Suppose that a 2x2 game has no saddle point. Suppose the game has the following pay-off matrix.

**Method
of solution of a 2x2 zero-sum game without saddle point**

Suppose that a 2x2 game has no saddle point. Suppose the game has the
following pay-off matrix.

**Since this game has no saddle
point, the following condition shall hold:** *Max *{*Min *{*a *,* b*},*
Min *{*c *,* d *}}* *≠* Min *{*Max *{*a
*,* c*},* Max *{*b*,* d*}} *In this case, the game is called a *mixed game.* No strategy of Player A can be called the best strategy for him.
Therefore A has to use both of his strategies. Similarly no strategy of Player
B can be called the best strategy for him and he has to use both of his
strategies.* *Let p be
the probability that Player A will use his first strategy. **Then the probability that Player
A will use his second strategy is 1-p.* __If Player B follows his
first strategy__

In the above equation, note that
the expected value is got as the product of the corresponding values of the
pay-off and the probability. __If Player B follows his
second strategy__

If the expected values in equations
(1) and (2) are different, Player B will prefer the minimum of the two expected
values that he has to give to player A. Thus B will have a pure strategy. This
contradicts our assumption that the game is a mixed one. Therefore the expected
values of the pay-offs to Player A in equations (1) and (2) should be equal.
Thus we have the condition

**To find the number of times that
B will use his first strategy and second strategy:** Let the probability that B will use his first strategy be r. Then the
probability that B will use his second strategy is 1-r. __When A use his first strategy__ The expected value of loss to
Player B with his first strategy = ar The expected value of loss to
Player B with his second strategy = b(1-r) Therefore
the expected value of loss to B = ar + b(1-r) | (3) |

__When A use his second strategy__ The expected value of loss to
Player B with his first strategy = cr The expected value of loss to
Player B with his second strategy = d(1-r) Therefore
the expected value of loss to B = cr + d(1-r) | (4) |

If the two expected values are
different then it results in a pure game, which is a contradiction. Therefore
the expected values of loss to Player B in equations (3) and (4) should be
equal. Hence we have the condition

**Problem 2** Solve the following game

**Solution** First
consider the row minima.

**Next consider the maximum of each column**

Therefore, out of 2 trials, player X will use his first strategy once
and his second strategy once.

Therefore, out of 3 trials, player Y will use his first strategy twice
and his second strategy once.

Tags : Operations Management - Game Theory, Goal Programming & Queuing Theory

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