bankroll question

[Popeye]

this doesn't have to be poker
but lets say i play a game where i have a very tiny 0.15% edge. how big does my bankroll y$ need to be for x$ bets and have a practical low risk of ruin?
anyone has a formula?
do i need to know the standard diviation or hourly rate to calculate that? or is this enough data to calculate?

[curious123]

[ QUOTE ]
do i need to know the standard diviation...to calculate that?

[/ QUOTE ]

Yes.

[gull]

Yes, you need to know the distribution of results.

Consider the following... (Bill Nye, hahaha)

1) You pay me $100. I pay you $100.15.

2) You pay me $100. I flip a coin. If it's heads, I pay you $200.30.

See the difference?

[AaronBrown]

I can give you some guidance. Suppose your chance of losing each bet is p. That means you must win 1.0015/(1-p) times the amount bet when you win (assuming there is a fixed payoff for winning).

If p is 0.5, as in a fair coin flip, you win 2.003 times the stake when you win. If p is 0.999999, as in a lottery, you have to win 1,001,500 times the stake when you win.

After k bets of b each time, your expected profit is 0.0015*k*b and the standard deviation is 1.0015*[k*p/(1-p)]^0.5. For two standard deviations (about 97.7%) protection, you need a bankroll of:

2.003*[k*p/(1-p)]^0.5 - 0.0015*k*p

This expression has a maximum when k = (1.0015/.0015)^2/[p*(1-p)]. At that point it equals 669/(1 - p). So you need a bankroll of about 1,338 bets for p = 0.5 (coin flip) or 669,000,000 for a 1 in a million lottery.

This is not an exact result, but it's a decent rule of thumb.