Teams named East and West are playing a best-2-out-of-3 series of games. This me

Question

Teams named East and West are playing a best-2-out-of-3 series of games. This means that the first team to win two games wins the series. (For example, in the NCAA College Baseball World Series, the two teams in the finals play a best-2-out-of-3 series.) If the first two games are won by the same team, the series is over. The 3rd game is played only if the teams are 1-1 after the first two games.

In each game that is played, the probability that East wins is 0.77. So the probability West wins is 0.23.

What is the probability that East wins the series?   
What is the probability that the series requires only two games?   
What is the probability that the team that wins the first game wins the series?   
What is the expected value for how many games are played?  

Explanation / Answer

Let E denote east winning and W denote west winning a game.

Then EWE means, E wins first and third games and W wins second game

P(East wins the series) = P(EE) + P(EWE) + P(WEE)

= 0.77x0.77 + 0.77x0.23x0.77 + 0.23x0.77x0.77

= 0.5929 + 0.1364 + 0.1364

= 0.8657

P(series requires only 2 games) = P(EE) + P(WW)

= 0.5929 + 0.23x0.23

= 0.5929 + 0.0529

= 0.6458

P(team wins first game wins the series) = 1 - P(EWW) - P(WEE)

= 1 - 0.77x0.23x0.23 - 0.23x0.77x0.77

= 1 - 0.0407 - 0.1364

= 0.8229

Expected number of games = 2xP(2 games) + 3xP(3games)

= 2x(0.6458) + 3(1 - 0.6458)

= 1.2916 + 1.0626

= 2.3542

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