors are empty.

    And that information is partially passed on when he opens one door AFTER you chose a door.

    And it’s that that makes the MHP solution “you’re better off changing your mind”.

    If the questioner were picking the son born on Tuesday, then the chance that their other child is a son is zero.

  37. #37 Another Matt
    November 17, 2011

    Although in the given TBP there is no apparent choice in telling you the one and only tuesday’s child. You have to put one in there.

    Yeah, that was a side issue which gave yet a fourth interpretation of the words. If you look at Tanya Khovanova’s analysis above, or my reframing using dice and coins, you’ll see that what questions elicited the “I have two kids, and one is a boy born on Tuesday” testimony affects the selection criteria for the population of fathers the one in question represents, and thus affects the probability that the other is a son. If you were placing a bet about whether the other child was a boy or a girl, and the probability it was a boy was 1/2, 1/3, or 13/27 (or 12/26 in the side-issue interpretation), if you didn’t know how to interpret the information, or didn’t know how it was gathered, you’d do well to bet on “girl.” In the analogy between MHP and TBP, “Monty KNOWS which doors are empty” is roughly equivalent to “The father KNOWS the sex and the day on which both of his children were born.”

  38. #38 John K.
    November 18, 2011

    There is a fatal mistake.

    You made order important in the case of different sex siblings and order unimportant in the case of same sex siblings. If order is unimportant there are 2 sets: BB, or BG. If order is important BB must be counted twice, once for each order of birth, the same as for different sex siblings. The order important set must then be BB, BB, BG, GB. The probability of 2 boys remains 2/4 or 1/2.

    This mistake goes forward to the days of the week. Order must be uniformly important or unimportant, so there is no prohibition on which day of the week the other child is born on. The probability remains ½.

  39. #39 Brent Ashley
    November 29, 2011

    Introducing the ordering of the children in the 1/3 scenario is a spurious requirement. One might also introduce the completely irrelevant requirement of handedness, with the unbalanced probability of 1/7 left to 6/7 right, further complicating things, however there was no requirement to do so.

  40. #40 Britton Smith
    November 29, 2011

    I was also unsure of the point raised in comment 38 about whether the ordering of the children is important and potentially reduces the final probability back to 50%.

    So I did the only thing I could think of to ease my mind, and coded up a brute force test of the Tuesday Birthday Problem. In this, I have a function that generates a child with a random sex and birthday. It has another function that generates them in pairs and only keeps a pair in which at least one of the children has the desired sex and birthday. I then do this 1,000,000 times and measure the number of times the second child was the sex we were looking for. You can download it here: http://paste.yt-project.org/show/1971/
    Assuming you have Python installed, you can run it with the following command: python tbp.py

    I tested this using a varying number of total matches, and 1,000,000 was well enough converged to be conclusive and I did indeed find that the answer is 13/27.

  41. #41 Britton Smith
    November 29, 2011

    Silly me, I had a bad character in a comment in the header. Here it is again with that fixed. This one also gives a running tally of the percentage as it goes:
    http://paste.yt-project.org/show/1972/

  42. #42 Bart
    November 30, 2011

    Disclaimer – NEVER CARRY A BOMB ON A PLANE (OR ANYWHERE ELSE), the following is only for academic purposes:

    You should always carry a bomb on a plane because the chances of TWO people carrying a bomb on a plane are much lower than one person carrying a bomb on a plane.

  43. #43 Richard Wein
    December 1, 2011

    To those discussing the wording of the original question…

    We need to distinguish between real situations and mathematical abstractions. If we were talking about a real situation (i.e. someone actually had the experience described) there would be a fact of the matter as to what the man meant and we could make some judgement (perhaps a probabilistic one) about his meaning, based on all the information available to us.

    The case here is not a real situation, but an abstraction invented for its mathematical interest. There is no fact of the matter as to what the man really meant. Mathematicians think about such artificial problems because they are of some theoretical interest. The original problem raises issues that are of mathematical interest, but, having raised the issues, mathematicians’ interest is in exploring those issues, and not necessarily in answering exactly the original question. So they formulate alternative questions which are better suited to clarifying the issues that interest them, as Tanya had done.

  44. #44 John K.
    December 2, 2011

    @ Britton Smith

    My programming skill is rudimentary at best, but having looked at the code you linked to, will your filter correctly handle the case of both siblings having the correct target birthday and sex? It seems like such a case would incorrectly trigger both if statements. I have no idea how Python handles sequential if statements, so I apologize if I am incorrect on this.

    I must otherwise admit that I am wrong on this. I am still baffled why not considering order will give a different answer than making order important does, if I am indeed incorrect.

    Thank you for putting your programming skills to task on this problem, by the way.

  45. #45 gr8hands
    December 6, 2011

    Thank you John K. @38 for making the point I was going to make.

