3 Rose and Colin repeatedly play a zero-sum game. The pay-off matrix shows the number of points won by Rose for each combination of strategies.
| \multirow{6}{*}{Rose's strategy} | Colin's strategy |
| | \(W\) | \(X\) | \(Y\) | \(Z\) |
| \(A\) | -1 | 4 | -3 | 2 |
| \(B\) | 5 | -2 | 5 | 6 |
| C | 3 | -4 | -1 | 0 |
| \(D\) | -5 | 6 | -4 | -2 |
- What is the greatest number of points that Colin can win when Rose plays strategy \(A\) and which strategy does Colin need to play to achieve this?
- Show that strategy \(B\) dominates strategy \(C\) and also that strategy \(Y\) dominates strategy \(Z\). Hence reduce the game to a \(3 \times 3\) pay-off matrix.
- Find the play-safe strategy for each player on the reduced game. Is the game stable?
Rose makes a random choice between the strategies, choosing strategy \(A\) with probability \(p _ { 1 }\), strategy \(B\) with probability \(p _ { 2 }\) and strategy \(D\) with probability \(p _ { 3 }\). She formulates the following LP problem to be solved using the Simplex algorithm:
$$\begin{array} { l l }
\text { maximise } & M = m - 5 ,
\text { subject to } & m \leqslant 4 p _ { 1 } + 10 p _ { 2 } ,
& m \leqslant 9 p _ { 1 } + 3 p _ { 2 } + 11 p _ { 3 } ,
& m \leqslant 2 p _ { 1 } + 10 p _ { 2 } + p _ { 3 } ,
& p _ { 1 } + p _ { 2 } + p _ { 3 } \leqslant 1 ,
\text { and } & p _ { 1 } \geqslant 0 , p _ { 2 } \geqslant 0 , p _ { 3 } \geqslant 0 , m \geqslant 0 .
\end{array}$$
(You are not required to solve this problem.) - Explain how \(9 p _ { 1 } + 3 p _ { 2 } + 11 p _ { 3 }\) was obtained.
A computer gives the solution to the LP problem as \(p _ { 1 } = \frac { 7 } { 48 } , p _ { 2 } = \frac { 27 } { 48 } , p _ { 3 } = \frac { 14 } { 48 }\).
- Calculate the value of \(M\) at this solution.