OCR D2 2010 June — Question 4 15 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Year2010
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeZero-sum game optimal mixed strategy
DifficultyModerate -0.3 This is a standard D2 game theory question covering fundamental concepts (zero-sum games, dominance, play-safe strategies, stability) and routine graphical solution for optimal mixed strategy. While it requires multiple steps and careful arithmetic, all techniques are textbook procedures with no novel insight required. The graphical method is algorithmic once expressions are found. Slightly easier than average A-level due to being methodical rather than conceptually challenging.
Spec7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation7.08e Mixed strategies: optimal strategy using equations or graphical method7.08f Mixed strategies via LP: reformulate as linear programming problem

4 Euan and Wai Mai play a zero-sum game. Each is trying to maximise the total number of points that they score in many repeats of the game. The table shows the number of points that Euan scores for each combination of strategies.
Wai Mai
\cline { 2 - 5 }\(X\)\(Y\)\(Z\)
\(A\)2- 53
\cline { 2 - 5 } \(E u a n\)- 1- 34
\cline { 1 - 5 } \(C\)3- 52
\(D\)3- 2- 1
  1. Explain what the term 'zero-sum game' means.
  2. How many points does Wai Mai score if she chooses \(X\) and Euan chooses \(A\) ?
  3. Why should Wai Mai never choose strategy \(Z\) ?
  4. Delete the column for \(Z\) and find the play-safe strategy for Euan and the play-safe strategy for Wai Mai on the table that remains. Is the resulting game stable or not? State how you know. The value 3 in the cell corresponding to Euan choosing \(D\) and Wai Mai choosing \(X\) is changed to - 5 ; otherwise the table is unchanged. Wai Mai decides that she will choose her strategy by making a random choice between \(X\) and \(Y\), choosing \(X\) with probability \(p\) and \(Y\) with probability \(1 - p\).
  5. Write down and simplify an expression for the expected score for Wai Mai when Euan chooses each of his four strategies.
  6. Using graph paper, draw a graph showing Wai Mai's expected score against \(p\) for each of Euan's four strategies and hence calculate the optimum value of \(p\).

Question 4:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
In each game, whatever combination of strategies is chosen, the total number of points won is zeroB1 Points won by Euan equals points lost by Wai Mai, and vice versa, in every case
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(-2\)B1 Loses 2
Part (iii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(Z\) is dominated by \(Y\); in each row she loses more by choosing \(Z\) than \(Y\): \(-3 < 5,\ -4 < 3,\ -2 < 5\) and \(1 < 2\) (or equivalent)M1, A1 Four valid comparisons and a convincing explanation (or equivalent in words)
Part (iv)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Determining row minima and column maxima, or equivalent (may be implied from both \(D\) and \(Y\) stated)M1
Play-safe for Euan is \(D\)A1 \(D\) stated (not just identified in table)
Play-safe for Wai Mai is \(Y\)A1 \(Y\) stated (not just identified in table)
Game is stable, since row maximin \(=\) col minimax, \(-2 = -2\)B1 Stable, with a valid reason attempted (numerical or in words) (www)
Part (v)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(A:\ -2p + 5(1-p) = 5 - 7p\)M1 Any one correct (or negative of correct), simplified or not
\(B:\ p + 3(1-p) = 3 - 2p\)A1 All four correct (or negative of correct) and simplified
\(C:\ -3p + 5(1-p) = 5 - 8p\)A1 All four correct and simplified
\(D:\ 5p + 2(1-p) = 2 + 3p\) (note: leaving \(DX\) as 3 gives \(D: 2-5p =\) M1A0A0)
Part (vi)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Graph plotted on graph paper with sensible scalesM1 Graph paper used with sensible scales
Their equations plotted correctlyA1 [2]
\(2 + 3p = 3 - 2p \Rightarrow p = 0.2\)M1 Solving correct pair, or from graph
\(p = 0.2\), cao, from correct equations used (algebraically or from graph) (www)A1 [2]
# Question 4:

