Moderate -1.0 This is a standard game theory question covering dominance, play-safe strategies, and LP formulation for mixed strategies. Parts (i)-(iii) require only basic comparison of payoffs, while (iv)-(vi) test understanding of standard LP setup for zero-sum games—all routine applications with no novel problem-solving required.
6 Lucy and Maria repeatedly play a zero-sum game. The pay-off matrix shows the number of points won by Lucy, who is playing rows, for each combination of strategies.
\cline { 2 - 5 }
\(X\)
\(Y\)
\(Z\)
\(A\)
2
- 3
4
\cline { 2 - 5 }
Lucy's
\(B\)
- 3
5
1
\cline { 2 - 5 }
strategyy
\(C\)
4
2
- 3
Show that strategy \(A\) does not dominate strategy \(B\) and also that strategy \(B\) does not dominate strategy \(A\).
Show that Maria will not choose strategy \(Y\) if she plays safe.
Give a reason why Lucy might choose to play strategy \(B\).
Lucy decides to play strategy \(A\) with probability \(p _ { 1 }\), strategy \(B\) with probability \(p _ { 2 }\) and strategy \(C\) with probability \(p _ { 3 }\). She formulates the following LP problem to be solved using the Simplex algorithm:
$$\begin{array} { l l }
\text { maximise } & M = m - 3 , \\
\text { subject to } & m \leqslant 5 p _ { 1 } + 7 p _ { 3 } , \\
& m \leqslant 8 p _ { 2 } + 5 p _ { 3 } , \\
& m \leqslant 7 p _ { 1 } + 4 p _ { 2 } , \\
& p _ { 1 } + p _ { 2 } + p _ { 3 } \leqslant 1 , \\
\text { and } & p _ { 1 } \geqslant 0 , p _ { 2 } \geqslant 0 , p _ { 3 } \geqslant 0 , m \geqslant 0 .
\end{array}$$
[You are not required to solve this problem.]
Explain why 3 has to be subtracted from \(m\) in the objective row.
Explain how \(5 p _ { 1 } + 7 p _ { 3 } , 8 p _ { 2 } + 5 p _ { 3 }\) and \(7 p _ { 1 } + 4 p _ { 2 }\) were obtained.
Explain why \(m\) has to be less than or equal to each of the expressions in part (v).
Lucy discovers that Maria does not intend ever to choose strategy \(Y\). Because of this she decides that she will never choose strategy \(B\). This means that \(p _ { 2 } = 0\).
Show that the expected number of points won by Lucy when Maria chooses strategy \(X\) is \(4 - 2 p _ { 1 }\) and find a similar expression for the number of points won by Lucy when Maria chooses strategy \(Z\).
Set your two expressions from part (vii) equal to each other and solve for \(p _ { 1 }\). Calculate the expected number of points won by Lucy with this value of \(p _ { 1 }\) and also when \(p _ { 1 } = 0\) and when \(p _ { 1 } = 1\). Use these values to decide how Lucy should choose between strategies \(A\) and \(C\) to maximise the expected number of points that she wins.
In column \(Y\): \(-3 < 5\) so \(A\) does not dominate \(B\); In column \(X\): \(-3 < 2\) (or column \(Z\): \(1 < 4\)) so \(B\) does not dominate \(A\)
B1
For \(-3 < 5\) or equivalent
B1
For \(-3 < 2\) or \(1 < 4\) or equivalent
(ii)
Answer
Marks
Guidance
The worst outcomes for Maria are: \(X\) lose 4, \(Y\) lose 5, \(Z\) lose 4
M1
For finding column maxima
A1
For rejecting 5 as being bigger than 4, or using a word like 'lose' or '-4, -5, -4'
(iii)
Answer
Marks
Guidance
If Lucy plays \(B\) she could win as much as 5.
B1
For '5 is the most she can win' or equivalent
(iv)
Answer
Marks
Guidance
Need to add 3 throughout matrix to make values non-negative, this removes the 3 again.
B1
For 'add 3 throughout matrix' or equivalent
(v)
Answer
Marks
Guidance
Having added 3 throughout, the expected number of points win by Lucy when Maria chooses strategy \(X\) is \(3p_1 + 0p_2 + 7p_3\), and similarly the second expression is the expected number of points win by Lucy when Maria chooses strategy \(Y\) and the third expression is the expected number of points won by Lucy if Maria chooses strategy \(Z\)
M1
For showing where one of the expressions came from, or for referring to 'when Maria plays each of her strategies' or equivalent in a non-specific way
A1
For specifically linking the expressions to Maria choosing strategy \(X\), strategy \(Y\) and strategy \(Z\) in that order
(vi)
Answer
Marks
Guidance
The number of points that Lucy can expect to win cannot be less than the worst of the three expressions, so it is less than or equal to each of them.
