[jay_shark]

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Well theres 7 losses in a row then theres losing 7 buyins during a stretch of matches, Im talking about the latter.

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Oh. The original comment (from another thread) was about 7 in a row. [img]/images/graemlins/smile.gif[/img] Being down 7 overall is obviously much more common.

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The probability you are down exactly 7 buy-ins in 100 hands is 0 . You cannot be down an odd number of buy-ins after playing an even number of matches . Now if you were interested in the probability of dropping 6 buy-ins or more in 100 matches as a 58% winner we proceed as follows :

We use the normal distribution to give a very accurate approximation .

To be down 6 buy-ins or more in 100 matches , means you won at most 47 matches and lost 53 matches .

var = 1^2*0.58 + (-1)^2*0.42 - 0.16^2
s.d = sqrt(0.9744)

sample s.d over 100 games = 9.871170143

z=(47-58)/9.871170143
z=-1.114356235

Using a z-score table we see that the probability of being down 6 buy-ins or more in 100 matches as a 58% winner is 50%-36.65% = 13.35%