30-sided die, 2-player game

Solution 1:

Let $x_1$ and $x_2$ be the player's choices. To maximize his chances of winning, player 2 will select a number a close as possible to player 1's but must choose whether to select a number above or below $x_1$. The expected payoff for choosing above $x_1$ is greater if:

$\frac{1}{30}\sum_{i=x_1+1}^{30}i\geq\frac{1}{30}\sum_{i=1}^{x_1-1}i$

Computation provides:

$x_1\leq21.56$

Thus, if player 1 chooses 21 or less, 2 chooses $x_1+1$. If player 1 chooses 22 or more, player 2 chooses $x_1-1$.

Given this behavior of player 2, choosing 21 gives player 1 an expected payoff of $\frac{77}{10}$ while choosing 22 gives player 1 an expected payoff of $\frac{78}{10}$. Thus, player 1 chooses 22 and player 2 chooses 21 (other choices give player 1 a lower payoff).

Player 1 with expected payoff $\frac{78}{10}$ does slightly better than player 2 with expected payoff $\frac{77}{10}$. Note that the sum of these expected values is $\frac{155}{10}=\frac{30*29}{2*30}$ which is the expected value of a dice roll.