Question 3:
3 | (a) | Beth
X Y Z row min
P 2 -3 c -3
Amir Q -3 b 4 -3
R a -1 -2 min(a, -2)
min(a, -2) > -3
-3 < a < 0 | B1
M1
A1
[3] | 2.1
1.1
2.2a | Row minima calculated (at least as far as showing that rows P and
Q each have row min = -3)
Ignore any other working
Appropriate comparison, may be implied from a > -3 or a ≥ -3 seen
cao
3 | (b) | Beth
X Y Z
P 2 -3 c
Amir Q -3 b 4
R a -1 -2
col max 2 b max(c, 4)
b < 2 and b < max(c, 4)
0 < b < 2 | B1
M1
A1
[3] | 2.1
1.1
2.2a | Col maxima calculated (or negatives of these values)
(at least as far as showing that col X has col max = 2)
Ignore any other working
Either of these seen, may be implied from b < 2 or b  2 seen
Accept 0 < b < min(2, c)
3 | (c) | Stable  min for row R = max for col Y
so min(a, -2) = b
But min(a, -2) < 0 and b > 0 so cannot be equal
Hence unstable | M1
M1d
A1 | 3.4
1.1
2.2a | Interpret stability in terms of min for row R and max for col Y
Using (a, -2) and b
Explain why condition fails, accept a < 0 < b
Written descriptions must be unambiguous
Hence not stable, from correct explanation seen
3 | (c) | Alternative method 1 | Interpret stability in terms of entry in cell (R, Y) = -1
Stable  -1 = min(a, -2) | M1 | Or Stable  -1 = b
But min(a, -2)  -2, so  -1 | But min(a, -2)  -2, so  -1 | M1d | Or But b > 0, so  -1
Explain why condition fails | Or But b > 0, so  -1
Hence unstable | A1 | Hence not stable, from correct explanation seen
Alternative method 2 | Finding Nash equilibrium cells
M1 | Need not explicitly show pay-offs for Beth
Amir: (P, X), (Q, Y), (P, Z), (Q, Z) depending on value of c
Beth: (P, Y), (Q, X), (R, X), (R, Z) depending on value of a
M1d | (R, Y) is not a Nash equilibrium
Hence unstable | A1 | Hence not stable, from correct explanation seen
Alternative method 3 | ‘Chasing’ best options
e.g. If Beth plays safe and chooses Y then
Amir’s best choice is Q | M1 | May use an informal approach with at least first step
Y → Q or R → ‘X or Z (depending on value of a)’ condone R → Z
Written descriptions must be unambiguous
But Q is not Amir’s play-safe strategy | M1d | Or equivalent, or similarly for Beth,
or (R, Y) is not a Nash equilibrium
Hence unstable | A1 | Hence not stable, from correct explanation seen
Alternative method 4 | Using minimax and maximin for whole table
Minimax for columns is min(2, b) > 0 | M1 | Minimax (2, b) or b and > 0 o.e.
Maximin for rows is max(-3, min(a, -2)) < 0 | M1 | Maximin (-3, a, -2) or (a, -2) < 0 o.e.
So minimax  maximin, hence unstable | A1 | Hence not stable, from correct explanation seen
[3]
Need not explicitly show pay-offs for Beth
Amir: (P, X), (Q, Y), (P, Z), (Q, Z) depending on value of c
Beth: (P, Y), (Q, X), (R, X), (R, Z) depending on value of a
e.g. If Beth plays safe and chooses Y then
Amir’s best choice is Q
May use an informal approach with at least first step
Y → Q or R → ‘X or Z (depending on value of a)’ condone R → Z
3 | 11