Question

Every time a certain basketball team wins, the players become over-confident. In that case, their chance of winning the next game is only 35%. Every time the team loses, the players become angry with themselves & more focused. In that case, their chance of winning the next game is 75%. Assuming that the team wins the first game, calculate the probability they win the fourth game. Find the equilibrium distribution as well.

Answer 1

P(win | win last game) = 0.35,
Therefore P( lose | win last game) = 1 - 0.35 = 0.65

Also, we are given here that:
P( win | lose last game) = 0.75,
Therefore P(lose | lose last game) = 1 - 0.75 = 0.25

Given that the team wins the first game probability that they win the fourth game is computed as:

• WWWW: Prob = 0.353 = 0.042875
• WWLW: Prob = 0.35*0.65*0.75 = 0.170625
• WLWW: Prob. = 0.65*0.75*0.35 = 0.170625
• WLLW: Prob = 0.65*0.25*0.75 = 0.121875

Therefore total probability:
= 0.042875 + 0.170625 + 0.170625 + 0.121875

= 0.506

therefore 0.506 is the required probability here.

Let the equilibrium distribution in the 2 states be W and L.

Then, we have here:
W = 0.35W + 0.75L
0.65W = 0.75L
W = (15/13)L

Also W + L = 1

(28/13)L = 1
L = 13/28, therefore W = 15/28

Therefore the steady state equilibrium probabilities here are given as: (15/28, 13/28)