Solution
Rock-Paper-Scissors with Stakes
Show the problem again
Two players play rock-paper-scissors once. The loser pays the winner an amount that depends on the winning hand: rock beating scissors is worth 1, paper beating rock is worth 2, and scissors beating paper is worth 3. Equal hands give 0. In the (unique) optimal mixed strategy, with what probability does a player throw rock? Give a reduced fraction.
Worked solution
The answer is 1/2. The game is symmetric and zero-sum, so its value is 0, and an optimal mix must leave the opponent indifferent among his three throws (each earning 0). Let r, p, s be the probabilities of rock, paper, scissors. Against opposing rock, paper wins 2 and scissors loses 1, so 2p − s = 0. Against opposing paper, rock loses 2 and scissors wins 3, so −2r + 3s = 0. Against opposing scissors, rock wins 1 and paper loses 3, so r − 3p = 0. Then p = r/3 and s = 2r/3, and r + r/3 + 2r/3 = 1 gives r = 1/2 (with p = 1/6 and s = 1/3).
Source: Weighted rock-paper-scissors is a standard variant in game-theory courses; stakes chosen for AxiomIQ. Statement written for AxiomIQ.