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View Full Version : Bankroll, risk of ruin, etc.

Guy McSucker
04-27-2005, 09:15 AM
Hey,

I guess this is known stuff, but anyway...

I'm familiar with risk of ruin calculations which assume a player reinvests all his winnings in his bankroll. I've lost the thread now but BruceZ gave a nice derivation of the same formula that Sileo got.

Most players don't want to keep their winnings in their roll forever: they want to spend some of it. A common plan is to have a cap on the bankroll and cash out (i.e. spend) anything above that cap. I got to wondering what the associated risk of ruin for this strategy would be.

WARNING: the answer is not nice, but it is sort of obvious once you think about it.

Consider a coin-flip game where you win 1 bet with probability p and lose 1 bet with probability 1-p. Once your bankroll reaches M bets, you will spend any further bets you win.

Let R(n) be your risk of ruin with a roll of n bets.

Clearly R(0) = 1. You're already broke.

For 0 &lt; n &lt; M, you either win one bet (probability p) or lose one, and then play on:

R(n) = pR(n+1) + (1-p)R(n-1).

When your roll is M bets, a bet you win gets spent, but a bet you lose is still a bet lost:

R(M) = pR(M) + (1-p)R(M-1).

I'm going to show that R(n) = R(n-1) for every n.

R(M) = pR(M) + (1-p)R(M-1), so

(1-p)R(M) = (1-p)R(M-1), so

R(M) = R(M-1), so my claim holds for M. (Not true if p = 1 but we're not dealin with sure things here...)

Now suppose that for some n, R(n+1) = R(n). Then

R(n) = pR(n+1) + (1-p)R(n-1), so

R(n) = pR(n) + (1-p)R(n-1), so

(1-p)R(n) = (1-p)R(n-1), so

R(n) = R(n-1).

So by induction, R(n) = R(n-1) for all n in 1 &lt; n &lt;= M.

Hence R(M) = 1: you will certainly go broke eventually.

Conclusion: if you do not let your bankroll grow indefinitely, you will go broke with probability 1 if you play forever. Which is a long time.

Question: what kind of playing strategies exist which allow for a small risk of ruin but allow you to spend some of your winnings? And how can we calculate risk of ruin/bankroll requirements for these strategies?

Guy.

jason1990
04-27-2005, 09:38 AM
If I spend x BB/100 and my true winrate is y BB/100, where y&gt;x, then I have the same risk of ruin as if I was spending nothing and my true winrate were y-x BB/100.

So suppose you're playing \$3/\$6 and you believe your true winrate is 3 BB/100. Then one thing you could do is triple your starting bankroll and spend (or withdraw) \$12 from your bankroll every 100 hands. This would give you the same risk of ruin. (But you better be right about your true winrate.)

Pokerscott
04-27-2005, 02:48 PM
If you play long enough you will eventually have a losing streak that will exceed any finite amount. Of course, if you are a +ROI player you have infinite winnings so it is not so bad /images/graemlins/smile.gif

True, but hard to translate into anything practical lol.

Pokerscott