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