Mixed strategies: the graphical methodEdexcel A-Level Further Maths: Revision notes

When a game has no stable solution, a player who always uses the same pure strategy can be exploited. A mixed strategy chooses each strategy with a fixed probability, picking at random each time. Rose plays row $i$ with probability $p_i$, with $\sum p_i=1$. Against any single column, Rose's expected pay-off is the sum of (probability of row $\times$ entry). For $\begin{pmatrix} 5 & 1 \\ 2 & 4 \end{pmatrix}$ with row 1 played with probability $p$:

• against column 1: $5p+2(1-p)=2+3p$
• against column 2: $p+4(1-p)=4-3p$. Rose's optimal mixed strategy maximises the worst of these. Setting $2+3p=4-3p$ gives $p=\frac13$ and a value of the game of $3$. The value is the expected pay-off to Rose each play, in the long run.

If Rose has two rows and Colin has $n$ columns ($n=2,3,4$), let Rose play row 1 with probability $p$, $0\le p\le1$. Each column gives a straight line for Rose's expected pay-off against that column. Draw all the lines for $0\le p\le1$. Whatever Rose does, Colin will choose the column that is worst for her, so her guaranteed pay-off is the lowest line at each $p$ (the lower boundary). She picks $p$ at the highest point of this lower boundary, which is where an increasing line meets a decreasing one. The height there is the value of the game. Example: $\begin{pmatrix} 6 & 1 & 4 \\ 2 & 5 & 3 \end{pmatrix}$ gives lines $2+4p$, $5-4p$ and $3+p$. The lower boundary peaks where the lines for columns 2 and 3 meet: $5-4p=3+p$, so $p=\frac25$ and the value is $\frac{17}{5}$. Column 1 is above the boundary there, so Colin never uses it.

The two lines that meet at the peak tell you which of Colin's columns he actually uses, and only those two columns are played. Solve the resulting $2\times2$ game for Colin: let him play one of the two columns with probability $q$, make Rose's expected pay-off the same for both rows, and solve for $q$. His other columns have probability $0$. Continuing the example (columns 2 and 3): row 1 gives $q+4(1-q)=4-3q$ and row 2 gives $5q+3(1-q)=3+2q$; equating gives $q=\frac15$. Colin plays column 2 with probability $\frac15$, column 3 with probability $\frac45$ and column 1 never. The value is $4-\frac35=\frac{17}{5}$, matching Rose's side.

If Rose has $n$ rows and Colin has two columns, work from Colin's side. Let Colin play column 1 with probability $q$. Each row gives a line for Rose's expected pay-off, and Colin picks $q$ to minimise the highest line: the lowest point of the upper boundary. Example: $\begin{pmatrix} 2 & 6 \\ 6 & 1 \\ 4 & 3 \end{pmatrix}$ gives $6-4q$, $1+5q$, $3+q$. The upper boundary is lowest where rows 1 and 2 meet: $6-4q=1+5q$, so $q=\frac59$, with value $\frac{34}{9}$ (row 3's line is only $\frac{32}{9}$ there). Rose then plays only rows 1 and 2, in the $2\times2$ game: $6-4p=1+5p$ gives $p=\frac59$.

1. Define the probability clearly ($p$ or $q$ and for which strategy).
2. Write each line in simplified form and plot or sketch them.
3. Identify the optimum (highest point of the lower boundary, or lowest point of the upper boundary) and solve by equating the two lines.
4. State both players' strategies, with zero probabilities, and the value with its sign: a positive value favours Rose.
5. Interpret in context: the value is the average gain per play in the long run, not what happens each time.