## One is a boy born on a Tuesday…

My colleague brought this intriguing probability question to my attention which, as the comments following the article show, has a controversial answer. I believe the reason for this is that there is more than one way to model the question in a random framework.

“I have two children. (At least) one is a boy born on a Tuesday. What is the probability I have two boys?”

The puzzle teller then asks, “What has Tuesday got to do with it?”

(We can assume that the probability of a baby being born a male is 1/2.)

Being pragmatic, to answer the above question, we need to think of a way of (at least conceptually) repeating the “experiment” many times, so that we can deduce what the long term average of there being two boys is.

**Model A**: Look at all families in this world who have two children, (at least) one of them a boy born on a Tuesday. How many of these families have two boys?

**Model B**: Look at all families in this world who have two children. Pick one of the children uniformly at random and state the sex and weekday of birth of that child. How many of these families will have both children of the same sex?

**Model C**: Look at all families in this world who have two children. Pick one of the children uniformly at random and state the sex and weekday of birth of that child. How many of these families will have two boys?

**Model D**: Look at all families in this world who have two children. Pick one of the children uniformly at random and state the sex and weekday of birth of that child. Out of only those families who stated they have a boy, how many will have two boys?

**Model E**: Look at all families in this world who have two children, (at least) one a boy. State the weekday of birth of the boy (or if there are two boys, pick one at random and state his weekday of birth). How many of these families will have two boys?

In Models B to D, the term “uniformly at random” is intended to mean that there is equal chance of picking either child. It should be compared with other choices, such as “pick the younger child” or “pick the boy if there is a boy”.

The puzzle teller is actually interested in Model A, and indeed, the answer is an interesting one in this case, as elaborated on below. The controversy over this puzzle arises because people, wittingly or otherwise, consider other models, especially Model E above. Indeed, *we need to ask for clarification from the puzzle teller on how and why he chose to tell us that one of his children was a son born on a Tuesday* before we can solve this puzzle without making any additional assumptions. (That said, the puzzle teller would presumably argue that Model A is the most sensible and therefore the default interpretation of the question. Note though that by being asked what the relevance of Tuesday is, the audience will start to think along the lines of Model E, thus fueling the controversy.)

In Model B, whether or not we learn the sex and weekday of birth of one of the children is irrelevant to the question being asked; one half of all two-children families will have both children of the same sex. Similarly, in Model C, the answer is 1/4 because we are considering the proportion of all families. In Model D, all families with two girls are discarded and there is 50% chance that families with one boy and one girl are discarded, while all families with two boys remain. Therefore, 1/2 of all remaining families will have two boys; it is irrelevant to the question being asked that we are told on which day of the week the boy was born. The answer to Model E is 1/3; learning the weekday of birth is again irrelevant to the question we have been asked.

The difference between Model A and E is that in Model A, we have discarded all families who do not have a son born on a Tuesday, whereas in Model E, we merely learn of the weekday of birth of the son. The distinction is crucial! Would we always be told the same information, or would the information we are told change from experiment to experiment?

### The Answer to Model A

The fail-safe method for answering any probability question concerning only a finite number of outcomes is to enumerate all the outcomes. It may not be illuminating on its own but it is rigorous. The first child could be a boy or a girl (2 possibilities) and it could have been born on any day of the week (7 possibilities). There are therefore 14 possible outcomes. Same for the second child, leading to a total of 14 times 14 equally likely outcomes (but we don’t need to know this number). What information does knowing that “one of the children is a boy born on Tuesday” tell us? Well, it fixes precisely the outcome of either the first child or the second child. To avoid double counting, we need a little care; we consider the three non-overlapping cases:

- First child is a boy born on a Tuesday, second child is not a boy born on a Tuesday
- First child is not a boy born on a Tuesday, second child is a boy born on a Tuesday
- Both children are boys born on a Tuesday

There are 13 possibilities for the first case with 6 being that the second child is a boy; 13 possibilities for the second case with 6 being that the first child is a boy; and only 1 possibility for the third case. So the probability that the other child is a boy is (6+6+1) / (13+13+1) = 13/27.

