The following probability question appeared in an earlier thread:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

The claim was that it is not actually a mathematical problem and it is only a language problem.

If one wanted to restate this problem formally the obvious way would be like so:

Definition: Sex is defined as an element of the set $\\{\text{boy},\text{girl}\\}$.

Definition: Birthday is defined as an element of the set $\\{\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday},\text{Sunday}\\}$

Definition: A Child is defined to be an ordered pair: (sex $\times$ birthday).

Let $(x,y)$ be a pair of children,

Define an auxiliary predicate $H(s,b) :\\!\\!\iff s = \text{boy} \text{ and } b = \text{Tuesday}$.

Calculate $P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y))$

I don't see any other sensible way to formalize this question.

To actually solve this problem now requires no thought (infact it is thinking which leads us to guess incorrect answers), we just compute

$$ \begin{align*} & P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y)) \\\\ =& \frac{P(x\text{ is a boy and }y\text{ is a boy and }(H(x)\text{ or }H(y)))} {P(H(x)\text{ or }H(y))} \\\\ =& \frac{P((x\text{ is a boy and }y\text{ is a boy and }H(x))\text{ or }(x\text{ is a boy and }y\text{ is a boy and }H(y)))} {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\ =& \frac{\begin{align*} &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday}) \\\\ + &P(x\text{ is a boy and }y\text{ is a boy and }y\text{ born on Tuesday}) \\\\ - &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday and }y\text{ born on Tuesday}) \\\\ \end{align*}} {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\ =& \frac{1/2 \cdot 1/2 \cdot 1/7 + 1/2 \cdot 1/2 \cdot 1/7 - 1/2 \cdot 1/2 \cdot 1/7 \cdot 1/7} {1/2 \cdot 1/7 + 1/2 \cdot 1/7 - 1/2 \cdot 1/7 \cdot 1/2 \cdot 1/7} \\\\ =& 13/27 \end{align*} $$

Now what I am wondering is, does this refute the claim that this puzzle is just a language problem or add to it? Was there a lot of room for misinterpreting the questions which I just missed?

  • $\begingroup$ The "born on a Tuesday" question also came up here: math.stackexchange.com/questions/3278/probability-of-a-given-b/… $\endgroup$ Sep 11, 2010 at 7:20
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    $\begingroup$ In case you haven't found it yet, there is a plethora of opinions here: sciencenews.org/view/generic/id/60598/title/… $\endgroup$ Sep 11, 2010 at 7:36
  • $\begingroup$ I was particularly hoping that people would directly address my derivation as written and the interpretations I used here. (Hence actually writing them out as opposed to just giving the value 13/27). $\endgroup$
    – anon
    Sep 11, 2010 at 7:38
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    $\begingroup$ So you are having a beer with a coworker. The coworker says "I have a sibling". What is the probability the sibling is a brother? Would the probability be different if I told you the coworker was female? I'm in the linguistic camp. (If it isn't obvious my co-worker scenerio is meant to parallel the two children problem. A family has two children; 1 is a boy, what is the probability the other is a boy. The answer is 2/3 but the coworker, the probability is 1/2) $\endgroup$
    – fleablood
    Jan 18, 2016 at 5:10
  • $\begingroup$ Yes, yes, yes it is a language trick. With some mathematical aspects. I'll say more, it is a language trick with several nuances specific to english language. $\endgroup$ Jun 7, 2018 at 0:34

12 Answers 12


There are even trickier aspects to this question. For example, what is the strategy of the guy telling you about his family? If he always mentions a boy first and not a daughter, we get one probability; if he talks about the sex of the first born child, we get a different probability. Your calculation makes a choice in this issue - you choose the version of "if the father has a boy and a girl, he'll mention the boy".

What I'm aiming to is this: the question is not well-defined mathematically. It has several possible interpretations, and as such the "problem" here is indeed of the language; or more correctly, the fact that a simple statement in English does not convey enough information to specify the precise model for the problem.

Let's look at a simplified version without days. The probability space for the make-up of the family is {BB, GB, BG, GG} (GB means "an older girl and a small boy", etc). We want to know what is $P(BB|A)$ where A is determined by the way we interpret the statement about the boys. Now let's look at different possible interpretations.

1) If there is a boy in the family, the statement will mention him. In this case A={BB,BG,GB} and so the probability is $1/3$.

2) If there is a girl in the family, the statement will mention her. In this case, since the statement talked about a boy, there are NO girls in the family. So A={BB} and so the probability is 1.

