[NLTRN] Maths question...

[ChicagoRy]

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Nice calculations Jay_Shark!

78% chance to win over 50 games (out of 100) with a winrate of 57%. That's actually pretty high I think, very nice to know [img]/images/graemlins/smile.gif[/img]

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It was more comforting to pretend I was a 57% winner who experiences a lot of downswings [img]/images/graemlins/wink.gif[/img]

[waxhax0rs]

Although winning 50/100 of games is still losing. So I'd guess (without doing any maths) that the probability that you at least break even (which is probably what my initial question should have been) is probably more like 70%). This means you probably lose money on over 30% of 100 game stretches which is actually kind of a lot. I just need to get over variance I guess :/

[jay_shark]

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Although winning 50/100 of games is still losing. So I'd guess (without doing any maths) that the probability that you at least break even (which is probably what my initial question should have been) is probably more like 70%). This means you probably lose money on over 30% of 100 game stretches which is actually kind of a lot. I just need to get over variance I guess :/

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Your standard deviation without the rake , which is how you would solve these problems is :

var(x) = 1^2*0.57 + (-1)^2*(0.43) - 0.14^2
var(x) = 0.9804
s.d(x)=sqrt(0.9804)

Or equivalently, your standard deviation is ~ 1 per sng . Your sample standard deviation over 100 games is sqrt(100*.9804)=9.804/100 games .

The probability you win 53 games out of 100 games as a 57% winner is :

z=(53-57)/9.804 = -0.407996736

The probability is ~ 50% + 15.7% = 65.7%

After 1000 games , the probability you win at least 525 games is ~ 42.22% + 50% = 92.22%