## Part (i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| In each game, whatever combination of strategies is chosen, the total number of points won is zero | B1 | Points won by Euan equals points lost by Wai Mai, and vice versa, in every case | [1] |

## Part (ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2$ | B1 | Loses 2 | [1] |

## Part (iii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Z$ is dominated by $Y$; in each row she loses more by choosing $Z$ than $Y$: $-3 < 5,\ -4 < 3,\ -2 < 5$ and $1 < 2$ (or equivalent) | M1, A1 | Four valid comparisons and a convincing explanation (or equivalent in words) | [2] |

## Part (iv)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Determining row minima and column maxima, or equivalent (may be implied from both $D$ and $Y$ stated) | M1 | |
| Play-safe for Euan is $D$ | A1 | $D$ stated (not just identified in table) |
| Play-safe for Wai Mai is $Y$ | A1 | $Y$ stated (not just identified in table) |
| Game is stable, since row maximin $=$ col minimax, $-2 = -2$ | B1 | Stable, with a valid reason attempted (numerical or in words) (www) | [4] |

## Part (v)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A:\ -2p + 5(1-p) = 5 - 7p$ | M1 | Any one correct (or negative of correct), simplified or not |
| $B:\ p + 3(1-p) = 3 - 2p$ | A1 | All four correct (or negative of correct) and simplified |
| $C:\ -3p + 5(1-p) = 5 - 8p$ | A1 | All four correct and simplified | [3] |
| $D:\ 5p + 2(1-p) = 2 + 3p$ | | (note: leaving $DX$ as 3 gives $D: 2-5p =$ M1A0A0) |

## Part (vi)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph plotted on graph paper with sensible scales | M1 | Graph paper used with sensible scales |
| Their equations plotted correctly | A1 | [2] |
| $2 + 3p = 3 - 2p \Rightarrow p = 0.2$ | M1 | Solving correct pair, or from graph |
| $p = 0.2$, cao, from correct equations used (algebraically or from graph) (www) | A1 | [2] |

---
4 Euan and Wai Mai play a zero-sum game. Each is trying to maximise the total number of points that they score in many repeats of the game. The table shows the number of points that Euan scores for each combination of strategies.

\begin{center}
\begin{tabular}{ l | r | r | r | r | }
 & \multicolumn{4}{c}{Wai Mai} \\
\cline { 2 - 5 }
 & \multicolumn{1}{|c|}{$X$} & \multicolumn{1}{c}{$Y$} & \multicolumn{1}{c}{$Z$} &  \\
\hline
$A$ & 2 & - 5 & 3 &  \\
\cline { 2 - 5 }
$E u a n$ & - 1 & - 3 & 4 &  \\
\cline { 1 - 5 }
$C$ & 3 & - 5 & 2 &  \\
\hline
$D$ & 3 & - 2 & - 1 &  \\
\hline
\end{tabular}
\end{center}

(i) Explain what the term 'zero-sum game' means.\\
(ii) How many points does Wai Mai score if she chooses $X$ and Euan chooses $A$ ?\\
(iii) Why should Wai Mai never choose strategy $Z$ ?\\
(iv) Delete the column for $Z$ and find the play-safe strategy for Euan and the play-safe strategy for Wai Mai on the table that remains. Is the resulting game stable or not? State how you know.

The value 3 in the cell corresponding to Euan choosing $D$ and Wai Mai choosing $X$ is changed to - 5 ; otherwise the table is unchanged.

Wai Mai decides that she will choose her strategy by making a random choice between $X$ and $Y$, choosing $X$ with probability $p$ and $Y$ with probability $1 - p$.\\
(v) Write down and simplify an expression for the expected score for Wai Mai when Euan chooses each of his four strategies.\\
(vi) Using graph paper, draw a graph showing Wai Mai's expected score against $p$ for each of Euan's four strategies and hence calculate the optimum value of $p$.

\hfill \mbox{\textit{OCR D2 2010 Q4 [15]}}