Question

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.

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)
=(_{4}C_{3})*(1/2)^{4}*(1/2)+(_{4}C_{3})*(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)
=(_{5}C_{3})*(1/2)^{4}*(1/2)^{2}+(_{5}C_{3})*(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)
=(_{6}C_{3})*(1/2)^{4}*(1/2)^{3}+(_{6}C_{3})*(1/2)^{4}*(1/2)^{3}=**0.3125**

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