I do not understand how to calculate the expected utility of $3$ or more players game.

For a $2$ player game, it is easy. Suppose I have two action $\{A, B\}$ and my opponent has two action $\{C, D\}$.

If I want to calculate my expected payoff if I play $A$ and my opponent mixes the play like this: with probability $p$, he plays $C$ and with probability $1-p$, he plays $D$, then I will get $\hat u=p\cdot u(A,C)+(1-p)\cdot u(A, D)$.

How to calculate the expected payoff for $3$ players game? Let say player $1$ has strategy $\{A, B, C\}$, player $2$ mix $\{C, D, E\}$ with probabilities $p$, $q$, and $1-p-q$ and player $3$ mix $\{F, G, H\}$ with probabilities $r$, $s$, and $1-r-s$.


Let's just say all the strategies are $s_{i}\in\{A,B,C\}\equiv\mathcal{S}_{i}$ for players $i=1,2,3$

Denote player $i's$ mixing probabilities over $\{A,B,C\}$ as $p_{i}^{A},p_{i}^{B},p_{i}^{C}$

Expected utility of player 1 playing strategy A is:


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  • $\begingroup$ So it is not that hard. Thanks. Is there a general formula for it? Maybe like: $E[u(A)]=\sum_{s}u(A, s)\prod_{s}\sigma(s)$ when other player $\{2, \cdots\}$ mix with probability $\sigma$? $\endgroup$ – drzbir Jun 23 '14 at 18:29
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    $\begingroup$ Denoting the mixed strategy profile by $\sigma\equiv(\sigma_{1},...,\sigma_{n})$, and the strategies for the $N$ players by $s\equiv(s_{1},...,s_{n})$, the expected utility of $\sigma$ to player $i$ is: $E\left(U_{i}(\sigma)\right)=\sum_{s\in\mathcal{S}}\prod_{k=1}^{N}\sigma_{k}(s_{k})U_{i}(s)$. Then getting the expected payoff to a particular pure strategy from player $i$ is just a simple restriction on that previous expression... $\endgroup$ – Newbacus Jun 23 '14 at 19:28

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