CTK Insights

In an early June issue of the NCTM newletter, President Mike Shaughnessy offered a Problem to Ponder:

Scenario: Students at your school have just finished competing in the qualifying round of a nationally sponsored contest on mathematical reasoning and sense making. When the work was scored, it turned out that four students at your school all had perfect preliminary papers—two girls and two boys. The school decided to hold a random drawing among these four students to select two of them to send to the national finals. The drawing takes place in the school auditorium. You show up late to the drawing, just as one of the winners—a girl—is leaving the stage amid cheers.

1. Suppose that the girl that you saw leaving the stage is the first winner. What is the probability that the second winner will also be a girl?
2. Suppose that the girl that you saw leaving the stage was the second winner. What is the probability that the first winner was also a girl?

The mid-July newsletter offered a solution, actually two of them. The two solutions for the second problem intentionally produce different answers, providing thus the food for thought.

Solution 1

If the girl that you saw on stage was actually the second winner, then the chance that the first winner was also a girl is 1/2, because before the first girl was picked, there were two girls and two boys, so the chance of a choosing a girl was then 2 in 4, or 1/2.

Solution 2

It doesn’t make any difference whether you saw the first girl or the second girl; the fact that you saw a girl winning at all means that there is only a 1 in 3 chance that the other winner was a girl, so, the probability that the first winner was also a girl if the girl that you see on the stage is the second winner is also 1/3.

Truth be told, when posting a solution at my site I have overlooked the first argument that led to the probability of 1/2. The reason I think was because I saw my task was to solve the problem, while Mike Shaughnessy - mindful to get math teachers excited - had the task of posing the problem. Coming up with two solutions that led to different answers should have certainly affected the level of interest of his readers.

As a result, I, too, has amended that page with two Bayesian approaches. The first solution of course should have raised a red flag for an attentive reader. Indeed, the argument never exploits the fact that the second winner was a girl, although it would have been strange if that information were just a red herring. Indeed, it is easily seen that it is not. Consider a situation where there are just 1 boy and 1 girl. They are selected in turns. You peek in, see a girl, and are told that this is the second selection. What is then the probability that the first selection was a girl? Zero, of course. This so, even if the a priori probability of a girl being selected on the first draw is obviously 1/2.

Here is a short argument leading to the probability of 1/3. There are 6 combinations of 2 items out of four, making the probability that both winners were the girls 1/6. If G and g are the events that the first/second selection was a girl then what we are looking for is the conditional probability P(G|g). But P(G|g) = P(Gg)/P(g). The a priori probabily that the second selection is a girl is the same as for a boy: P(g) = P(b) = 1/2. It then follows that P(G|g) = 1/6 ÷ 1/2 = 1/3.

Related posts:

1. Probability Wonders
2. Probability of Divisibility