# Question 6:

## Part (a)
Total pay-off is 16, subtract 8 from each pay-off | **M1** [1.1] | Converting to zero-sum, or implied from total = 16 (for each cell) or from working; Each player puts 8 into 'pot'; May use differences
Pay-off matrix for Drew all correct (or a positive multiple of this): Drew P: $-5, 6, 3$; Q: $4, -4, -1$; R: $3, -4, -2$ | **A1** [1.1] |

## Part (b)
Strategy Q (weakly) dominates strategy R so matrix can be reduced | **M1** [1.1] | Identifying dominance; Row (from their zero-sum table) removed by dominance; If an A-level candidate chooses to use simplex — see guidance
Reduced matrix: Drew P: $-5, 6, 3$; Q: $4, -4, -1$ | | |
Let Drew choose randomly between P and Q so that the probability of P is $p$ and the probability of Q is $1-p$ | **A1** [2.1] | Describing (prob of) $P = p$ and (prob of) $Q = 1 - p$, or equivalent, in terms of one variable only
If Emma plays X, Drew expects $-5p + 4(1-p)$; If Emma plays Y, Drew expects $6p - 4(1-p)$; If Emma plays Z, Drew expects $3p - (1-p)$ | **B1** [1.1] | All three correct (in any form) for the values in their $2 \times 3$ reduced table; Or implied from graph if no expressions are given
$4 - 9p = 10p - 4$ | **M1** [1.1] | Solving simultaneous equations when $X = Y$
$p = \frac{8}{19}$ | | |
P with probability $\frac{8}{19}$ and Q with probability $\frac{11}{19}$ | **A1** [2.1] | Both probabilities correct in context

## Part (c)
From graph, if Drew plays optimally then Emma should not choose Z | **M1** [2.2a] | Or equivalent
Let Emma choose randomly between X and Y so that the probability of X is $x$ and the probability of Y is $1 - x$ | **A1** [2.1] | Description including symbols used for probabilities (one unknown only)
$5x - 6(1-x) = -4x + 4(1-x)$ or $= -\frac{4}{19}$ | **B1** [1.1] | Pay-offs for Emma are $\times(-1)$
Valid method | **M1** [1.1] |
X with probability $\frac{10}{19}$ and Y with probability $\frac{9}{19}$ | **A1** [2.1] | Either probability correct

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