[BruceZ]

The formula I posted is from Blackjack Attack by Don Schlesinger. It is a highly respected formula used by the blackjack community, and I have demonstrated earlier that it gives the exact result when applied to a problem that can be done by a direct method. This formula gives the risk of ruin for a given bankroll if you play forever. We can use it for this problem by noting that if the risk of losing 50 bets before we get to 300 bets is p, and the risk of losing a 50 bet bankroll if we play forever is p(50) and the risk of losing a 300 bet bankroll if we play forever is p(300), then we have:

p(50) = p + (1-p)p(300)

That is, the probability of losing a 50 bet bankroll if we play forever is the probability of losing a 50 bet bankroll before we get to 300, plus the probability that we get to 300 (which is 1-p) times the probability that we lose a 300 bet bankroll before we play forever. Solving for p gives:

p = [p(50)-p(300)]/[(1-p(300)]

Schlesinger's forumla gives us p(50) and p(300) as follows:

P(50) = [(1 - 1.05/13.11)/(1 + 1.05/13.11)]^(50/13.11) = 54.2%.

p(300) = [(1 - 1.05/13.11)/(1 + 1.05/13.11)]^(300/13.11) = 2.53%.

So p = 53%.

So you will lose the 50 bets before you get to 300 bets 53% of the time, and you will get to 300 bets 47% of the time. Perfect agreement with irchans approximate formula (irchans: care to give your derivation?).

If you want a 10% risk of ruin for a bankroll of size B, then working backwards:

.1 = [p(B) - p(300)]/[1-p(300)]

since p(300) = .0253, p(B) = .123. This is the chance of going broke if you play forever with a bankroll of B. Now calculate what B is from the inverse ruin equation:

B = [-(13.11)^2/(2*1.05)]ln(.123) = 172 bets.