Bayes' Theorem and a Driver's Test Probability Puzzle

I'm Sawyer, and welcome to Real Numbers, where we solve real-world problems using mathematical techniques. Our sequence of probability puzzles leads into Bayes' theorem, a foundational result in probability theory.

Evelyn administers written tests at the local DMV and asks each applicant whether they studied. She finds that 60% studied, the overall pass rate is 75%, and only half of those who did not study pass. One applicant just passed, but Evelyn forgot to ask whether they studied. What is the probability that they did?

We know the conditional probability of passing given that an applicant did not study: the proportion passing within that subset. We want the probability of having studied given that the applicant passed. Notice that we have changed both the event of interest and the condition we know occurred.

Bayes' theorem relates P(A|B) to P(B|A). At Jane Street, one application is Bayesian inference, which assesses the relative accuracy of models given empirical data. The theorem states P(A|B) = P(B|A) × P(A) / P(B), provided P(B) is nonzero.

The proof follows from the conditional probability multiplication rule: P(A and B) = P(A) × P(B|A). But A and B is the same event as B and A; ‘and’ is commutative. If a muffin is warm and delicious, it is also delicious and warm.

Writing the intersection in the other order gives P(B and A) = P(B) × P(A|B). Equating the two expressions and dividing by P(B) yields Bayes' theorem. Try using it to solve the driver's test problem.

Now let's solve last week's War challenge. Clara has the 13 clubs shuffled and Damian has the 13 diamonds shuffled. We want the probability that Damian loses his ace of diamonds after their first move is settled.

An ace wins any ordinary battle, so Damian can lose it only if there is a war: the first cards must have the same rank. Whatever Damian turns over, exactly one of Clara's 13 equally likely cards matches it, so the probability of a war is 1/13.

There are three disjoint ways Clara can win Damian's ace. They can tie initially with aces and Clara can win the ensuing war; they can tie with non-aces, Damian's ace can be among the next three cards, and Clara can win; or they can tie with non-aces, then tie with aces in the war, and Clara can win the coin flip. The cases are disjoint because Damian's single ace occurs in different places in his deck.

For the first case, multiply the chance that both players turn over their aces by Clara's chance of winning the subsequent war. Once the aces are set aside, their remaining decks are equal, so symmetry gives each player a 1/2 chance. The result is (1/13) × (1/13) × (1/2) = 1/338.

For the second case, the factors are 12/13 for Damian's first card being a non-ace, 1/13 for the initial match, 3/12 for his ace being among the next three cards, and Clara's conditional chance of winning the war. If Clara draws her ace she wins; otherwise her remaining deck is symmetric with Damian's and she wins with probability 1/2. The resulting probability is 1/104.

In the final case, the players initially tie with non-aces, then tie with aces in the war, and Clara wins the coin flip. Its probability is 1/4056. Adding the three disjoint cases gives 1/338 + 1/104 + 1/4056 = 1/78. The simple result suggests there might be a more direct solution; perhaps you can find one.

Returning to this week's problem: 60% of applicants studied, 75% passed, and someone who did not study passes with probability 50%. Given that an applicant passed, what is the probability they studied? I invite you to work it out and share your reasoning; the solution comes next week.