Question

Solve the following game

x y
x 2 5
y 4 1

Answer 1

Answer on Question #56227 - Math - Other

Solve the following game

x y

x 2 5

y 4 1

Solution

We associate $x$ with 1, $y$ with 2 in the next part of question.

If A chooses the first row with probability $p_1$ (i.e. uses the mixed strategy $(p_1, p_2 = 1 - p_1)$ ), we equate his average return when B uses columns 1 and 2.

$$a _ {1 1} p _ {1} + a _ {2 1} (1 - p _ {1}) = a _ {1 2} p _ {1} + a _ {2 2} (1 - p _ {1}).$$

Solving for $p_1$ , we find

$$p _ {1} = \frac {a _ {2 2} - a _ {2 1}}{\left(a _ {1 1} + a _ {2 2}\right) - \left(a _ {2 1} + a _ {1 2}\right)}$$

Player A's average return using this strategy is

$$V = a _ {1 1} p _ {1} + a _ {2 1} (1 - p _ {1}) = \frac {a _ {1 1} a _ {2 2} - a _ {2 1} a _ {1 2}}{\left(a _ {1 1} + a _ {2 2}\right) - \left(a _ {2 1} + a _ {1 2}\right)}$$

If B chooses the first column with probability $q_{1}$ (i.e. uses the strategy $(q_{1}, q_{2} = 1 - q_{1})$ ), we equate his average losses when A uses rows 1 and 2.

$$a _ {1 1} q _ {1} + a _ {2 1} (1 - q _ {1}) = a _ {1 2} q _ {1} + a _ {2 2} (1 - q _ {1}).$$

Hence,

$$q _ {1} = \frac {a _ {2 2} - a _ {1 2}}{\left(a _ {1 1} + a _ {2 2}\right) - \left(a _ {2 1} + a _ {1 2}\right)}$$

Player B's average loss using this strategy is

$$a _ {1 1} q _ {1} + a _ {2 1} (1 - q _ {1}) = \frac {a _ {1 1} a _ {2 2} - a _ {2 1} a _ {1 2}}{\left(a _ {1 1} + a _ {2 2}\right) - \left(a _ {2 1} + a _ {1 2}\right)}$$

The following formulae are used to find the value of the game and optimum strategies:

$$p _ {1} = \frac {a _ {2 2} - a _ {2 1}}{\left(a _ {1 1} + a _ {2 2}\right) - \left(a _ {2 1} + a _ {1 2}\right)}; p _ {2} = 1 - p _ {1}$$ $$q _ {1} = \frac {a _ {2 2} - a _ {1 2}}{(a _ {1 1} + a _ {2 2}) - (a _ {2 1} + a _ {1 2})}; q _ {2} = 1 - q _ {1}$$

and the value of the game is

$$V = \frac {a _ {1 1} a _ {2 2} - a _ {2 1} a _ {1 2}}{(a _ {1 1} + a _ {2 2}) - (a _ {2 1} + a _ {1 2})}.$$

Therefore,

$$p _ {1} = \frac {1 - 4}{(2 + 1) - (4 + 5)} = \frac {1}{2}; p _ {2} = 1 - \frac {1}{2} = \frac {1}{2}.$$ $$q _ {1} = \frac {1 - 5}{(2 + 1) - (4 + 5)} = \frac {2}{3}; q _ {2} = 1 - \frac {2}{3} = \frac {1}{3}$$

and the value of the game is

$$V = \frac {2 \cdot 1 - 5 \cdot 4}{(2 + 1) - (4 + 5)} = 3.$$

The value of the game is 3, the optimal strategy for A is $\left(\frac{1}{2},\frac{1}{2}\right)$ and the optimal strategy for B is $\left(\frac{2}{3},\frac{1}{3}\right)$ .

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