Solutions — Decision Mathematics Applications

Fluency Matrix [[3,−1],[−2,4]]: maximin, minimax, saddle point?

Row minima: A1 = min(3, −1) = −1 • A2 = min(−2, 4) = −2
Maximin = max(−1, −2) = −1 (Row player chooses A1)

Column maxima: B1 = max(3, −2) = 3 • B2 = max(−1, 4) = 4
Minimax = min(3, 4) = 3 (Column player chooses B1)

Maximin (−1) ≠ Minimax (3) → No saddle point exists. Mixed strategy analysis is required.

Fluency Find saddle point of 3×3 matrix [[2,4,3],[5,6,4],[1,3,2]].

Row minima: A1 = 2 • A2 = 4 • A3 = 1
Maximin = max(2, 4, 1) = 4 (Row player chooses A2)

Column maxima: B1 = max(2,5,1) = 5 • B2 = max(4,6,3) = 6 • B3 = max(3,4,2) = 4
Minimax = min(5, 6, 4) = 4 (Column player chooses B3)

Maximin = Minimax = 4 → Saddle point at (A2, B3).

Optimal pure strategies: A plays A2; B plays B3.
Value of the game = 4 (payoff to A at the saddle point entry).

Fluency Identify dominated strategies: A1(3,5), A2(1,2), A3(4,3).

Compare each pair of rows (the row player wants to maximise, so a higher value is always better):

A1 vs A2: B1: A1=3 > A2=1 ✓ B2: A1=5 > A2=2 ✓
A1 > A2 in every column → A2 is dominated by A1. A2 should be eliminated.

A1 vs A3: B1: A1=3 < A3=4 B2: A1=5 > A3=3. Neither dominates.
A2 vs A3: B1: A2=1 < A3=4 B2: A2=2 < A3=3. A2 is also dominated by A3.

Eliminate A2. A3 does not dominate A1 (and vice versa), so only A2 is eliminated.

Fluency Mixed strategy for matrix [[1,4],[3,2]].

a=1, b=4, c=3, d=2.
D = (a+d) − (b+c) = (1+2) − (4+3) = 3 − 7 = −4

Row player (A):
p = (d−c)/D = (2−3)/(−4) = (−1)/(−4) = 1/4
A plays A1 with probability 1/4, A2 with probability 3/4.

Column player (B):
q = (d−b)/D = (2−4)/(−4) = (−2)/(−4) = 1/2
B plays B1 with probability 1/2, B2 with probability 1/2.

Value of the game:
V = (ad−bc)/D = (1×2 − 4×3)/(−4) = (2−12)/(−4) = (−10)/(−4) = 2.5

Understanding Complete analysis of matrix [[5,−1],[2,3]].

Saddle point check:
Row minima: A1 = −1, A2 = 2. Maximin = 2 (A2).
Column maxima: B1 = 5, B2 = 3. Minimax = 3 (B2).
2 ≠ 3 → No saddle point.

Mixed strategy:
a=5, b=−1, c=2, d=3.
D = (5+3) − (−1+2) = 8 − 1 = 7

Row player: p = (3−2)/7 = 1/7
A plays A1 with prob 1/7, A2 with prob 6/7.

Column player: q = (3−(−1))/7 = 4/7
B plays B1 with prob 4/7, B2 with prob 3/7.

Value of the game: V = (5×3−(−1)×2)/7 = (15+2)/7 = 17/7 ≈ 2.43
The game favours A (positive value): A gains approximately 2.43 units on average.

Understanding Café competition matrix [[6,4],[2,8]].

(a) Saddle point check:
Row minima: Pasta=4, Salad=2. Maximin = 4 (A chooses Pasta).
Column maxima: B:Pasta=6, B:Salad=8. Minimax = 6 (B chooses Pasta).
4 ≠ 6 → No saddle point.

(b) Mixed strategy:
a=6, b=4, c=2, d=8.
D = (6+8) − (4+2) = 14 − 6 = 8

Café A: p = (8−2)/8 = 6/8 = 3/4
A offers Pasta with prob 3/4, Salad with prob 1/4.

Café B: q = (8−4)/8 = 4/8 = 1/2
B offers Pasta with prob 1/2, Salad with prob 1/2.

(c) Value of the game: V = (6×8−4×2)/8 = (48−8)/8 = 40/8 = 5
In the long run, Café A gains an expected 5 extra customers per day from Café B.

Understanding 3×3 dominance reduction: [[2,5,3],[4,6,1],[1,3,2]].