3) The statement talks about the sex of the firstborn. In this case A={BB,BG} and so the probability is $1/2$.

The bottom line: The statement about the family looks "constant" to us, but it must be looked as a function from the random state of the family - and there are several different possible functions, from which you must choose one otherwise no probabilistic analysis of the situation will make sense.

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    $\begingroup$ Nope. I'm saying that in order to give a probabilistic meaning to this statement, you need to attach probabilistic assumptions to it. In your case it seems you've chosen the assumption of "if there is a boy, a boy will be given in the statement". Other possible interpretations: "If there is a girl, a girl will be said" (so since a boy was said, there are no girls 100%), and "If there is a boy and a girl, one will be said at random" and "The sex of x is said" and so on. Forget about days - try to do the probability for these cases and see what happens. $\endgroup$
    – Gadi A
    Sep 11, 2010 at 6:11
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    $\begingroup$ This is of course correct, but you must also take into account the question of how the statement "I have a boy" comes to be. I've tried to elaborate in my answer. $\endgroup$
    – Gadi A
    Sep 11, 2010 at 6:40
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    $\begingroup$ The statement (ignoring the day red herring) is "I have two children. One is a boy." As I said before, this statement should be considered as part of the probabilistic scenario otherwise the problem is simply not well-defined. $\endgroup$
    – Gadi A
    Sep 11, 2010 at 7:37
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    $\begingroup$ What I meant was that adding the day does not affect the "strangeness" of the situation. However, I'm not sure if it's the strangeness that interests you. The main point is that it's called "a language problem" because different mathematical interpretations of the same English statement yield different mathematical models, and hence different mathematical results - and so for every intuition about the problem there is a interpretation in which the result is counter-intuitive. $\endgroup$
    – Gadi A
    Sep 11, 2010 at 11:52
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    $\begingroup$ Exactly; this is why this question, in its ambiguous phrasing, would make a poor exam question. Since there are several acceptable formulations "from first principles" of this problem. $\endgroup$
    – Gadi A
    Sep 15, 2010 at 15:45

It is actually impossible to have a unique and unambiguous answer to the puzzle without explicitly articulating a probability model for how the information on gender and birthday is generated. The reason is that (1) for the problem to have a unique answer some random process is required, and (2) the answer is a function of which random model is used.

  1. The problem assumes that a unique probability can be deduced as the answer. This requires that the set of children described is chosen by a random process, otherwise the number of boys is a deterministic quantity and the probability would be 0 or 1 but with no ability to determine which is the case. More generally one can consider random processes that produce the complete set of information referenced in the problem: choose a parent, then choose what to reveal about the number, gender, and birth days of its children.

  2. The answer depends on which random process is used. If the Tuesday birth is disclosed only when there are two boys, the probability of two boys is 1. If Tuesday birth is disclosed only when there is a sister, the probability of two boys is 0. The answer could be any number between 0 or 1 depending on what process is assumed to produce the data.

There is also a linguistic question of how to interpret "one is a boy born on Tuesday". It could mean that the number of Tuesday-born males is exactly one, or at least one child.

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    $\begingroup$ +1. We have a reasonably standard jargon for specifying simple random processes in English. People come up with "puzzles" like this by exploiting the impreciseness of English for describing more complicated situations. If the person posing the question had directly stated the random process, nobody would be confused. The reason this is a "puzzle" is that it's worded too vaguely to convey the intended meaning. So @muad: any time someone asks you a probability "puzzle", you should simply press them about exactly what random process they have in mind. Then it will be easy. $\endgroup$ Sep 12, 2010 at 12:07
  • $\begingroup$ What you call "a linguistic question" over what "one is a boy born on Tuesday" means is actually the observation that people are capable of misunderstanding even unambiguous language. I do not think this question is about that sort of misunderstanding, and it would be unchanged by substituting "at least one is a boy born on a Tuesday" for the above phrase. (comment reposted to correct an incorrect quote.) $\endgroup$
    – sdenham
    Aug 7, 2017 at 15:28
  • $\begingroup$ @sdenham, no, that has absolutely nothing to do with it. Consider this puzzle (a real one): I am typing this comment on a laptop. I have worked in graphic design and I have worked as a sysadmin. What is the probability that I am using a Mac? This question, just like the original, is impossible to answer without a probability model being set up. It's actually a meaningless question. $\endgroup$
    – Wildcard
    Sep 27, 2017 at 5:45

I guess the following two versions of framing the question yield two different probabilities:

  1. Dave has two children. Is atleast one of them a boy who is born on Tuesday? Dave answers Yes.

  2. Dave has two children. I ask him to first choose and fix one child at random, and tell me if it is a boy who was born on Tuesday. Dave answers yes he is a boy born on Tuesday.

