<p>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.</p>

<p>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.</p>

CC BY-SA 4.0
2019-11-08T07:41:01.810Z
2019-11-08T07:41:01.810Z
induction_is_a_laddah
<h2>Repeated games and nonlinear utility</h2>

<p>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.</p>

<p>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.</p>

<p>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 <em>not</em> 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.</p>

CC BY-SA 4.0
2019-11-08T19:06:34.557Z
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.
Giskard
CC BY-SA 4.0
2019-11-08T19:16:15.007Z
@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.
Peteris
CC BY-SA 4.0
2019-11-08T20:26:32.627Z
@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).
Steven Jackson
CC BY-SA 4.0
2019-11-08T21:17:26.640Z
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.
Giskard
CC BY-SA 4.0
2019-11-08T18:40:06.893Z
2019-11-08T19:16:26.500Z
2019-11-08T19:16:26.500Z
Peteris
<p>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?</p>

<game-theory><nash-equilibrium><dominant-strategy>
Is a mixed strategy ever the best response to a pure strategy?
Keshav Srinivasan
# Is a mixed strategy ever the best response to a pure strategy?

(Asked by: username_0 on 2019-11-08)

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?

**Tags:** game-theory, nash-equilibrium, dominant-strategy


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## 2 answers

### Answer by username_1 on 2019-11-08. Score: 8

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.


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### Answer by username_2 on 2019-11-08. Score: 2

## 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.

> **Comments:**
> **username_3**: 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.  
> **username_2**: @username_3 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.  
> **username_4**: @username_3, 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).  
> **username_3**: 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.  

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