Step 1 — Row dominance (row player maximises):
Compare A1(2,5,3) and A3(1,3,2): A1 > A3 in every column (2>1, 5>3, 3>2).
A3 is dominated by A1. Eliminate A3.

Reduced matrix:

	B1	B2	B3
A1	2	5	3
A2	4	6	1

Step 2 — Column dominance (column player minimises):
Compare B3(3,1) and B2(5,6): B3 < B2 in every row (3<5, 1<6).
B2 is dominated by B3 (B3 gives A lower payoffs, so B prefers B3 over B2). Eliminate B2.

Resulting 2×2 matrix:

	B1	B3
A1	2	3
A2	4	1

Row mins: A1=2, A2=1. Maximin=2. Col maxs: B1=4, B3=3. Minimax=3. 2≠3, no saddle point.

a=2, b=3, c=4, d=1. D = (2+1)−(3+4) = 3−7 = −4
Row player: p = (1−4)/(−4) = 3/4 (A plays A1 with prob 3/4, A2 with prob 1/4)
Column player: q = (1−3)/(−4) = 1/2 (B plays B1 with prob 1/2, B3 with prob 1/2)
Value of the game: V = (2×1−3×4)/(−4) = (2−12)/(−4) = 2.5

Understanding Interpret a game with value V = −2.

A game value of V = −2 means that when both players use their optimal strategies, the expected payoff to the row player (A) is −2 per game — that is, A pays 2 units to B on average each time the game is played.

This is not a fair game. A fair game has value V = 0.

Player B has the advantage: in the long run, B gains 2 units per game. Player A cannot avoid this loss (on average) if B plays optimally, regardless of the strategy A chooses. A’s optimal mixed strategy limits the loss to exactly 2 per game; any deviation makes it worse for A.

Problem Solving Company pricing matrix [[4,2],[1,6]].

(a) Saddle point check:
Row minima: A:Low = 2, A:High = 1. Maximin = 2 (Low pricing).
Column maxima: B:Low = 4, B:High = 6. Minimax = 4 (Low pricing).
2 ≠ 4 → No saddle point.

(b) Mixed strategy:
a=4, b=2, c=1, d=6. D = (4+6)−(2+1) = 10−3 = 7

Company A: p = (6−1)/7 = 5/7
A uses Low pricing 5/7 of the time, High pricing 2/7 of the time.

Company B: q = (6−2)/7 = 4/7
B uses Low pricing 4/7 of the time, High pricing 3/7 of the time.

(c) Value of the game: V = (4×6−2×1)/7 = (24−2)/7 = 22/7 ≈ 3.14
Company A gains approximately 3.14 market share units per period on average in the long run.

(d) If B always plays Low:
Company A’s payoffs are: A:Low→4 (vs B:Low), A:High→1 (vs B:Low).
Best response: Company A plays Low (payoff 4 > 1). A should always choose Low pricing if B is committed to Low.

Problem Solving 3×3 dominance then solve: [[2,5,3],[4,6,2],[1,3,2]].

(a) Dominance elimination:

Step 1 — Row dominance:
Compare A1(2,5,3) and A3(1,3,2): A1 > A3 in every column (2>1, 5>3, 3>2).
A3 is dominated by A1. Eliminate A3.

Reduced matrix:

	B1	B2	B3
A1	2	5	3
A2	4	6	2

Step 2 — Column dominance (minimiser prefers lower payoffs):
Compare B3(3,2) and B2(5,6): B3 < B2 in every row (3<5, 2<6).
B2 is dominated by B3. Eliminate B2.

Resulting 2×2 matrix:

	B1	B3
A1	2	3
A2	4	2

(b) Solve the reduced game:
Row mins: A1=2, A2=2. Maximin=2. Col maxs: B1=4, B3=3. Minimax=3. 2≠3, no saddle point.

a=2, b=3, c=4, d=2. D = (2+2)−(3+4) = 4−7 = −3
Row player: p = (2−4)/(−3) = (−2)/(−3) = 2/3
A plays A1 with probability 2/3, A2 with probability 1/3.

Column player: q = (2−3)/(−3) = (−1)/(−3) = 1/3
B plays B1 with probability 1/3, B3 with probability 2/3 (B2 eliminated).

Value of the game: V = (2×2−3×4)/(−3) = (4−12)/(−3) = (−8)/(−3) = 8/3 ≈ 2.67

(c) Interpretation:
The value V = 8/3 ≈ 2.67 > 0 means the game favours Player A. On average, A gains approximately 2.67 units per game when both players use their optimal mixed strategies. Player B cannot do better than giving away 2.67 units per game on average, regardless of the strategy chosen.