For 1st the probability (of both being boys) is 13/27, while for the second the probability is 1/2.

The way in which the question is asked, it's in line with 1st, hence the answer should be 13/27.

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    $\begingroup$ +1 for correct answer. More specifically, your first example is equivalent to the case where Dave knows the sex/birthdate of both children, while the second example is equivalent to the case where Dave doesn't know the sex/birthdate of both children; so what we have here is yet another question about knowledge-of-knowledge. See this answer for more information. $\endgroup$ Dec 21, 2010 at 20:11
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    $\begingroup$ Also, since there is - presumably - no ambiguity over whether or not Dave knows the sex of his children, there is indeed no ambiguity in the quesiton; the answer is correctly 13/27 (the accepted answer, by Gadi, is incorrect). $\endgroup$ Dec 21, 2010 at 20:13
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    $\begingroup$ @BlueRaja-DannyPflughoeft: The issue isn't whether Dave knows his children, but under what circumstances he made his statement. In other words if Dave says "I have two children, and at least one of them is a boy born on a Tuesday", the information conveyed by his statement depends on the space of possible ones. (You're assuming that the information conveyed is the content of his statement: and yes, the answer to "if we are given the information that Dave has a boy born on a Tuesday..." is indeed 13/27.) See Tanya Khovanova's article "Martin Gardner's Mistake". $\endgroup$ Jan 17, 2014 at 5:38
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    $\begingroup$ In my brain, I have an implicit prior distribution over the possible kids someone has and whether they are male or female. Upon hearing they have 2 kids, I update on them having 2 kids. Upon hearing that one of them is a boy, I update not on the fact that at least one is a boy, but rather on the fact that a random child of theirs was a boy -- in other words, mentally, I assume WLOG they are talking about a random child when they make the statement. Thus I get that the other child is equally likely a boy or a girl. "Boy born on tuesday" additionally does not affect it :) $\endgroup$ Apr 8, 2017 at 21:25
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    $\begingroup$ I agree with (1) and (2) in your answer. The thing I object to is: "The way in which the question is asked, it's in line with 1st." That makes a nontrivial probabilistic assumption and sweeps the details under the rug. Namely, that the probability of me SAYING X is exactly the same as the probability that, had I been ASKED X, I would have answered yes. This assumption, by the way, is patently false. $\endgroup$ Jun 6, 2017 at 11:05

Well, given the unstated assumption that the writer is a mathematician and therefore not using regular english, then I agree with the 13/27 answer.

But in everyday english, from "there are two fleems, one is a glarp" we all infer that the other is not a glarp.

From "there are two fleems, one is a glarp, which is snibble" we would still infer that the other is not a glarp. Whereas from "there are two fleems, one is a glarp which is snibble" (absence of comma, or when spoken, difference in intonation) we would infer that the other is not a snibble glarp, but it could still be an unsnibble glarp.

  • $\begingroup$ I see what you mean, Whether one interprets "One is ..." to mean "Exactly one (and no more) is ..." or "At least one ...". This would change the answer from 13/27 to 12/26. I don't think this ambiguity is an intentional part of the problem though - just an unfortunate problem of the language. $\endgroup$
    – anon
    Sep 12, 2010 at 7:57
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    $\begingroup$ I think "But in everyday english, from "there are two fleems, one is a glarp" we all infer that the other is not a glarp." is debatable. $\endgroup$
    – Chris Card
    Sep 12, 2010 at 8:31

The Tuesday is a red herring. It's stated as a fact, thus the probability is 1. Also, it doesn't say "only one boy is born on a Tuesday". But indeed, this could be a language thing.

With 2 children you have the following possible combinations:
1. two girls
2. a boy and a girl
3. a girl and a boy
4. two boys

If at least 1 is a boy we only have to consider the last three combinations. That gives us one in three that both are boys.
The error which is often made is to consider 2. and 3. as a single combination.

I find it completely counter-intuitive that the outcome is influenced by the day, and I simulated the problem for one million families with 2 kids. And lo and behold, the outcome is 12.99 in 27. I was wrong.

