# How do we prove that e = RPC, in game theory?

Here e is the expected value of the game for the row player, P is the payoff matrix from the perspective of the row player, R is the row matrix containing the probabilities for each of the row player's strategies, and C is the column player's matrix showing the probabilities for his strategies.

This page provides a more detailed statement of the theorem, sans proof, under the heading "Mixed Strategies, Expected Value"

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Let the payoff matrix be given by $$P=(p_{ij})=\pmatrix{p_{11} & \cdots & p_{1n} \cr \vdots & & \vdots \cr p_{m1} & \cdots & p_{mn}}$$ Here $p_{ij}$ is the payoff to the row player if the row player chooses row $i$ and the column player chooses column $j$.
Let the matrices (vectors) $R$ and $C$ be given by $$R=(r_i)=\pmatrix{r_1 & \cdots & r_m}, \qquad C=(c_j)=\pmatrix{c_1 \cr \vdots \cr c_n}$$ Here $r_i$ is the probability that the row player chooses row $i$, and $c_j$ is the probability that the column player chooses column $j$. The index $i$ runs from $1$ to $m$ (the number of rows), and the index $j$ runs from $1$ to $n$ (the number of columns).
The probability that the row player will receive the payoff $p_{ij}$ is $r_ic_j$, since we assume that the two players choose their strategies independently of each other. You get the expected value of the game for the row player by multiplying each possible payoff with the probability for that payoff to occur, and to sum all the products. This will give you the double sum $$e=\sum_{i=1}^m\sum_{j=1}^n r_ic_jp_{ij}$$ Now it only remains to verify that the matrix product $RPC$ will give you the same double sum. You might want to write this out in two steps, if you haven't worked with this kind of expressions before. We first compute $$PC=\pmatrix{p_{11} & \cdots & p_{1n} \cr \vdots & & \vdots \cr p_{m1} & \cdots & p_{mn}}\pmatrix{c_1 \cr \vdots \cr c_n}=\pmatrix{p_{11}c_1+\cdots+p_{1n}c_n \cr \vdots \cr p_{m1}c_1+\cdots+p_{mn}c_n}=\pmatrix{\sum_{j=1}^np_{1j}c_j\cr \vdots \cr \sum_{j=1}^np_{mj}c_j}$$ Then you can multiply by $R$ from the left: $$RPC=\pmatrix{r_1 & \cdots & r_m}\pmatrix{\sum_{j=1}^np_{1j}c_j\cr \vdots \cr \sum_{j=1}^np_{mj}c_j}=r_1\sum_{j=1}^np_{1j}c_j+\cdots+r_m\sum_{j=1}^np_{mj}c_j$$ $$=\sum_{i=1}^m r_i \sum_{j=1}^n p_{ij}c_j=\sum_{i=1}^m\sum_{j=1}^n r_ip_{ij}c_j$$ This is the same that you got from computing the expected value $e$ above, and hence $$e=RPC$$