[BruceZ]

[ QUOTE ]
As always, you are the man. We were missing the idea of subtracting your risk of ruin at the upper end, which was resulting in numbers that we knew defied reason. Your method results in numbers that "feel" more accurate.

If you'll indulge me, I think this results in:

.5/1 -> 1/2 .787
1/2 -> 2/4 .787
2/4 -> 3/6 .884 (only need 30 BBs to advance)
3/6 -> 5/10 .849
5/10 -> 10/20 .787
10/20 -> 15/30 .884
Total = 32.3 chance of success. This seems much more like I thought it would be. Maybe even a bit higher than I expected.

However, when I try to adjust for 2 bb/100, something seems wrong. I'm doing:

exp((-2 * 2 * 20) / (18^2)) = 0.781 - exp((-2 * 2 * 40) / (18^2)) = 0.610
which says I have only a 17.1% chance of busting if my earn-rate is lower.

Can this be correct or have I missed something fundamental?

Thanks for your help.

[/ QUOTE ]

Sorry, I made a mistake. The formula is more complicated than that. Let ror(B) mean the risk of ruin for a B bet bankroll if you play forever, and let r(B') mean the risk of going broke before the bankroll has grown to B'. Then we have:

ror(B) = r(B') + [1 - r(B')]*ror(B')

That is, to go broke with bankroll B, we can go broke before we ever reach B', which has probability r(B'), or we can reach B', which has probability [1 - r(B')], and then lose B', which has probability ror(B'). I was leaving out the factor of [1 - r(B')] before.

Solving for r(B') gives:

r(B') = [ror(B) - ror(B')] / [1 - ror(B')]

where as always, ror(B) = exp(-2*EV*B/SD^2).


So if we start with a 20 bet bankroll, with a win rate of 2 bb/100, and SD = 18 bb for 100 hands, then ror(B) = ror(20) = 69%. Then ror(B') = ror(40) = 47.6%, and the probability that we go broke before we reach 40 is:

r(40) = (69% - 47.6%) / (1 - 47.6%) =~ 40.8%.

So you need to divide the risk numbers you computed earlier by the probability of success. With that correction, the probability of going broke before doubling your bankroll is higher with a win rate of 2 bb/100 than with 3 bb/100, as you would expect.

Note that the probability of reaching the next level is 1 - r(B') or

P(success) = [1 - ror(B)] / [1 - ror(B')].