Suppose you are playing a game against an opponent whom you know only uses pure strategies. My question is, is there any such game in which using a mixed strategy in response is better than all the pure strategies you have at your disposal?


Fixing the strategy of the opponent, a mixed strategy never yields a strictly higher utility if you are expected utility-maximizing. The reason is that the expected utility from a mixed strategy is at most as high as the highest utility from the pure strategies which this mixed strategy plays with positive probability. That is not to say that a mixed strategy cannot be a best response, though. Consider for example rock-paper-scissors and a imagine your opponent playing each of the three actions with equal probability. This leaves you indifferent between all three actions and hence any mixed strategy is a best response.

Things can be different when players are not expected-utility maximizers. If that's what you're interested in, you could take a look at the paper "Equilibrium without independence" by V. Crawford, 1990, as well as some of the references therein.


Yes. For example, consider the rather trivial game

\begin{array}{|c|c|}\hline &A&B\\\hline A&0,0&0,0\\\hline B&0,0&0,0&\\\hline \end{array}

Clearly, playing a mixed strategy is a best response to either of the pure strategies (since all strategies, mixed or otherwise, yield a payoff of 0).

Of course, as pointed out by a previous commenter, playing a mixed strategy can never be strictly better than playing a pure strategy: mixed strategies can only be weakly better than pure strategies.

Finally, it seems plausible (to me) that if all payoff entries in a (two player) game are distinct, the game will never have an equilibrium in which one player mixes but the other does not. However, I won't try to prove this here.

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    $\begingroup$ "My question is, is there any such game in which using a mixed strategy in response is better than all the pure strategies you have at your disposal?" (The question in the title was indeed slightly different.) $\endgroup$ – Giskard Nov 8 '19 at 13:58
  • $\begingroup$ Exactly, and the answer is 'yes' if we interpret 'better' as 'weakly better' (but 'no' if we interpret 'better' as 'strictly better') $\endgroup$ – user17900 Nov 11 '19 at 17:47

Repeated games and nonlinear utility

Let's assume a trivial two-player game where each player has two options A and B; and the payout is +1/-1 if players pick the same and -1/+1 if players pick differently.

Let's assume that the game is repeated 100 times with the strategies chosen and committed to beforehand. This means that if your opponent picks a fixed strategy, and you choose a fixed strategy, then the expected result is 50% chance of +100 and 50% chance of -100, depending on what strategy they've chosen. However, if you choose a 50/50 mixed strategy (random), then the expected result is close to 0 - between -10 and +10 with a ~95% probability or so, and if you choose a 50/50 mixed strategy that's strictly alternating, then you get an expected result of exactly 0 with absolute certainty.

Thing is, many theories of utility (and behavioral economics experiments) expect that it can be nonlinear, and that "50% chance of +100 and 50% chance of -100" and "certain 0" are not equivalent, there can be a strong preference for one or the other depending on circumstances. So, if the utility of one player "prefers" reducing variance and avoiding the risk of a large loss, then a mixed strategy is strongly preferrable.

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    $\begingroup$ What about the pure strategy of playing A 50 times then playing B 50 times? This guarantues a payoff of zero, so it is even better than your close to zero yielding mixed strategy. $\endgroup$ – Giskard Nov 8 '19 at 19:06
  • $\begingroup$ @Giskard yes, equivalent to "you choose a 50/50 mixed strategy that's strictly alternating". I guess it comes down to whether we'd consider that as a mixed or a fixed strategy in this repeated game scenario. $\endgroup$ – Peteris Nov 8 '19 at 19:16
  • $\begingroup$ @Giskard, if that does count as a pure strategy, similar options are also available to your opponent (play B 50 times then A 50 times), which means such a strategy does not yield a guaranteed payoff of zero (it can be anywhere from -100 to 100). A mixed strategy will tend towards 0 no matter what your opponents pure strategy is, and the rest of the analysis in the answer holds up (except the part about 50/50 strictly alternating being a certain 0 is invalidated if a "50/50 alternation" is a pure strategy available to the opponent). $\endgroup$ – Steven Jackson Nov 8 '19 at 20:26
  • $\begingroup$ This is definitely a pure strategy in the repeated game. Choosing a strategy/action in a stage game is not entire an strategy in the repeated game, just an element of it. I guess this also means that given the opponent's pure strategy of repeating A or repeating B the 50-50 response is never best, mirroring the opponent is best. $\endgroup$ – Giskard Nov 8 '19 at 21:17

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