Buy in requirements

[jay_shark]

I made the assumption that the probability you win every individual game is constant at x% , for x=55%,60%,65% .

If you want to be precise , you should use the formula that includes your standard deviation and win rates to calculate your ror . It's usually the case that you may win x % of your games on average but you don't necessarily win each game x % of the time . There is a subtle difference but one that should be pointed out .

Nycballer , the number of buy ins depends on how comfortable you are with a certain risk of busting . Some players may want a 1% risk while others wouldn't mind a 5% risk .Every player is different , so there isn't one correct number of buy ins for every single player .

[ORTOM]

[ QUOTE ]

I made the assumption that the probability you win every individual game is constant at x% , for x=55%,60%,65% .

If you want to be precise , you should use the formula that includes your standard deviation and win rates to calculate your ror . It's usually the case that you may win x % of your games on average but you don't necessarily win each game x % of the time . There is a subtle difference but one that should be pointed out .

Nycballer , the number of buy ins depends on how comfortable you are with a certain risk of busting . Some players may want a 1% risk while others wouldn't mind a 5% risk .Every player is different , so there isn't one correct number of buy ins for every single player .

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I addressed this issue in my other post in this topic. If you'd like to see the results for other hypothetical players, let me know.

[jay_shark]

Good post Ortom .

You may use the formula that I gave earlier about calculating your ror depending on your s.d and win rate .


r0r= e^(-2uB/sigma^2)

ror= risk of ruin or the probability you go broke
u= hourly rate which is $/h over time
B=bankroll
sigma=standard deviation

This is a better formula to use because it tells you your ror for various win rates and s.d's .This formula has been posted extensively in the probability forum which is very neat .

[jay_shark]

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 .

[donkeykong2]

[ QUOTE ]
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.

[ORTOM]

[ QUOTE ]
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.