M1
For reference to 'number of points won by Lucy' or equivalent
A1
For reference to 'the worst outcome' or equivalent
For solving \(4 - 2p_1 = \) their expression to get a probability
M1
For evaluating \(4-2p_1\) at their \(p_1\) and the values \(-3\) and \(2\)
A1
For reference to maximin, or equivalent, leading to selection of \(p = \frac{7}{9}\), or in context
**(i)**
In column $Y$: $-3 < 5$ so $A$ does not dominate $B$; In column $X$: $-3 < 2$ (or column $Z$: $1 < 4$) so $B$ does not dominate $A$ | B1 | For $-3 < 5$ or equivalent
| B1 | For $-3 < 2$ or $1 < 4$ or equivalent
**(ii)**
The worst outcomes for Maria are: $X$ lose 4, $Y$ lose 5, $Z$ lose 4 | M1 | For finding column maxima
| A1 | For rejecting 5 as being bigger than 4, or using a word like 'lose' or '-4, -5, -4'
**(iii)**
If Lucy plays $B$ she could win as much as 5. | B1 | For '5 is the most she can win' or equivalent
**(iv)**
Need to add 3 throughout matrix to make values non-negative, this removes the 3 again. | B1 | For 'add 3 throughout matrix' or equivalent
**(v)**
Having added 3 throughout, the expected number of points win by Lucy when Maria chooses strategy $X$ is $3p_1 + 0p_2 + 7p_3$, and similarly the second expression is the expected number of points win by Lucy when Maria chooses strategy $Y$ and the third expression is the expected number of points won by Lucy if Maria chooses strategy $Z$ | M1 | For showing where one of the expressions came from, or for referring to 'when Maria plays each of her strategies' or equivalent in a non-specific way
| A1 | For specifically linking the expressions to Maria choosing strategy $X$, strategy $Y$ and strategy $Z$ in that order
**(vi)**
The number of points that Lucy can expect to win cannot be less than the worst of the three expressions, so it is less than or equal to each of them. | M1 | For reference to 'number of points won by Lucy' or equivalent
| A1 | For reference to 'the worst outcome' or equivalent
**(vii)**
$2(p_1) + 4(p_3) = 2p_1 + 4(1-p_1) = 4-2p_1$ (given); $4p_1 - 3(1-p_1) = 7p_1 - 3$ | B1 | For $2p + 4(1-p)$
| B1 | For $7p - 3$
**(viii)**
$4 - 2p_1 = 7p_1 - 3 \Rightarrow p_1 = \frac{7}{9}$; $p_1 = \frac{7}{9} \Rightarrow 2 \frac{5}{9}$, $p_1 = 0 \Rightarrow \min(4, -3) = -3$, $p_1 = 1 \Rightarrow \min(2, 4) = 2$ | B1 | For solving $4 - 2p_1 = $ their expression to get a probability
| M1 | For evaluating $4-2p_1$ at their $p_1$ and the values $-3$ and $2$
| A1 | For reference to maximin, or equivalent, leading to selection of $p = \frac{7}{9}$, or in context
6 Lucy and Maria repeatedly play a zero-sum game. The pay-off matrix shows the number of points won by Lucy, who is playing rows, for each combination of strategies.
\begin{center}
\begin{tabular}{ l | r | r | r | r | }
\multicolumn{4}{c}{\begin{tabular}{ c }
Maria's \\
strategy \\
\end{tabular}} & \\
\cline { 2 - 5 }
& \multicolumn{1}{l}{$X$} & \multicolumn{1}{c}{$Y$} & $Z$ & \\
\hline
$A$ & 2 & - 3 & 4 & \\
\cline { 2 - 5 }
Lucy's & $B$ & - 3 & 5 & 1 \\
\cline { 2 - 5 }
strategyy & $C$ & 4 & 2 & - 3 \\
\hline
\end{tabular}
\end{center}
(i) Show that strategy $A$ does not dominate strategy $B$ and also that strategy $B$ does not dominate strategy $A$.\\
(ii) Show that Maria will not choose strategy $Y$ if she plays safe.\\
(iii) Give a reason why Lucy might choose to play strategy $B$.
Lucy decides to play strategy $A$ with probability $p _ { 1 }$, strategy $B$ with probability $p _ { 2 }$ and strategy $C$ with probability $p _ { 3 }$. She formulates the following LP problem to be solved using the Simplex algorithm:
$$\begin{array} { l l }
\text { maximise } & M = m - 3 , \\
\text { subject to } & m \leqslant 5 p _ { 1 } + 7 p _ { 3 } , \\
& m \leqslant 8 p _ { 2 } + 5 p _ { 3 } , \\
& m \leqslant 7 p _ { 1 } + 4 p _ { 2 } , \\
& p _ { 1 } + p _ { 2 } + p _ { 3 } \leqslant 1 , \\
\text { and } & p _ { 1 } \geqslant 0 , p _ { 2 } \geqslant 0 , p _ { 3 } \geqslant 0 , m \geqslant 0 .
\end{array}$$
[You are not required to solve this problem.]\\
(iv) Explain why 3 has to be subtracted from $m$ in the objective row.\\
(v) Explain how $5 p _ { 1 } + 7 p _ { 3 } , 8 p _ { 2 } + 5 p _ { 3 }$ and $7 p _ { 1 } + 4 p _ { 2 }$ were obtained.\\
(vi) Explain why $m$ has to be less than or equal to each of the expressions in part (v).
Lucy discovers that Maria does not intend ever to choose strategy $Y$. Because of this she decides that she will never choose strategy $B$. This means that $p _ { 2 } = 0$.\\
(vii) Show that the expected number of points won by Lucy when Maria chooses strategy $X$ is $4 - 2 p _ { 1 }$ and find a similar expression for the number of points won by Lucy when Maria chooses strategy $Z$.\\
(viii) Set your two expressions from part (vii) equal to each other and solve for $p _ { 1 }$. Calculate the expected number of points won by Lucy with this value of $p _ { 1 }$ and also when $p _ { 1 } = 0$ and when $p _ { 1 } = 1$. Use these values to decide how Lucy should choose between strategies $A$ and $C$ to maximise the expected number of points that she wins.
\hfill \mbox{\textit{OCR D2 2006 Q6 [15]}}