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 < n < 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 < n <= 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.
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 < n < 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 < n <= 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.