[donkeykong2]

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One other thing Ortom .

I'd like to clarify is that your answer may still vary depending on your variance or s.d . For instance , take two players with a mean of 60 % but one player has a higher variance and plays more aggressively . His variance is higher and consequently his risk of going broke is higher . If you really want to be precise then you should use the formula I gave in the preceding post .

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this is not true imo, if u assume 60% wins for each player their variance/sd is exactly the same, the ultra aggro player does not have higher variance with the same winrate, because in this case the variance is a function of the winrate (and number of games).
variance for binomial distribution is =n*winrate*(1-winrate) n is number of games.

off topic: variance ist highest for breakeven players.
doesnt make much of a difference though: var is n*0.25 for breakeven players n*0.24 for 60% winners.
so for 100 games the variance is approximately 25.
more interesting is sd which is var^1/2=5 because u can create confidence intervalls with sd.
a breakeven player should be within 40 and 60 wins of 100 for about 95%. 60% winner same within 50 and 70.
assuming 1000 games, var is 250 sd is around 16 so 50% winner should be within 500+- 2*sd=468,532 for 95% and 60% winner between 568,632 for 95%.
u can also see that your real winrate should be within +-3% of your observed winrate at 1000 games, and +-10% at 100 assuming 0,95 confidence level. u need 4x more games for double precision (half as big intervall), u would need 10k games to get under +-1, which sucks cause conditions will likely change or u might get better in that time.