Well, no, those two events are not equally likely to occur, unless those sets have the same size, which only occurs for \(x=10\). It's not about the probability of winning, so much as it is about the players' *expected* winnings. For instance, if player 1 plays 12, then player 2 will play 13 because this maximizes his expected earnings; however, it does not maximize his *probability* of winning; he wins with probability 8/20 if he plays 13, but with probability of 11/20 if he plays 11. In most cases, however, the e.v.-maximizing play also maximizes the probability of winning.

Now, if risk aversion comes into play, which is better, going first or second? The variance of player 1's e.v.-optimizing strategy of playing 14 first is \(\frac{1^2+\cdots+14^2}{20}-\large(\frac{105}{20}\right)^2\), smaller than player 2's variance of \(\frac{15^2+\cdots+20^2}{20}-\large(\frac{105}{20}\right)^2\). Obviously roles are then reversed if Player 1 instead plays 15, even though 15 gets the same e.v. So from a risk perspective, you should actually choose to go first and play 14.