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Old 12-05-2005, 12:15 AM
pzhon pzhon is offline
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Join Date: Mar 2004
Posts: 66
Default Re: RoR -- What % loss before double?

If you want a rough 95% confidence interval, you have about a 95% chance of having a maximum drop before you double up of between B1 and B2, where B1 is the bankroll level at which you would have a 97.5% risk of ruin, and B2 is the bankroll level at which you would have a 2.5% risk of ruin.

B1 = B log(.025)/Log(.001) = 0.534 B.
B2 = B log(.975)/log(.001) = 0.00367 B.

That gives you a confidence interval for the worst drawdown ever. It's not very different from your minimum before doubling your bankroll. The latter problem can be solved exactly, but I don't think that is what you really want.

Here is a useful tool for seeing the difference: The probability you lose L units before you win W units is roughly

(c^L-c^(L+W))
-------------- .
(1-c^(L+W))

Let B be the current size of your bankroll. When you set W=infinity and L=B, you should get the risk of ruin of 0.1% you assume, and the above simplifies to c^B, so c=0.001^(1/B).
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