3 ms·
That's not an equilibrium; as soon as one player deviates from a true 1/3 random strategy, the other player has a theoretical ability to punish. But note the r
by smallnamespace 9y ago
That's not an equilibrium; as soon as one player deviates from a true 1/3 random strategy, the other player has a theoretical ability to punish.
But note the relationship between computation and exploitation here: if the deviating player makes its deviation difficult to detect (e.g. using a PRNG stream cipher, which is completely deterministic but may look random), then the cost of making good predictions to take advantage of the deviation may exceed the payoff.
- bluetwo 9y agoSure, but the ability to punish doesn't exist in the formula.
- vilhelm_s 9y agoA Nash equilibrium says that each player plays optimally given what all the other players do. If one player plays deterministically, it's no longer optimal for the other one to play randomly, so it's not an equilibrium.
- bluetwo 9y agoSetup your own experiment and let me know what you find.
- qubex 9y agoThe point is that if your strategy becomes a known quantity, it can be strictly dominated by your opponent choosing the strategy that wins against it. This would push you both back towards random choice (whether iterated or one-off).