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Rock beats scissors, scissors beats paper, paper beats rock. No move is safe, and any pattern can be read and punished. So is there any way to play that a clever opponent cannot beat?

Existence of Nash equilibrium

Matching Pennies had no equilibrium in fixed moves, but it had one in probabilities. Rock-Paper-Scissors makes you wonder whether the trick always works. It does. Nash proved that every finite game has at least one equilibrium, as long as you allow mixed strategies.

From earlier lessons

Two quick ones before we start.

You mix between two moves in order to...

Matching Pennies has how many pure-strategy Nash equilibria?

Predict, then play

Declare a mix, then the bot sees the odds and chooses its best reply before your move is drawn. Try to find a mix that leaves it nothing to exploit.

Why did that happen?

Take a guess, then reveal the reason

Any lopsided mix gives the bot a target. Put too much probability on Rock and it chooses Paper. The only mix that gives it no preferred counter is exactly one-third each, because then every pure response has the same expected payoff. That even mix is the mixed-strategy equilibrium of the game.

The math

Rock-Paper-Scissors is symmetric and zero-sum, and it has no pure-strategy equilibrium. Try the symmetric mix: each player plays every move with probability $\tfrac13$ .

Against that mix, any single move you make wins one-third of the time, loses one-third, and ties one-third, for an expected payoff of

\tfrac13(1) + \tfrac13(-1) + \tfrac13(0) = 0.

Every pure move earns $0$ against the opponent’s uniform mix, so you are indifferent among all of them. Each player’s uniform mix is a best response to the other player’s uniform mix. That is the equilibrium.

Definition, Nash's existence theorem

Every finite game (finitely many players, each with finitely many pure strategies) has at least one Nash equilibrium in mixed strategies. The proof looks for a mixed profile in which each player’s strategy is a best response to the other players’ strategies. Best response is a correspondence, not a function (ties leave several best replies), so Nash’s 1950 proof applies Kakutani’s fixed-point theorem for correspondences; his 1951 proof instead builds a continuous map with the same fixed points and applies Brouwer’s.

The theorem guarantees existence. It says nothing about how many equilibria there are, or which one gets played.

Work it out

Verify the uniform mix is an equilibrium

Fix the opponent at $\tfrac13$ each. Compute your payoff to Rock: it beats Scissors ( $+1$ , probability $\tfrac13$ ), loses to Paper ( $-1$ , probability $\tfrac13$ ), ties Rock ( $0$ , probability $\tfrac13$ ), so the expected payoff is $0$ . The same holds for Paper and for Scissors, by symmetry.

Since all three pure moves pay $0$ , any mix of them also pays $0$ , so you cannot do better than the uniform mix. Neither can the opponent. Both playing $\tfrac13$ each is therefore a Nash equilibrium.

It is also the only one, and symmetry alone does not show that. Suppose you play Rock, Paper, Scissors with probabilities $r, p, s$ . The opponent’s payoff to Rock is $s - p$ , to Paper is $r - s$ , to Scissors is $p - r$ . For your mix to be part of an equilibrium it must hold them to the game’s value of $0$ , so all three must be at most zero: $s \le p$ , $r \le s$ , $p \le r$ . That chain closes on itself, forcing $r = p = s = \tfrac13$ .

Key idea

Every finite game has at least one mixed-strategy equilibrium. Existence does not tell us whether it is easy to compute, unique, or likely to be played.

Check yourself

Check yourselfrecognise

A finite two-player game has no cell where both players are best-responding. What does Nash's theorem tell you?

Check yourselfapply

At the 1/3-1/3-1/3 mix, what is each player's expected payoff in Rock-Paper-Scissors?

Check yourselfbrings back: nash equilibrium

Nash's theorem guarantees an equilibrium exists. What does it not tell you?

Sources

Nash, J. F. (1950). Equilibrium points in n-person games. PNAS, 36(1), 48-49.
Osborne, M. J. (2004). An Introduction to Game Theory. Oxford University Press.