DF5 - Mixed strategies

Mixed strategy

A mixed strategy assigns a nonnegative probability to each pure strategy, with probabilities summing to 1. For two rows, write $P(A)=p$, $P(B)=1-p$, where $0\leq p\leq1$. A pure strategy is the boundary case $p=0$ or $p=1$.

Worked example 1: a proposed row mixture

Question. For the displayed matrix, find the expected pay-off against each pure column if the row player chooses A with probability $1/2$. State the worst expected pay-off.

Plan. Weight each column's two entries by the row probabilities, then take the smaller result.

Answer. The row player guarantees expected pay-off $-1/2$, since the opponent can choose Y. This improves the pure-strategy guarantee $-2$, but is not yet proved optimal.

Check. Each result lies between its two possible pay-offs. The opponent's mixture gives a weighted average of 1 and $-1/2$, so cannot reduce expectation below $-1/2$.

Worked example 2: both players randomise

Question. For the same matrix, the row player uses $P(A)=1/2$, and the column player uses $P(X)=2/3$. Find the expected pay-off. Does it replace the guarantee found above?

Plan. Average the two conditional column expectations using the column probabilities.

Answer. Against this particular mixture the expected pay-off is $1/2$. The row player's guarantee remains $-1/2$, because Y alone is still available to the opponent.

Check. The four joint probabilities are $1/3,1/6,1/3,1/6$, in row order. Directly summing gives $4/3-2/6-2/3+1/6=1/2$, agreeing with the conditional calculation.

Watch out

Use probabilities, not the ratio of the two pay-offs. Against each opposing pure strategy, calculate a separate expectation. A favourable expectation against one chosen mixture is not a worst-case guarantee.

Worked example 3: answering the opening question

Question. Find both optimal mixtures and the value for $\begin{pmatrix}4&-2\\-2&1\end{pmatrix}$.

Plan. Equalise the two column expectations for $p$, then the two row expectations for $q$, and verify both guarantees.

For the column player, the row expectations are $6q-2$ and $1-3q$. Equalising gives $q=1/3$, also with pay-off zero.

Answer. Each player chooses their first strategy with probability $1/3$ and second with probability $2/3$. The value for both is zero.

Check. The row mixture gives zero against either column, hence against every column mixture. The column mixture gives zero against either row, hence against every row mixture. These opposing guarantees prove optimality. At $p=0,1$, the row guarantees are both $-2$, so the interior mixture improves on both endpoints.

Worked example 4: one pay-off changes the balance

Question. Find both optimal mixtures and the value for $\begin{pmatrix}7&-2\\-2&1\end{pmatrix}$.

Plan. Rebuild the affected expectation, then use the same equalisation and guarantee checks.

Row expectations against the column mixture are $9q-2$ and $1-3q$, so $q=1/4$.

Answer. Both players use probabilities $(1/4,3/4)$. The value is $1/4$ for the row player and $-1/4$ for the column player.

Check. Substitution into all four expectation expressions gives $1/4$. The row mixture guarantees that value against all columns, and the column mixture caps it against all rows. The pure row guarantees are $-2$, so the endpoints cannot be better.

Interior two-by-two solution

Let $D=a-b-c+d\ne0$. The equalising candidates are

Use these as an interior mixed solution only when $0<p<1$, $0<q<1$, and the corresponding opposing guarantees are verified. Boundary, parallel and tied cases require direct envelope or saddle analysis. Not in the formula booklet: learn the expectation method.

Watch out

Define which strategy receives $p$ and which receives $q$. Do not use a probability outside $[0,1]$, or divide by $D=0$. Equalising only one player's expressions does not by itself state both players' optimal strategies.