It seems like there are pathological counterexamples. If you're player three, and player one chooses 1 always, and player two chooses 2 always, then the player using this strategy can never win.
Player 2 wins with probability .45 (only wins when player 1 gets "knocked out"), and player 1 with probability 1-.45, and player 3 never wins.
andrewla|3 years ago
Player 2 wins with probability .45 (only wins when player 1 gets "knocked out"), and player 1 with probability 1-.45, and player 3 never wins.