I got this problem from Rustan Leino, who was inspired to create it by a puzzle Aistis Simaitis gave him.
I solved it and wrote up my solution.
In a room are three boxes that on the outside look identical. One of the boxes contains a car, one contains a key, and one contains nothing. You and a partner get to decide amongst yourselves to each point to two boxes. When you've made your decision, the boxes are opened and their contents revealed. If one of the boxes your partner is pointing to contains the car and one of the boxes you're pointing to contains the key, then you both win. What strategy maximizes the probability of winning, and what's the probability that you'll win?
The optimal strategy is for you to point to the box your partner isn't pointing to, along with one of the boxes they aren't pointing to. Following that strategy gives you a $1/2$ probability of winning. We now discuss why.
Consider the possible outcomes as permutations of the letters C, K, and N, where C stands for car, K stands for key, and N stands for nothing. The first letter denotes the contents of the box both of you are pointing to, the second letter denotes the contents of the other box your partner is pointing to, and the last letter denotes the contents of the other box you're pointing to.
The winning permutations for you are CNK, KCN, and NCK. There are only six possible permutations, so your probability of winning is $3/6 = 1/2.$
The only other strategy available is for you and your partner to point to the same pair of boxes. You win with this alternate strategy if and only if the box neither of you points to contains nothing, which occurs with probability $1/3.$