### An Explanation

In Model A, we are filtering out families that do not meet the criteria of i) at least one boy; and ii) at least one boy born on a Tuesday. What is the difference between filtering out just on (i) versus filtering out on both (i) and (ii)? They key point is that (ii) will filter out more families with just one boy than it will filter out families with two boys. For every family with one and only one boy, we will discard 6 out of 7 of them (equivalently, 42 out of 49) because the boy will fail to be born on a Tuesday. For families with two boys though, they have a greater chance at passing the Tuesday test because, roughly speaking, they get two goes at it! We will discard only 36 out of 49 such families.

Consider a pool of families having at least one boy. We may assume that 2/3 of them have exactly one boy, that is 98 families have exactly one boy and 49 have two boys. Now, we discard from the 98, 42 out of every 49, that is, we discard 84. So by discarding families not having a son born on a Tuesday, we are left with 14 families having exactly one boy. From the two-boy families, we discard 36 out of 49, so we are left with 13 two-boy families. Therefore, 13 out of families having at least one boy born on a Tuesday will have two boys.

### Moral

Probability (and mathematics in general) sometimes gives counter-intuitive results. We choose to accept these results because they are founded on axioms and logic which experience has shown to be extremely useful and conceptually very intuitive. The trick then is to improve our intuition based on such examples so we are better prepared in the future.

Another moral is that English is imprecise. The puzzle is controversial because it can be interpreted in different ways, and in particular, I speculate that although we feel that we understand the question, our subconscious mind doesn’t quite know which way is best to understand it, so it sporadically jumps from one interpretation to another the more we think about it, just as it does in the young woman / old woman optical illusion where the mind sporadically jumps between seeing a young woman or seeing an old woman. (Rather than say “I have two children..”, the puzzle teller should have said, “Pick at random a two-child family who has a son born on a Tuesday.” It is not as catchy though this way…)

Very impressive puzzle. Thank you for explaining this so clear.

The controversy does not come from ambiguity in English, although there is plenty of ambiguity there. Probability is actually a measure of ambiguity: when more than one option could apply to a situation, probability tells us the relative likelihoods that each applies. The controversy comes from ignoring the ambiguity that is present, which should be modeled with probability.

Let me explain by example. Suppose, instead of telling you about his child, the person in the problem hands you a sealed envelope and says “I have two children. Inside this envelope is information that applies to at least one of them, although it may apply to both. The information consists of the child’s gender, and the day of the week upon which he or she was born. What is the probability that both of my children share the same gender?”

If you believe that Model A should apply to the question you asked, then you have a problem. If you were to open the envelope and read the information, you would have to say the probability I asked for is 13/27 regardless of what that information is. And since your answer would be 13/27 in every case, it must be 13/27 even if you don’t open the envelope at all! But if the person hadn’t described the contents of the envelope, which actually tells you nothing, you would say the answer was 1/2.

This contradiction – that merely compiling some information about one child would change the probability – is a form of Bertrand’s Box Paradox. The common property that makes it so is that there are two sets of similar information that could be compiled, but only one actually was. This is an ambiguity that is a valid part of the problem statement, not a weakness in it. A probability needs to be assigned to the two cases where the information is about child A, or child B (however you want to distinguish them – most people do it by age, but I prefer to alphabetize their names), and those probabilities must sum to 1. And since you can’t assume that they are different based on the problem statement, you can only assume each is 1/2. The result is your Model B (which, with the assumption of equal probabilities, is also Model D).

When solving a real-world problem mathematically, there are three steps: 1) Convert the real-world problem into a mathematical statement of the problem (that you believe suitably approximates the real-world problem); 2) Solve the mathematical problem; 3) By reversing the steps in 1, convert the mathematical solution to a real-world solution.

The given problem is sufficiently simple that, once a mathematical statement has been given, there is a clear solution. Hence if different people obtain different answers, it is because they converted the problem into different mathematical statements.

Your example of a sealed envelope has converted the original real-world problem into a different real-world problem. To see this, consider repeating the experiment many times. First a person is picked. Then they write down some information and place it in the envelope. This is Model B.

“When solving a real-world problem mathematically, there are three steps: 1) Convert the real-world problem into a mathematical statement of the problem …” I agree 100%.

“Your example of a sealed envelope has converted the original real-world problem into a different real-world problem.” Here, I must disagree 100%. The “real world problem” expressed by your colleague’s question is that someone told you he had a Tuesday Boy, not that someone with a Tuesday Boy was picked to talk to you. In the real world, which is more likely:

(A) If a parent of two, including a Tuesday Boy, makes a statement of the form “One is a born on a ,” it will always be about the Tuesday Boy.

