What is the equilibrium probability that Charles will shoot

Use the following information to answer questions 10-14. Basketball power forward Charles Duncan has two post moves: the jump hook with the left hand and the newly learned drop step. His defender must anticipate one move or the other. The probability that Charles makes the shot is given in the table. (Remember that Charles prefers a high probability, while his defender prefers a low probability!) Defender Anticipate Jump hook Anticipate Drop step Jump hook 0.2 0.5 Charles Drop step 0.6 0.3

Solution

The probability given here is conditional probability, let us define the actions as given by notations

Defender Anticipate the jump hook- AJK, Defender Anticipate the drop step – ADS

Charles Shoot Jump Hook- JK , Charles Drop Step- DS

Now P(AJK/JK)=0.2 ,

P(ADS/JK)= 0.5,

P(AJK/DS)=0.6,P(ADS)/DS=0.3

What is the equilibrium probability that Charles will shoot a jump hook?

THIS IS EQUAL TO THE SUM OF THE P(AJK/JK)+P(ADS/JK) = 0.7

        What is the equilibrium probability that the defender plays anticipate a jump hook?

= P(AJK/JK)+P(AJK/DS)=0.6 =0.8

What is the equilibrium probability that Charles will hit the shot?

= P( NO NOT ANTICIPATE JUMP HOOK/ JUMP HOOK) + P( DO NOT ANTICIPATE DROP STEP/DROP STEP) = 0.2/0.7+0.2/0.9= 0.507