Let’s consider a symmetrical, two-player, zero-sum game, where each player must simultaneously choose one of three moves: A, B, or C. After each player choses their move, a winner is randomly selected with odds based on what the two players selected. (Note that because its symmetrical, if both players choose the same strategy, it is guaranteed to be a 50⁄50 coin flip for who wins).
In the simplest case, one of the moves wins more than 50% of the time against both of the other two moves. In this case, the best strategy is to always select this move.
Now let’s consider a rock-paper-scissors type game. Say A beats B with probability 0.6, B beats C with probability 0.6, and C beats A with probability 0.6. In this game, the best strategy is to chose each of the moves A, B, and C with probability 1⁄3. So far, so good.
Now let’s improve move C by making it beat move A with probability 1, leaving the other two fixed (so A beats B 60% of the time, and B beats C 60% of the time). What does the optimal play look like now?
Even if you don’t want to fully compute it, notice what your intuition is telling you should happen.
As it turns out, the optimal strategy is to play move A 1⁄7 of the time, move B 5⁄7 of the time, and move C 1⁄7 of the time. Against this strategy, move A wins (0.5)(1/7) + (0.6)(5/7) + (0)(1/7) = 1⁄2 of the time, move B wins (0.4)(1/7) + (0.5)(5/7) + (0.6)(1/7) = 1⁄2 of the time, and move C wins (1)(1/7) + (0.4)(5/7) + 0.5(1/7) = 1⁄2 of the time, so there is no way to beat this strategy: it is optimal.
This was really surprising to me! We improved move C, and somehow people want to play is less. Is there a name for this effect?
Rock-Paper-Scissors Can Be Weird
Let’s consider a symmetrical, two-player, zero-sum game, where each player must simultaneously choose one of three moves: A, B, or C. After each player choses their move, a winner is randomly selected with odds based on what the two players selected. (Note that because its symmetrical, if both players choose the same strategy, it is guaranteed to be a 50⁄50 coin flip for who wins).
In the simplest case, one of the moves wins more than 50% of the time against both of the other two moves. In this case, the best strategy is to always select this move.
Now let’s consider a rock-paper-scissors type game. Say A beats B with probability 0.6, B beats C with probability 0.6, and C beats A with probability 0.6. In this game, the best strategy is to chose each of the moves A, B, and C with probability 1⁄3. So far, so good.
Now let’s improve move C by making it beat move A with probability 1, leaving the other two fixed (so A beats B 60% of the time, and B beats C 60% of the time). What does the optimal play look like now?
Even if you don’t want to fully compute it, notice what your intuition is telling you should happen.
As it turns out, the optimal strategy is to play move A 1⁄7 of the time, move B 5⁄7 of the time, and move C 1⁄7 of the time. Against this strategy, move A wins (0.5)(1/7) + (0.6)(5/7) + (0)(1/7) = 1⁄2 of the time, move B wins (0.4)(1/7) + (0.5)(5/7) + (0.6)(1/7) = 1⁄2 of the time, and move C wins (1)(1/7) + (0.4)(5/7) + 0.5(1/7) = 1⁄2 of the time, so there is no way to beat this strategy: it is optimal.
This was really surprising to me! We improved move C, and somehow people want to play is less. Is there a name for this effect?