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April 20, 2014

April 20, 2014

Posted by **Urgent!!** on Wednesday, February 13, 2013 at 8:49pm.

- Math -
**Reiny**, Wednesday, February 13, 2013 at 9:53pmBe "equally matched" I will assume

P(win) = 1/2 = prob(loss)

winning in 4 games = (1/2)^4 = 1/16

in 5 games, has to lose once)

LWWWW

WLWWW

WWLWW

WWWLW ----- each of these has a prob of (1/2)^5

but there are 4 cases,

so prob(5games) = 4/32 = 1/8

6 games , 2 losses, 4 wins

number of ways = 6!/(2!4!) = 15 , but that includes the case of ending with a loss, which can't happen

so number of 6 games is 14

prob(6games) = 14/(1/2)^6 = 14/64 = 7/32

7 games, 4 wins, 3 losses

number of cases = 7!/(3!4!) = 35 , less the case of ending in L

we have 34 such cases

prob(7 games) = 34/(1/2)^7 = 34/128 = 17/64

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