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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?

PS - I'm not sure if I should be assigning some percentage basis
on how often I will lose a given portion of my roll. (I'm talking about confidence levels when I say this.)

 AaronBrown 12-04-2005 11:44 PM

Re: RoR -- What % loss before double?

I'm not sure I understand the situation. I think you're making a series of identical bets. Your expected profit from each one is fixed at 0.1% of your initial bankroll. To make it concrete, you start with \$1,000 and make bets on which you have an expected profit of \$1. You want to know the expected minimum point before you get up to \$2,000.

Obviously this depends on the standard deviation of the individual bets. If that standard deviation is zero, if you just get \$1 each time, then \$1,000 in your certain minimum. If, instead, you're betting \$1,000 for one chance in 1 million of getting \$1,001,000,000, then you've got a 0.999999 chance of hitting zero before going above \$2,000.

Let's say you have 0.55 chance of winning \$10 and 0.45 chance of losing \$10 each time. Your average minimum point is about \$955. 18% of the time you'll never be behind, and 70% of the time, you'll never be below \$950. 89% of the time you'll never be below \$900 and 98% of the time you'll never be below \$800.

 pzhon 12-05-2005 12:15 AM

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).

 DavidC 12-05-2005 01:57 AM

Re: RoR -- What % loss before double?

[ QUOTE ]
I'm not sure I understand the situation. I think you're making a series of identical bets. Your expected profit from each one is fixed at 0.1% of your initial bankroll.

[/ QUOTE ]

And realized that I'd screwwed up. I'm sorry. [img]/images/graemlins/blush.gif[/img]

I was thinking from the standpoint of Kelly-Betting, how they increase their betsize based on their EV per hand, and that they take on a big ROR for their bankroll. I think that they can expect swings to be something like 90% of their roll sometimes before they double their roll, but I can't say how often this happens.

I'm just kinda curious as to what sorts of swings I should expect if I'm making a bunch of bets at a low baknroll ror (0.1%), but it's possible that I'm not giving you enough info here.

 DavidC 12-05-2005 02:03 AM

Re: RoR -- What % loss before double?

Thanks a lot, pzhon. This is perfect.

If yourself, Aaron, and Bruce would be willing to post up a bit about who you guys are / what you guys do for a living / where you learned this stuff / how you made it to this site, that'd be pretty sweet. [img]/images/graemlins/smile.gif[/img]

If you'd prefer not to, that's alright.

Take care,
Dave.

 BruceZ 12-05-2005 04:09 AM

Re: RoR -- What % loss before double?

[ QUOTE ]
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?

[/ QUOTE ]

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>

 BruceZ 12-06-2005 03:31 PM

Re: RoR -- What % loss before double?

I corrected an error in my computation of the mean fractional loss. Since P(f) is actually the probability of losing f or more, we need the density function -P'(f) to compute the expected value of f. This has the effect of multiplying the computed mean by -ln(ror), which for a ror of 0.1% makes the mean fractional loss 14.4%, which is closer to the median value of 10%. The changes are marked in red.

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