  46. #46 JeffJo
    December 11, 2011

    @Richard Wein

    You meet a man on the street and he says, “I have two children. What is the probability my two children share the same gender?” This seems easy; 1/2.

    You meet a woman on the street and she says, “One of my two children is a son. What is the probability my two children share the same gender?” By your reply above, it seems you would say 1/3. But what if the woman had said “One of my two children is a daughter?” If this is really just a matter of being a “mathematical abstraction,” your answer would have to be 1/3 also. But if it is 1/3 regardless of what the woman says, how is the problem different than the one about the man?

    It isn’t. This fact was first recognized by Joseph Bertrand in 1889 when he published his famous Box Paradox, which I tweaked only by adding a fourth case. Other than that, the two problems are identical. With that fourth case, the “wrong” answer is 1/3, and the “right” one is 1/2. As a “mathematical abstraction,” you have to include the probability of obtaining the information you have, not just the probability that it is true, or else you will get a paradox.

  47. #47 Stuart H
    June 29, 2012

    Sorry, but no. The Tuesday is irrelevant.

    If the son who was born on a Tuesday is the older son, then there are 2*7 = 14 ways to choose the second son, and 7 of those ways are boys, so its a 50% chance.

    If the son who was born on a Tuesday is the younger son, then there are 2*7 = 14 ways to choose the second son, and 7 of those ways are boys, so its a 50% chance.

    You DO NOT adsd 14 + 14 to get 28 total configurations (minus 1 for overlap)…rather, you see that the answer is 50% regardless of this unknown information.

    Based on the logic of this paper, you could always arbitrarily condition your numbers based on additional information and change the result. Not the case.

  48. #48 JeffJo
    July 20, 2012

    And with the third half of the pie … :)

    Stuart H, your solution is wrong because it counts the family of two boys, both born on a Tuesday, twice. In that family, we do not know if the either son was being specifically referred to.

    However, the answer (the number 50%) you give is right, for different reasons.

    Of the 196 possible combinations:

    169 do not include a boy born on a Tuesday. We can only assume the man you met is truthful, so he can’t be one of these 169 combinations.

    14 have a boy born on a Tuesday and a girl; 7 where the boy is the older child, and 7 where he is the younger. But the father of such a combination could have told you he has a daughter who was born on any day of the week. So we can only count 7 of these combinations, even though 14 exist.

    12 have a boy born on a Tuesday and a boy born on a different day. But the father of such a combination could have told you he has a son who was born on a day of the week other than Tuesday. So we can only count 6 of these combinations, even though 12 exist.

    1 has two boys born on a Tuesday. The father of this combination could not have said something else.

    So the answer is (6+1)/(7+6+1)=7/14=1/2. You get the incorrect answer if you count all of the combinations that exist, (12+1)/(14+12+1)=13/27. That can only be the correct answer only the man said yes to the question “Is one of your children a boy born on a Tuesday?”

  49. #49 wow
    July 20, 2012

    Jeff, stuart doesn’t count two boys on tuesday twice.

    Though he should have used child rather than son.

  50. #50 Sune Foldager
    Denmark
    July 25, 2012

    What I find increadible is that even after the right analysis has been posted over and over, and even in many different varieties. Even after computer simulations have been shown to agree with the correct answer in numerous blogs, that there are still people who refuse to belive the correct answer.

    That every blog on this trigger my “crank alert” with posts starting with “sorry, but you’re all wrong”, “there is a fatal flaw” or “epic fail”. What, do you guys really think that you alone have discovered a problem in countless peoples combined reasoning? It’s one thing to not understand a problem, but another to claim everyone is wrong.

    To be totally clear, the problem in precise terms is “Out of all pairs of children where at least one is a boy born on a tuesday, what proportion of them consist of two boys?”. We can argue semantics all day, but I believe this is the most natural interpretation of the problem GIVEN that it’s supposed to be a math puzzle and not a casual conversation between friends. In math puzzles, generally all pieces of information should be used for something, and everything not stated should be assumed equally distributed.

    Now, I won’t reiterate the many different ways to arrive at the correct 13/27 answer, but I can point to this page (below) which does a nice graphical proof. This may also help to convince people who somehow think “BB should be counted twice”. I don’t know how you can think that two random variables, A and B, each with two outcomes, B and G with 50% for each, can result in anything other than (ordered A then B): BB, GG, BG, GB outcomes. 2 * 2 = 4 (outcomes) after all. and 50% * 50% = 25% (for each outcome).

    For those who still don’t believe it, please write a computer simulation. You’ll see that the answer is correct :).

    http://mikeschiraldi.blogspot.dk/2011/11/tuesday-boy-problem-in-under-300-words.html

    (sorry if I sound a bit arrogant above; I don’t mean any disrespect to people who don’t understand the problem, or interpret it a different way; I just dislike claims to authority by people who aren’t knowledgable in the field of probability.)

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