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[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.

[jay_shark]

Donkeykong , it is true .

Your variance is NOT equal if you have the same win rate . You're assuming that the probability you win each individual game is the same when it's not . If this were the case , that is , the probability you win each individual game is always y% , then two players with the same win rate have the same s.d . Two players can have different styles and still end up with the same win rate over time , or close enough . Every players s.d is different and it's not a function of your win rate . They are two completely different things .

The variance of a binomial distribution is n*P*(1-p)
s.d = sqrt[n*P*(1-p)] but this assumes p is constant for each game .

[donkeykong2]

ah i get your point, but does it real make a big difference?
the variance for 1 game only reduces drastically if u win with over 80% oder lose with over 80% probability 0.16 vs 0.25 for breakeven players i just dont think u get over these values often enough to make that a big factor especially at higher stakes with few complete retards.
if u assume the ultra aggro player has almost always the same chance to win regardless of his opponent and the other player often wins or loses almost for sure than it does of course make a difference.

[CallYNotRaise06]

i have a question, lets say your playing a heads up cash game. if you feel your a 65% favorite vs that player..would that make you a 65% winner? and would those calulations work here?

[ORTOM]

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The variance of a binomial distribution is n*P*(1-p)
s.d = sqrt[n*P*(1-p)] but this assumes p is constant for each game .

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I didnt' assume that p was constant for each game. It varied between 50 and 70 percent. In my simulation, I first picked what the winrate would be, then, based on that, decided if the player won.

[ORTOM]

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i have a question, lets say your playing a heads up cash game. if you feel your a 65% favorite vs that player..would that make you a 65% winner? and would those calulations work here?

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In other words
1. Probability of winning is .65 aka
2. In the long run, you win 65 out of every 100 games.

[donkeykong2]

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The variance of a binomial distribution is n*P*(1-p)
s.d = sqrt[n*P*(1-p)] but this assumes p is constant for each game .

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I didnt' assume that p was constant for each game. It varied between 50 and 70 percent. In my simulation, I first picked what the winrate would be, then, based on that, decided if the player won.

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but assuming u sit down first in a sng, u should calculate with p=0.6 as u dont know anything about your random opponent, dont u? sitting there with a good or bad opponent is part of luck, so your variance shouldnt be reduced if u do well vs certain opps and do bad vs others. pls correct me if i m wrong.

[cwar]

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The variance of a binomial distribution is n*P*(1-p)
s.d = sqrt[n*P*(1-p)] but this assumes p is constant for each game .

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I didnt' assume that p was constant for each game. It varied between 50 and 70 percent. In my simulation, I first picked what the winrate would be, then, based on that, decided if the player won.

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but assuming u sit down first in a sng, u should calculate with p=0.6 as u dont know anything about your random opponent, dont u? sitting there with a good or bad opponent is part of luck, so your variance shouldnt be reduced if u do well vs certain opps and do bad vs others. pls correct me if i m wrong.

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Taking the average of your winrate does seems like it would reduce variance IMO but much much easier.

[solinar]

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i have a question, lets say your playing a heads up cash game. if you feel your a 65% favorite vs that player..would that make you a 65% winner? and would those calulations work here?

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I would say not. The whole point of winning $ at poker is not winning the higher percentage of pots you possibly can, its winning big pots when you win, and losing small pots when you lose.

[ORTOM]

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I would say not. The whole point of winning $ at poker is not winning the higher percentage of pots you possibly can, its winning big pots when you win, and losing small pots when you lose.

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We're talking about sit and gos. Also, the entire point of poker is to force your opponent to make mistakes.