  • $\begingroup$ Note that this too may seem counter-intuitive: It means that if someone tells you "I have two children. One of them is a boy" the probability the the other one is a girl is 2/3, but you might expect 1/2 since the sex of the children is independent. $\endgroup$
    – Gadi A
    Sep 11, 2010 at 7:00
  • $\begingroup$ What you are saying here contradicts my derivation of 13/27 (which I cannot see any flaw in) - can you tell me if you think there is a mistake in my derivation? $\endgroup$
    – anon
    Sep 11, 2010 at 7:03
  • $\begingroup$ @muad: there doesn't seem a mistake in your derivation; the error was in my intuition that the day couldn't possibly have anything to do with it. $\endgroup$
    – stevenvh
    Sep 12, 2010 at 7:04
  • $\begingroup$ I also had to write a computer simulation before I could believe the answer! $\endgroup$
    – anon
    Sep 12, 2010 at 7:49
  • $\begingroup$ Indeed the 13/27 answer is counter-intuitive (+1 for your edit). FYI I have provided a new answer that gives an intuitive argument why it should be between $0.333$ and $0.5$. $\endgroup$
    – Michael
    Nov 8, 2021 at 17:47

There is always room for misinterpreting a question when one does not fully understand the language in which it is written. I think that the way mathematics and mathematicians use conditional probability is clear:

$$P(A|B)=P(A \cap B)/P(B).$$

So I believe that this is the interpretation that one should take, and thus arrive at your answer of 13/27, and not search for further nuances, which are not too difficult to find.

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    $\begingroup$ Conditional probability is not the issue. The problem is in defining what "P" is implied in the problem. Once a probability measure is specified ,of course it is then possible to use the formula for conditional probability. $\endgroup$
    – T..
    Sep 11, 2010 at 19:08
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    $\begingroup$ So A = [Dave has 2 boys] and B = [Dave says "I have 2 children. One is a boy born on a Tuesday."]. The problem is that we don't know P(B). What space does Dave's statement come from? The only way to get the 13/27 answer is to make the unjustified unreasonable assumption that Dave is boy-centric & Tuesday-centric: if he has two sons born on Tue and Sun he will mention Tue; if he has a son & daughter both born on Tue he will mention the son, etc. See [this article](arxiv.org/abs/1102.0173). The information in [Dave says X] is not the same as [X]; it depends on what all Dave could have said. $\endgroup$ May 5, 2011 at 11:08

This, in my opinion, is why the intuitive approach fails:

One has a tendency to think that the probability of 7*P(b AND d1) = P(b AND d1) + P(b AND d2) + ... + P(b AND d7) = P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)) = P(b AND (d1 OR d2 OR ... OR d7)) = P(b).

However, the flaw here is that, in reality, P(b AND d1) + P(b AND d2) + ... + P(b AND d7) is NOT equal to P((b AND d1) OR (b AND d2) OR ... OR (b AND d7)). This means that mentioning independent (and one might think irrelevant) information alongside with relevant information actually changes the resulting probabilities.

One interesting consequence: if I say something like "I have two children. One of them is a boy who was born at 10:24 PM on February 10th," The probability that I have two boys is now almost exactly the same as as the probability that I have a girl and a boy. Adding a unique or almost unique piece of information makes the stuff I want to know about the other child independent of the information I have on the first child. If I took this to the extreme and said that I have a firstborn boy, won't know anything additional about the other child.

  • $\begingroup$ What is an example of the "... is NOT equal ..."? (A well-specified probability model where the two sides of the equation are not the same.) $\endgroup$
    – T..
    Sep 13, 2010 at 1:26

The VERY simple answer is that you can get rid of all ambiguity if you can clarify from what pool the parent was chosen. This allows you to restate the "probability" question as a "percentage" question, taking idealized percentages.


Given a parent randomly selected from the pool of all parents who have two children where at least one of the children is a boy born on Tuesday, what is the probability that both children are boys?

This can be stated as a percentage question:

What percentage of (parents who have two children where at least one of the children is a boy born on Tuesday) have two boys?

By contrast, if the day was a matter of chance (not a restriction on the pool size), we get a different question:

Given a parent randomly selected from the pool of all parents who have two children at least one of whom is a boy, where after the random selection is made we shall be told the day of the week on which the boy was born (if there is only one boy) or if there are two boys we shall be told the day of the week on which a randomly chosen one of the two boys was born, what is the probability that both children are boys if the day we are told is Tuesday?

As you can see, in this setup the "Tuesday" part has absolutely no influence on the selection process, and can be entirely disregarded.

This is what is meant by Wikipedia's statement:

The moral of the story is that these probabilities do not just depend on the known information, but on how that information was obtained.

