| Exam Board | OCR |
|---|---|
| Module | Further Discrete AS (Further Discrete AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Challenging +1.2 This is a standard game theory question requiring application of the mixed strategy formula for 2×2 games. While it involves multiple steps (finding expected values, solving equations, interpreting results), the method is algorithmic and well-practiced. The conceptual demand is moderate—understanding what makes a strategy optimal—but no novel insight is required beyond applying the standard technique. |
| Spec | 7.08e Mixed strategies: optimal strategy using equations or graphical method |
| Player \(\boldsymbol { Y }\) | ||
| Strategy \(\boldsymbol { R }\) | Strategy \(\boldsymbol { S }\) | |
| Strategy \(\boldsymbol { P }\) | 4 | - 2 |
| Strategy \(\boldsymbol { Q }\) | - 3 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | Let X play strategy P with probability p and |
| Answer | Marks |
|---|---|
| and Q so that P has prob 0.4 and Q has prob 0.6 | M1 |
| Answer | Marks |
|---|---|
| [5] | i3.3 |
| Answer | Marks |
|---|---|
| 3.4 | m |
| Answer | Marks |
|---|---|
| Interpretation of p(cid:32)0.4 in context | May use p for P(cid:11)Xplays Q(cid:12) |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (ii) | E.g. You must assume that Y is not going to play |
| Answer | Marks | Guidance |
|---|---|---|
| pure strategy | E1 | |
| [1] | 3.5b | For one limitation of the model with |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1 | 0 |
| 3 | 0 | 1 |
Question 3:
3 | (i) | Let X play strategy P with probability p and
strategy Q with probability 1(cid:16) p
If Y plays strategy R, X can expect to win
4p(cid:16)3(cid:11)1(cid:16) p(cid:12)(cid:32)7p(cid:16)3
If Y plays S, X can expect (cid:16)2p(cid:14)(cid:11)1(cid:16) pp (cid:12)(cid:32)1(cid:16)3p
p(cid:32)0(cid:159)minE(cid:11)win(cid:12)(cid:32)(cid:16)3
S
p(cid:32)1(cid:159)minE(cid:11)win(cid:12)(cid:32)(cid:16)2
7p(cid:16)3(cid:32)1(cid:16)3p(cid:159) p(cid:32)0.4 and
minE(cid:11)win(cid:12)(cid:32)(cid:16)0.2
X should choose randomly between strategies P
and Q so that P has prob 0.4 and Q has prob 0.6 | M1
e
A1
E1
M1
A1
[5] | i3.3
c
1.1
3.4
1.1a
3.4 | m
Calculating expected winnings
if Y plays R or S
Both correct
Explicitly considering extreme points,
or using a sketch graph to show that
optimum point is at intersection
Solving their expressions
simultaneously to achieve their p
Interpretation of p(cid:32)0.4 in context | May use p for P(cid:11)Xplays Q(cid:12)
May add a constant throughout
Need not simplify expressions
BC
3 | (ii) | E.g. You must assume that Y is not going to play
predictably
E.g. You must assume that Y is not going to use a
pure strategy | E1
[1] | 3.5b | For one limitation of the model with
reference to Y’s limited behaviour
3 | 1 | 0 | 7.00
3 | 0 | 1 | 8.50
3
n3 A zero-sum game is being played between two players, $X$ and $Y$. The pay-off matrix for $X$ is given below.
\section*{Player X}
\begin{center}
\begin{tabular}{ l | r | r }
& \multicolumn{2}{|c}{Player $\boldsymbol { Y }$} \\
& Strategy $\boldsymbol { R }$ & Strategy $\boldsymbol { S }$ \\
\hline
Strategy $\boldsymbol { P }$ & 4 & - 2 \\
\hline
Strategy $\boldsymbol { Q }$ & - 3 & 1 \\
\hline
\end{tabular}
\end{center}
(i) Find an optimal mixed strategy for player $X$.\\
(ii) Give one assumption that must be made about the behaviour of $Y$ in order to make the mixed strategy of Player $X$ valid.
\hfill \mbox{\textit{OCR Further Discrete AS Q3 [6]}}