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For the mixed strategies, the expected utility (or payoff) is used to find the mixed strategy Nash Equilibrium. The main assumption is players try to maximize their expected payoffs. However, I think we should also observe the risks of selecting a strategy, because the assumption of risk aversion is also important I guess. If we assume that both players try to minimize their risks, can we find a Nash Equilibrium by using variance rather than expected utility?

$$\mathrm{Var}(U_1)=\sum^i_{n=1} (\mathrm{E_{1i}-\mathrm{E_1}})^2 \times p_i$$ $$U_1= \textrm{Utility of first player}$$ $$\mathrm{E_{1i}}=\textrm{Expected utility of first player for i$^{th}$ strategy}$$ $$\mathrm{E_1}=\textrm{Expected utility of first player}$$ $$p_i=\textrm{Probability of selecting a strategy for first player}$$ Is it reasonable to apply this solution where both players are risk averse and will minimize the variance of their utilities(they do not look the utility)?

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There's no need to consider the variance of payoffs when playing mixed strategies. Players' attitudes towards risk is already taken account of. This is because the "payoff" they receive is not a monetary payment. Instead, payoff functions are von Neumann-Morgenstern utility functions, meaning that they represent players' preferences over lotteries over outcomes (which could be monetary payments).

On top of that, even when we consider monetary payments it wouldn't make sense to assume that players try to "minimize their risk" in the sense of minimizing the variance of their income. Note that such a player would prefer a sure zero payment to any lottery, even to one which pays 1 million when a flipped coin comes up heads and 2 millions when it comes up tails ...

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