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Pushing Probability to Punishing Exhaustion

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by dtweah » Wed May 13, 2009 5:02 pm
A and B play a series of games; the winner of each game has probability 0.8 of winning the next game. If A wins the first game, what is the probability A will win the third game?

(a) .68
(b) .76
(c) .72
(d) .60
(e) .64
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by ontherocks27 » Wed May 13, 2009 11:29 pm
There are two ways in which A can win the 3rd game after winning the first game

1st way: A wins 2nd game and A wins 3rd game
Probablity = 0.8 * 0.8 = 0.64

or

2nd way: B wins 2nd game and A wins 3rd game
Probablity = (1-0.8) * (0.2) = 0.2*0.2 = 0.04

Overall probability of A winning the 3rd game = 0.64+0.04=0.68

Hence (A)
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by sanju09 » Thu May 14, 2009 4:54 am
After A wins the first game, A can win the third game in two situations

1. A loses second and wins the third: probability = 0.2 × 0.2 = 0.04. OR
2. A wins second and also the third: probability = 0.8 × 0.8 = 0.64.

Total probability = 0.04 + 0.64 + 0.68. My [spoiler]A[/spoiler].

Explanation: In case 1, A loses second means B wins, and A wins third means B loses. See how the probabilities change as per the condition given. Hope no doubts.
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