The Efron dice are laid out before you:

The game is played by two people. The first chooses one of the four dice to roll, and the opponent chooses from the three that remain. The player who rolls the higher number wins. Your opponent graciously invites you to pick a die first. What should you do?

Solution.

You should decline the invitation to pick first, since for any die that the first player chooses, the second player can choose a die which gives him two chances out of three of winning.

We received solutions from

Xing Chiu is in the sixth grade, and used common sense to see that each die has a weakness against at least one of the remaining dice: Against die A, die D has a 50% chance of winning automatically by rolling a 5, and can still win by rolling a 1 against a 0, which gives die D better than a 50% chance of winning. Similarly, against die D, die C has a 50% chance of winning automatically because die D rolls a 1, and can still win by rolling a 6 against a 5, which gives die C better than a 50% chance of winning. Obviously, die C, with a majority of faces showing 2 is the underdog against die B with all faces showing 3, and the latter is the underdog against die A which has a majority of faces showing 4. This completes the cycle: there is no good first pick.

Alexander Potapenko computed the exact probabilities in each match above, and concluded that the first player loses with probability 2/3 in any case. For instance in the first case, die D will roll a 5 with probability 1/2, to which we add the (1/2)*(1/3) = 1/6 probability that die D rolls a 1 and die A rolls a 0. A similar argument works in all cases.

Jeffrey Hansen, Norman Laliberté, and Juan Mir Pieras all completed a table of probabilities for each matchup:

It was unclear to some solvers that it might be possible to decline the invitation to pick first. (You may look up the question again and see how you feel about this.) In any case, we must also account for the case where the opponent also refuses to pick first, and you end up losing a coin toss over the issue. Now, an opponent who knows the strategies of the game would always pick a die that has a 2/3 probability of winning against your first pick, so the only hope is that the opponent does not know the game, and picks a die at random.

With this interpretation, some solvers computed the expected value of each die, and got

To settle this issue, suppose that we modify the Efron dice by adding 1000 to every face value greater than or equal to 3. We get the following configurations:

Since the ranking of the face values is not changed, the probabilities of winning remain the same for each pair of dice. However the ranking of the expected values is now changed, with B being first, followed by A, D and C. Therefore the die with the highest expected value is not necessarily the best pick in this game. (It would be relevant only if the winner won cash equal to the number shown on the die.)

Against a player who chooses randomly, it is the average probability of winning that determines which die is a best first choice. Die C wins 56/108 of the time against a random opponent, which is just over 50% and as good as you can do. If forced to pick first, the best option is, as Mir puts it, to pick die C and hope that the opponent doesn't know what he is doing.