What are the Nash equilibria of the “aim for the middle” game?

Question

Consider the following three-player game: Alice chooses an integer congruent to $0$ mod $3$, Bob chooses an integer congruent to $1$ mod $3$, and Chris chooses an integer congruent to $2$ mod $3$. (Here, “integer” means “element of $\mathbb{Z}$, of course.) All choices are made simultaneously, i.e., without knowing the other players' choice. There is only one round. Whoever chose the middle integer (i.e., the one which is neither the smallest nor the largest) wins — for definiteness, say they win $1$ and the others win $0$.

Question: does this game admit a Nash equilibrium other than the pure ones where the three players play three consecutive integers? If so, can we describe the Nash equilibria?

(Just to be entirely clear what the question is: does there exist a triplet consisting of probability distributions on $3\mathbb{Z}$, $1+3\mathbb{Z}$, and $2+3\mathbb{Z}$, such that each each one maximizes the probability of drawing the middle integer against the other two, other than by having each distribution pick a specific integer — necessarily consecutive — with probability $1$?)

Games with infinitely many option can, of course, fail to have Nash equilibria: for example, the two player game in which Alice chooses an even integer and Bob chooses an odd integer, and whoever chooses the largest one wins, clearly doesn't have a Nash equilibrium (because for every probability distribution one can find one which does strictly better against a fixed opponent's strategy).

Answer 1

Yes, there are some other strategies. However, one can prove that in every Nash equilibrium there is at least one player which always loses with the other two players always choosing the same adjacent numbers.

Suppose there is a Nash equilibrium in which every player has a non-zero chance of winning. Given the strategies of Alice and Bob, Chris will always choose a number with maximal probability of winning. If this probability is zero, Chris has no chance of winning, i.e. the strategies of Alice and Bob are adjacent numbers. Otherwise, the mixed strategy of Chris is a combination of finite number (at most one over the probability) of pure strategies.

By symmetry the same holds for Alice and Bob. Now, consider the numbers which get chosen with non-zero probability by Alice, Bob or Chris together and take the maximum. This number is in the middle with probability zero, which is a contradiction. Thus there is always a player which loses with probability one.

If Chris is the loser, the strategies must have the following form.

• Alice: $x$ with probability 1
• Bob: $x + 1$ with probability 1
• Chris: $y$ with $P[y < x] \geq P[y > x + 2]$ and $P[y < x - 1] \leq P[y > x + 1]$

Answer 2

The answer of 1001 extends to mixed strategies with infinite support.

If there is a positive probability that the other players play non-adjacent numbers, a player has a pure best response that gives a positive payoff, and every mixed best response must be a mixture of pure best replies.

So, assume that no two players play adjacent numbers for sure. Then every player must have a strictly positive payoff from a number in the support of their strategies. Moreover, no player's support can be bounded, because the largest or smallest number played must lose for sure. So assume that some pure best reply of Alice gives her a positive payoff of $\pi$. There must be some $n$ such that the probability of both Bob and Chris choosing a number in $[-n,n]$ is larger than $1-\pi$. But then Alice playing a number outside $[-n,n]$ will give her a payoff of less than $\pi$, and her best reply must have bounded support.