(B) If a parent of two, including a Tuesday Boy, makes a statement of the form “One is a born on a ,” could be about a Thursday Girl.”

My point is that, IN THE REAL WORLD, (A) is absurd, because (B) is possible. The information you have is based on what someone says about a family, not the family itself. So Model A fails your step (1) because it overlooks one essential part of the correct mathematical model for this problem. That statement requires a term for P(tell “Tuesday Boy”|”Tuesday Boy and Thursday Girl”), as well as 26 similar terms for similar conditional probabilities.

Model A implicitly assumes that every such term is 1. This is absurd because, IN THE REAL WORLD, P(tell “Tuesday Boy”|”Tuesday Boy and Thursday Girl”) and P(tell ” Thursday Girl “|”Tuesday Boy and Thursday Girl”) can’t both be 1 like model A assumes. Model B (or D) assumes that all but the terms where the two children match each other are 1/2.

When Martin Gardner popularized the simpler (without “born on Tuesday,” and incidentally with “you know that” in place of “someone tells you that”) version of this problem in May 1959, his original answer was 1/3, which corresponds to Model A. But he retracted it the next October, essentially because either model A or model B could apply to his description. But when you include the “someone tells you part, you exclude Model A unless you provide a reason why the person is biased.

We might actually be in agreement.

My position has not been to say Model A or Model B is correct, only to say “we need to ask for clarification from the puzzle teller on how and why he chose to tell us that one of his children was a son born on a Tuesday before we can solve this puzzle without making any additional assumptions.” And only because this is a good lesson that holds when solving real problems in the real world.

I agree with you that it is “most unlikely” that out of all puzzle tellers in the world, the only ones that presented this puzzle were the ones with two children, one a boy born on a Tuesday. And I agree that if we were not allowed to ask for clarification and had to make an assumption, then the more sensible assumption is the one you made.

🙂

My position is that Model A not only is “Most Unlikely,” it is “Impossible” for the problem as stated. We simply can’t make that assumption. Further, that without additional information (which I agree would clarify the situation), the Principle of Indifference applies and Model B is the only assumption that we can make.

That doesn’t mean it is right, based on any actual circumstances. It means it is right based on the information we have. There is no contradiction in two people – like us, and someone who knows which model was used – assigning different probabilities to the same instance. That’s what conditional probability is, a probability based on the information at hand.

As evidence for Model B over Model A, I give you the famous Monty Hall Problem, with which I hope you are familiar. As an example, say you choose door #1 and Monty Hall opened door #3. Most explanations go something like “Door #1 had a 1/3 chance to be right originally. Since Monty Hall can always open a door without the prize, that can’t change. So you still have a 1/3 chance, and door #2 has a 2/3 chance.” While this seems valid, and does get the right answer, it is incorrect reasoning.

If the prize is behind door #2 (P(D2)=1/3), Monty Hall must open door #3 (P(O3|D2)=1).

If the prize is behind door #3 (P(D3)=1/3), Monty Hall can’t open door #3 (P(O3|D3)=0).

If the prize is behind door #1 (P(D1)=1/3), Monty Hall must choose between door #3 (P(O3|D1)=X) and door #2 (P(O2|D1)=1-X).

Given that he opened door #3, the probability the prize is behind door #1 is:

P(D1|O3) = P(O3|D1)*P(D1)/[P(O3|D1)*P(D1)+P(O3|D2)*P(D2)+P(O3|D3)*P(D3)]

= (1/3)*(X)/[(1/3)*(X)+(1/3)*(1)+(1/3)*(0)]

= X/(1+X).

Similarly, P(D1|O2)=(1-X)/(2-X). Model A says X=1 and so P(D1|O3)=1/2. But then P(D1|O2)=0; and in fact, they are different for most X’s. The reasoning I quoted above is incorrect because it can’t reproduce this difference.

Both are the same, 1/3, if and only if X=1/2 which is Model B. This is why we apply the Principle of Indifference. It’s not that we are assuming Monty Hall is unbiased, it is because our information does not allow us to get a different answer if the question is reversed but our assumptions remain the same. Our *knowledge* of Monty Hall allows no bias. So we must assume X=1/2.

The same holds in the Two Child Problem. Given no information about how the information was determined, Model B is required and the answer is 1/2. Whether or not you are told about “Tuesday.”