As a further note, you can't "prove" anything about the "actual" meaning by using computer simulations, because in order to program a computer simulation in the first place you must first disambiguate which scenario you are actually talking about. So the only thing a computer simulation can "prove" is how the programmer interpreted the question.

  • $\begingroup$ Wikipedia link please $\endgroup$
    – Roland
    Jun 11, 2020 at 11:49
  • $\begingroup$ @Roland edited it in. $\endgroup$
    – Wildcard
    Jun 11, 2020 at 17:12

The $13/27\approx 0.481$ answer is an interesting counter-intuitive result. Here I give an intuitive explanation (which I do not think has been given in the above answers):

Intuition of why it is between $0.333$ and $0.5$:

We can distinguish the two children by firstborn and secondborn. If I tell you the firstborn is a boy, it tells you nothing about the secondborn, so the chances the secondborn is a boy is $0.5$.

If we know one was born on Tuesday, we can "roughly" distinguish the two children by "born on Tuesday" and "not born on Tuesday" (with a small probability that both were born on Tuesday). So the answer is "almost" $0.5$, since it is similar to distinguishing between firstborn and secondborn. This is the intuition why the probability changes from $0.333$ to $0.481$ based on the Tuesday info (but does not go all the way up to $0.5$).

Formal derivation of $13/27$

I did not understand the $H(x)$ notation given by the OP so I simply rewrite the argument in standard notation. Let $B_1, B_2$ be events that the firstborn and secondborn are boys, respectively. Let $T_1, T_2$ be events that the firstborn and secondborn were on Tuesday, respectively. For events $A, B$ we define $AB=A \cap B$.

\begin{align} P[B_1B_2 | B_1 T_1 \cup B_2 T_2]&=\frac{P[B_1 B_2(B_1 T_1 \cup B_2T_2)]}{P[B_1T_1 \cup B_2 T_2]}\\ &=\frac{P[B_1B_2T_1 \cup B_1B_2T_2]}{P[B_1T_1 \cup B_2 T_2]}\\ &=\frac{P[B_1B_2T_1]+P[B_1B_2T_2]- P[B_1B_2T_1T_2]}{P[B_1T_1] + P[B_2T_2] - P[B_1B_2T_1T_2]}\\ &=\frac{(1/2)^2(1/7) + (1/2)^2(1/7) - (1/2)^2(1/7)^2}{(1/2)(1/7) + (1/2)(1/7)-(1/2)^2(1/7)^2} = \frac{13}{27} \end{align} where the last line assumes independence and uniformity over boy/girl and days of week.

Of course without the Tuesday info we get $$ P[B_1B_2|B_1 \cup B_2] = \frac{P[B_1B_2]}{P[B_1\cup B_2]} = \frac{1}{3}$$


It's not a language trick, and here's a better example to illustrate the phenomenon. Let's take 4 types of families in equal numbers:

  1. 10 boys
  2. 9 girls, then 1 boy were born
  3. 1 boy, and then 9 girls were born
  4. 10 girls

When we say "family has at least 1 boy", we end up with the same probability to get the 10-boy family: $1/3$.

But when we add a condition "family has at least 1 boy born on Tue", this filters out most of 1-boy families. But many 10-boy families will satisfy this additional condition.

So any additional condition will favour the group which has more potential to have that condition. If we change the condition to "at least one boy and none of the boys were born on Tue", this would favour the families with just one boy.


Generically, one could say that there are two halves to any math problem: First, you establish the mathematical relationships between the objects in the problem statement. Then, using mathematical theorems, you solve for the missing element. The only "language trick" here occurs when people decide that an problem statement, which is ambiguous, has an incontrovertible meaning. For example, in geometry, say we are told that two sides of a triangle have lengths 2 and 3, and that the angle between these sides is 30 degrees. There are actually two triangles that satisfy these conditions, one with three acute angles and one with an obtuse angle. But a beginning student might consider only the acute one.

The issue with probability problems like this one, is that such ambiguous details can very easily exist in a hidden manner. And even an experienced mathematician can get confused by taking the stated details too literally. For example, in the famous Monty Hall Problem, if one looks at only (1) the car was equally likely to be behind any of the three doors and (2) we know it wasn't door #3, then the correct solution is that it is still equally to be behind #1 or #2. The correct solution is that Monty Hall had to open door #3 if the car is behind door #2 (the contestant's first choice was #1). We aren't told, but have to assume, that he would pick randomly between #2 and #3 if the car is behind #1. This eliminates half of the cases where it is behind #1, but none of the cases where it is behind #2, making it now twice as likely that the car is behind #2.

