Sunday, December 27, 2015

One of two children is a boy, what's the probability the other is a boy?

It's one-half. Not one-third, one-half.

And no, none of that "both solutions work" or "it depends" or "the Bayesian or frequentist solution" nonsense. It's one-half for Bayesians, for frequentists, and for people who don't know what these categories mean. The only thing the one-third "solution" is good for is an example of the importance of understanding the difference between states of the world and observed events.

I go over the Bayesian solution and explain what's wrong with one-third in the following video:



I had a frequentist video, as well, but I deleted it in the big online cleaning of 2012, so here's a simple version.

There are four possible states of the world $\{(B_1,B_2),(G_1,B_2),(B_1,G_2),(G_1,G_2)\}$, where $B,G$ is the sex and the subscript is the birth order. From each of these states there are two possible events, observing the first or the second child, leading to eight possible state-event pairs:
\[\begin{array}{rlc}
(B_1,B_2) & \rightarrow & B_1 \\
(B_1,B_2) & \rightarrow & B_2 \\
(G_1,B_2) & \rightarrow & G_1 \\
(G_1,B_2) & \rightarrow & B_2 \\
(B_1,G_2) & \rightarrow & B_1 \\
(B_1,G_2) & \rightarrow & G_2 \\
(G_1,G_2) & \rightarrow & G_1 \\
(G_1,G_2) & \rightarrow & G_2
\end{array}\]
The event "one is a boy" means that only four of these state-event pairs are feasible in the universe of possibilities:
\[\begin{array}{rlc}
(B_1,B_2) & \rightarrow & B_1 \\
(B_1,B_2) & \rightarrow & B_2 \\
(G_1,B_2) & \rightarrow & B_2 \\
(B_1,G_2) & \rightarrow & B_1
\end{array}\]
The frequentist answer to "how likely is the state $(B_1,B_2)$?" is computed as the number of favorable pairs, that is pairs including $(B_1,B_2)$, two, divided by the total number of feasible pairs in the universe of possibilities, four. Two divided by four is one-half.

Why do people fall for the one-third "solution"? Two main reasons, I believe:

1. Understanding the difference between states and events and how to relate information to changes in probabilities is not a simple matter; most people think that they know how to do this better than they actually do.

2. The one-half solution sounds too simple, and therefore doesn't allow the person to affect sophistication. That, and most people's interest in STEM as an identity product only, is one of the most destructive mental attitudes one can have: it blocks learning.

Here's a different probability puzzle that suffers from the same problems:



In all fairness, in illo tempore when I saw the one-third "solution" I believed it correct, but my much smarter classmate Dave Godes immediately showed me it was wrong. To my credit, I didn't argue – when I'm shown to be wrong, I change my mind. I know, crazy.

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PS: Yes, I know that the frequentist computation is the total probability (not number) of the favorable pairs divided by the total probability of the feasible pairs. Since this is a simple example and all pairs are equiprobable, counting is equivalent to computing total probability in this computation.