Chances of going broke

[Ra_]

I originialy posted this on the sports betting forum, but didn't get much help. anyone know of any formulas on th chances of going broke?

below is a problem i am going to simulate, just wonder if it could be solved.

Given:
-we are only taking standard spread bets
-x% vig,
-y% winrate, (assume the average winrate is the probability each individual bet will win, although in reality im sure its not)
-z% percent of bankroll bet as unit.
-You never lower your bet (ex. if you start with $100 bet %1 as unit, if you win your first bet your next bet will be $1.01 but if you lose it will stay $1)

what is the chance you will go broke before doubling up?

The only way i know to do this is to simulate it. is there a formula anyone could come up with? Are there any other formulas that deal with chances of going broke? Any help is appreciated.

[AaronBrown]

Let me make sure I understand.

You start with $N and bet z%*$N. y% of the time you win z%*$N*(1-x%), 1-y% of the time you lose z%*$N.

If you win, you have $N*[1+z%*(1-x%)], in that case you will raise your bet to z%*$N*[1+z%*(1-x%)]. If you lose, you will keep your bet at z%*$N. You'll keep betting until you are either broke or have 2*$N.

Schemes like this can be solved, but it's easier to simulate them. If you did want to solve it, you would enumerate all possible states (combinations of bankroll and last bet) and construct the transition matrix.

There is an approximate method that's easier. Obviously if your expectation is zero for every bet (that is, if the game is fair), you have a 50% chance of getting to 2*$N before you go broke. Suppose you have a positive expectation of 1% per bet. After $K have been bet (total handle), you expect to have $N + 0.01*$K. Therefore, if you know you are either at 2*$N or $0, the probability of being at 2*$N is 0.5 + 0.005*$K/$N. If you use $N/z for $K, that reduces to 0.5 + 0.005/z. That doesn't work for all z, but it's not a bad guess for reasonably large z relative to your edge.

[pzhon]

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-You never lower your bet (ex. if you start with $100 bet %1 as unit, if you win your first bet your next bet will be $1.01 but if you lose it will stay $1)

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This is a funny condition which is different from the assumptions behind standard ROR formulas. It looks like this means you will never be more than 100 (or 100/z) losses away from busting out. The probability p that you reach your next peak before busting out should be easy to compute*. You need to accomplish this some number of times n without failing (roughly 70 ~ ln(2)/1% if you stand to gain 1% of your original bankroll), so the probability that you do this is p^n.

* I'm not being specific at this step since I don't know how the juice is taken. However, a standard gambler's ruin formula where you start with 100 of the 101 chips (100/z of the 1 + 100/z) in play should be a good approximation.