I mention the Monty Hall Problem, because it belongs to the same family of problems as this one. And has, essentially, the same issues and the same solution. There are two possible ways that Gary Foshee could have arrived statement in the problem:

  1. The producers of the G4G conference could have formulated the conditions "Two children, at least one born on a Tuesday" and then sought a presenter who met the criteria. Since 27 out of every 196 parent of two children meet these requirements, and 13 of those have two boys, the answer is 13/27.
  2. The producers asked Gary Foshee to present a problem. Since he had two children, he decided to present a variation of a famous one made famous by the conference's namesake, Martin Gardner. He wanted to alter the statement to one in the form "I have two children, and at least one is a who was born on ." There were probably two such statements that apply to his children, but there is at least one. There is one case where he has no choice except "Tuesday Boy" (and so he has two boys) But there are 26 where he could choose "Tuesday Boy" but could also choose something else. If he chooses randomly, only 13 remain. In 6 of those, he has two boys. The answer is (1+6)/(1+13)=1/2.

I must stress that both are possible. But the first requires an unreasonable interpretation of the ambiguous problem.


At the annual meeting of the Society for Parents of Exactly Two Children, each attendee is given a name tag pre-printed with their name and this sentence: "I have at least one who was born on a ." They are asked to fill in the blank spaces as indicated. (BTW, I would write "Boy" and "Saturday" on mine.)

Say the first person you meet at this conference has written "Boy" and "Tuesday." What is the probability that this person has two boys? If you think it is 13/27, then a similar answer - adjusted for the gender they wrote - applies to every attendee at the meeting. That is, you should expect that 14/27 of the attendees have a boy and a girl. Yet we know this isn't true - you should expect it to be 1/2.

This paradox has a name, "Bertrand's Box Paradox." That name does not refer to the problem Joseph Bertrand posed (three boxes, which contain two gold coins, two silver coins, and one of each, respectively). It refers to how he demonstrated that choosing one box, and one coin from it, "eliminates from the sample space" more than just the outcome where you picked the box with two of the other kind of coin. It also eliminates the outcome that you picked the mixed box, and picked the other kind of coin from it.

The problem asked about here is not a language trick, it is a logic trick. The naive solution "eliminates" just the cases that can't produce the observed result. The correct solution first accounts for all possible outcomes, and then "eliminates" all that don't produce the observed outcome. That includes some that could.

Let's say there are 392 members of this society (yes, that's twice the number other answers here considered).

  • 98 have two boys.
  • 26 have two boys and at least one Tuesday Boy.
  • 14 have two boys, at least one Tuesday Boy, and would write "Tuesday Boy." It is one more than 26/2, to account for families with two of them.
  • 196 have one boy and one girl.
  • 28 have a Tuesday Boy and a girl.
  • 14 have a Tuesday Boy, a girl, and would write "Tuesday Boy."

So, of the 28 attendees who would write "Tuesday Boy," 14 have two boys. The answer to the original question is 1/2. Not 13/27.


Bertrand's Box Paradox can be used in many famous paradoxes. In some, it confirms the canonical answer. In others, like this one and the Two Child Problem it is based on, it contradicts it.

In the Monty Hall Problem, after you pick door #1 and the host opens door #3 to reveal a goat, there are two places left for the car. The naive answer says eliminating door #3 increases your chances from 1/3 to 1/2. Bertrand's Box Paradox says that if opening door #3 changes the probability to 1/2, so would opening door #2. Since the host will always open one of those two doors, we don't need to see him do it. That fact makes the answer 1/2. The correct solution is that we also need to eliminate the case where both doors #2 and #3 have goats, and the host chooses #2 to open. That makes it twice as likely that door #2 has the car, than door #1.

In the Two Child Problem ("Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?"), the very naive answer says that "the other child has a 1/2 chance to be a boy." The naive answer says that we eliminate GG families, and keep BB, BG, and GB, making the answer 1/3. Bertrand's Box Paradox says that if this is true and we are told that Mr. Smith has at least one girl, then the probability of two girls is also 1/3. But that means that without being told a gender, the chances of two of the same gender is 1/3. The correct answer is that in half of the BG and GB families, we would expect to be told about the girl. So the answer actually is 1/2.

In the real world, this solution method is known as the Principle of Restricted Choice. It is used by Contract Bridge masters when opponents hold two functionally-equivalent cards, and one has been played.


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