The Boy or Girl Paradox: A Martingale Perspective.

Mr Smith always tells the truth.

Interviewer: Mr Smith, how many children do you have?
Mr Smith: I have two.
Interviewer: Mr Smith, do you have a son?
Mr Smith: Yes

Q1: What is the probability that Mr Smith's other child is also a boy?

This is a clean version of the boy or girl paradox. The correct answer is 1 in 3, as a simple application of Bayes' Rule establishes. Given forty families, thirty will have at least one boy. Of those thirty, ten will have two boys.

Slightly different evidence would result in an answer of 1 in 2. For example:

Interviewer: Mr Smith, how many children do you have?
Mr Smith: I have two.
Interviewer: Mr Smith, is your oldest child a boy?
Mr Smith: Yes

Q2: What is the probability that Mr Smith's other child is also a boy?

The answer to the second question is clearly 1 in 2.

The boy or girl paradox rises to the status of a controversy because many people will answer 1 in 2 in circumstances other than this as well (as pointed out by a reader - see comments below). However both Q1 and Q2 are, I think, entirely clear. The first situation results in a lower posterior probability.

Now don't ask me why I was looking at this right now but I got to wondering, is there an even easier way to see the inequality between the two answers?  Bayes Rule is pretty simple when you write it down but humans are notoriously bad at using it semi-consciously, if you know what I mean. Scanning the wikipedia explanation left me a bit unsatisfied. So here's another angle.

                                The Martingale Argument (Informal)

Suppose you had wagered that Mr Smith had two boys. You paid $1 and you'll receive $4 if he has two boys. It's an investment. What evidence would make you happier about your investment? Learning that at least child of two is a boy, or learning that at least one child of one is a boy? The latter is a priori less likely, and therefore better news.

Now to polish it off, note that in the former case, betting directly on Mr Smith's first answer would turn $1 into $4/3 because fair odds are 3:1 on. But had the news been bad (i.e., none of the two children were boys) our parlay would be dead in the water. Since we are betting with fair odds our wealth is a martingale. To get to $4 you we still have to have to increase your wealth threefold. So it must be that the odds of betting on two children being boys given that you already know that at least one is a boy correspond to a threefold increase in wealth. Thus the probability is 1 in 3.


                                 The Martingale Argument (Formalized)

A reader has asked me to tighten the argument and relate to the martingale notion in mathematics. My title references the fact that when we face a bookmaker who offers fair odds the expected value of our wealth will not change at the moment we enter any bet with him. Our wealth is a martingale no matter what betting strategy we employ. Try as we might, we cannot win or lose money on average.

A martingale is defined with respect to both a stochastic process (here our wealth) and a sequence of information sets (representing information arriving, such as whether Mr Smith has at least one boy). If \(\cal{F}_{t}\) represents the information available to us at time \(t\) and \(\cal{F}_{\tau}\) the information at an earlier time then the statement that our wealth is a martingale is written: \begin{equation} \label{martingale} W_{\tau} = E \left[ W_{t} | \cal{F}_{\tau} \right] \end{equation} In general both sides of the equation are random variables but to warm up with a special case, take \(\tau=T\) where \(T\) is the time we learn whether Mr Smith has two boys or not, and set \(t\) equal to the time before we know anything. Our initial wealth is not random so the left hand side of \begin{equation} W_t = E \left[ W_T | \cal{F}_{t} \right] \end{equation} is just 1. If we further assume we make an all-or-nothing bet with only two outcomes this reads \begin{equation} 1 = p W^{win} \end{equation} where \(W^{win}\) is our wealth conditional on winning (i.e. the payout). This is just the statement that for a fair bet the payout is inversely proportional to the probability of winning. Since \(p=0.25\) we have \(W^{win}=4\).

A somewhat more interesting use of the martingale property considers a time \(\tau\) just after we learn some information. If we again take \(T\) to be the "terminal" time just after we learn whether Mr Smith has two boys or not we have \begin{equation} W_{\tau}(\omega) = E \left[ W_T | \cal{F}_{\tau} \right](\omega) \end{equation} where I have emphasized that both sides are random variables. If we take an expectation with respect to \(\cal{F}_{t} \), the information we have when we make the bet, we'll get the third inequality below: \begin{equation} 1 = W_t = E \left[ W_{\tau} | \cal{F}_{t} \right] = E \left[ E \left[ W_T | \cal{F}_{\tau} \right]| \cal{F}_{t} \right] \end{equation} Here the first equality and the second we have seen. The first is our assumption that we start with one dollar. The second equality is the martingale property.

These expressions become quite trivial in the case where the first revelation of information will result in one of two outcomes and one of these outcomes is fatal for our bet (if we learn that none of Mr Smith's children are boys, or if we learn that the first child is not a boy then obviously we are out of the running). Indeed we have \begin{eqnarray} E \left[ E \left[ W_T | \cal{F}_{\tau} \right]| \cal{F}_{t} \right] & = & p^A E \left[ W_T | \cal{F}_{\tau} \right] \\ & = & p^A p^B W^{win} \end{eqnarray} where \(p^A\) is the probability of good news and \(p^B\) is the conditional probability of winning our bet at time \(T\) given that we received good partial information at time \(\tau\) (this second probability is what we are asked for). And noting the chain of equalities we therefore have \begin{equation} 1 = p^A p^B W^{win} \end{equation} and thus \begin{equation} p^B = \frac{1}{p^A W^{win} } \end{equation} My informal argument merely noted that \(W^{win}\) is a known quantity (=4) and thus \(p^B\) is inversely proportional to \(p^A\).

To make this explicit consider two scenarios. In scenario \(1\) we learn that at least one child is a boy. In case \(2\) we learn that the oldest child is a boy. Denote the corresponding probabilities of this news (good in both cases) by \(p^A_1\) and \(p^A_2\) respectively. Trivially \(p^A_2 = 0.5\). On the other hand \(p^A_1\) is the probability that at least one child out of two is a boy. This is true in 3 out of 4 cases so \(p^A_1 = 0.75\). It follows that \begin{eqnarray} p^B_1 & = & \frac{1}{p^A_1 W^{win} } = \frac{1}{ \frac{3}{4} 4 } = \frac{1}{3} \\ p^A_2 & = & \frac{1}{p^B_2 W^{win} } = \frac{1}{ \frac{1}{2} 4 } = \frac{1}{2} \\ \end{eqnarray} Thus the answer to the first question about Mr Smith is 1 in 3, whereas the answer to the second is 1 in 2.

Really it is trivial as the statement \(1 = p^A p^B 4\) is just a statement about fair odds for a parlay bet (where the winnings from a first bet are invested in another). You can always think of the expected value of your wealth for the original bet as equal to the actual cash you hold in the absolutely equivalent parlay bet.