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Old 12-05-2005, 04:09 AM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default Re: RoR -- What % loss before double?

DavidC asked me to post this question after a chat tonight.

If I'm betting at ROR = 0.1%, and I don't ever adjust my betsize, (therefore if I go on an initial downswing I will, during that downswing, be betting at greater than RoR= 0.1%), how much of my roll can I assume that I will lose before I will double my roll on average?

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I set about answering your question literally. I found that your average loss before doubling your bankroll (including the times you fail to double) is about <font color="red">14.4%</font> of your bankroll. However, I also found the probability distribution of this loss which decreases as an exponential from 0% to 100%, and the median loss is about 10% of your bankroll. That is the loss you exceed 50% of the time.

If ror is the probability that you will lose your whole bankroll if you play forever, then the probability that you lose a fraction f of your bankroll is ror^f. Let P(f) be the probability that you lose a fraction f of your bankroll before you double your bankroll. Then P(f) satisfies:

ror^f = P(f) + [1 - P(f)]*ror^(1 + 1/f)

That is, since ror^f is the probability that you lose a fraction f over all time, this can be written as a sum of the probability that you lose f before you double, which is P(f), plus the probability that you double before losing f, which is 1 - P(f), times the probability that you then lose the fraction f of your original bankroll after that, and this has probability ror^[(1+f)/f] = ror^(1 + 1/f) since we must now lose the amount of the original bankroll plus f. For example, if f is 1/2, then we must now lose 3/2 of the original bankroll, and this has probability ror^(1 + 1/2) = ror^(3/2). Solving the above for P(f) gives:

P(f) = [ror^f - ror^(1 + 1/f)] / [1 - ror^(1 + 1/f)]

This is the exact probability distribution for losing a fraction f <font color="red">or more</font>, and by substituting your value of 0.1% for ror, you can set this to 0.5 and solve for f to find the median fraction lost by using the Excel goal search function. However, for your small ror of 0.1%, the term ror^f dominates this expression, and we can simply use this this as P(f):

P(f) =~ ror^f = 0.001^f = 0.5

f = ln(0.5) / ln(0.001) =~ 10%.

This is the median loss. The approximation used is equivalent to ignoring those times that we lose f after doubling, since this is rare. In this case it makes no practical difference to the final answer.

We can now use P(f) to find the expected value of the loss fraction f. <font color="red">P(f) is actually the probabilty of losing f or more, so 1 - P(f) is the cumulative distribution of f, so the derivative of this -P'(f) is the density function which we use for the expected value of f.</font>

E(f) = integral{f = 0 to 1} f*<font color="red">-P'(f)</font> df

Again we will only use the dominant term ror^f for P(f).

<font color="red">-P'(f) =~ -d/df (ror^f) = -d/df exp[f*ln(ror)] = -ln(ror)*exp[f*ln(ror)]</font>

E(f) =~ integral{f = 0 to 1} f*<font color="red">-ln(ror)</font>*exp[f*ln(ror)] df

If you don't know this integral, you can integrate it by parts, or simply guess the answer:

E(f) = <font color="red">-ln(ror)*{</font>f/ln(ror) * exp[f*ln(ror)] - 1/[ln(ror)]^2 * exp[f*ln(ror)] <font color="red">}</font> evaluated between 1 and 0.

You can verify this antiderivative using the chain rule. Evaluating from 1 to 0 gives the final result:

E(f) = <font color="red">-ln(ror)*{</font>ror/ln(ror) - ror/[ln(ror)]^2 + 1/[ln(ror)]^2 <font color="red">}

E(f) = (ror-1)/ln(ror) - ror

In this case, the -1/[ln(ror) term dominates the calculation. Evaluating for ror = 0.1% gives the mean fraction loss of about 14.4% of the bankroll.</font>
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