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Probability : Games Won

1. Suppose that two teams play a series of games that ends when one of the teams has one i number of games. Suppose that each game played is, independently, won by team A with probability p. Find the expected number of games that are played when (a) i = 2 and when (b) i = 3. Show also in both cases that this number is maximized when p = ½.

2. Consider problem 1 with i = 2. Find the variance of the number of games played and show that this number is maximized when p = ½.

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The two teams play a series of games that ends when one of them has won i games
P(A) = p, then P(B) = 1-p

(a) when i = 2, we need at the most 3 rounds.
In the first round, the probability of A winning is p
In the second round, the probability of A winning again is p, when A won the whole game.
Then probability for A to finish the game in two rounds is p*p = p^2
("^" is the power of)
However, the same story applies to team B, probability for B to finish the game in two rounds is (1-p)^2
the total probability of finishing the game in two rounds is p^2+(1-p)^2

If the game goes to the third round, the probability is then 1-( p^2+(1-p)^2)
Then the expected number of games (rounds) that are played is
N = 2*( p^2+(1-p)^2) + 3*(1-( p^2+(1-p)^2))
= 3 - (p^2+(1-p)^2)
= 3 - (2 p^2 - 2p +1)

To maximize N, we write:
N = 3 - (2 p^2 - 2p +1)
= 3 -2 (p^2 - p +0.5)
= 3 -2 (p^2 - p +0.25) - 0.5
= 2.5 -2 (p- 0.5)^2

To maximize N, we need p-0.5=0, then p = 0.5
And N(max) = 2.5 - 0 =2.5

[ also we can deduct the following caculation:
when i = 2, we need at the most 3 rounds.
In the first round, the probability of A winning is 0.5
In the second round, the ...

Solution Summary

Probability of games won is investigated. The solution is detailed and well presented.

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