Question

Suppose that two teams are playing a series of games, each team independently wins each game...

Suppose that two teams are playing a series of games, each team independently wins each game with 1/2 probability. The final winner of the series is the first team to win four games. Let X be the number of games that the two teams have played. Find the distribution of X.

Math   Statistics-And-Probability   
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Answer #1

below is probability distribution of X:

P(X=4) =P(1st win 1st four games)+P(2nd wins first 4 games) =(1/2)4+(1/2)4=0.1250

P(X=5)=P(1st win 3 out of 1st four games and wins 5th game)+P(2nd win 3 out of 1st four games and wins 5th game) =(4C3)*(1/2)4*(1/2)+(4C3)*(1/2)4*(1/2)=0.2500

P(X=6)=P(1st win 3 out of 1st five games and wins 6th game)+P(2nd win 3 out of 1st five games and wins 6th game) =(5C3)*(1/2)4*(1/2)2+(5C3)*(1/2)4*(1/2)2=0.3125

P(X=7)=P(1st win 3 out of 1st six games and wins 7th game)+P(2nd win 3 out of 1st six games and wins 7th game) =(6C3)*(1/2)4*(1/2)3+(6C3)*(1/2)4*(1/2)3=0.3125

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