# Thread: Karel: Calculating ROR given Kelly fraction \$p\$

1. ## Cacarulo: Re: Playing bankroll

> It is psychologically useful to have a
> playing bankroll, although it is
> theoretically redundant.

This is more or less what I do although I understand that your definition is more precise: I play "full" kelly of my playing bankroll (PBR, say it is \$10,000). If the PBR goes down I don't resize. If I happen to double it then I double my unit size.
Of course, my theoretically kelly number is 1/n being "n" the quotient between my "Net Worth" and my "PBR". If my NW is \$100,000 then my kelly number is 1/10.

Sincerely,
Cacarulo

2. ## Karel: Except

You should not double the unit size after doubling the bankroll, since then your theoretical Kelly number would increase quite a bit. If the theoretical number was originally 1/n, after doubling the bankroll you have (n+1)/n, so the optimal new unit size should be (n+1)/n times larger.

Regards,

Karel

> If the PBR goes down I don't resize. If I
> happen to double it then I double my unit
> size.
> Of course, my theoretically kelly number is
> 1/n being "n" the quotient between
> my "Net Worth" and my
> "PBR". If my NW is \$100,000 then
> my kelly number is 1/10.

> Sincerely,
> Cacarulo

3. ## Karel: Re: Playing bankroll

Karel could you please tell me what your ROR would be for a fixed bank playing to a 1.5% ROR when, after doubling it, you withdrew the winnings and reverted back to the original bank using an identical playing/betting strategy?

I am not 100% sure I undestand your problem properly, but let me give it a try. Are you asking the probability of ruin if you "try the ruin twice"? If this is the case, you can use a trick to calculate the number.

Let me do this in general for \$n\$ trials. The proper question is: What is the probability of NOT experiencing the ruin in \$n\$ trials. If ROR per trial is \$r\$, then probability of no ROR in one trial is \$1-r\$. Thus, due to independence of trails, the probability of no ROR in \$n\$ trials is \$(1-r)^n\$. The probability of at least one ROR in \$n\$ trials is then the complement, i.e.,
\$\$ 1 - (1-r)^n.\$\$

In your case, the probability would be
\$\$1 - (1-1.5%)^2 = 2.978%.\$\$

Regards,

Karel

4. ## Cacarulo: Agree

> You should not double the unit size after
> doubling the bankroll, since then your
> theoretical Kelly number would increase
> quite a bit. If the theoretical number was
> originally 1/n, after doubling the bankroll
> you have (n+1)/n, so the optimal new unit
> size should be (n+1)/n times larger.

The problem usually is in determining "n". I think an increase of 10%-20% would be in order.

Thanks.

Sincerely,
Cacarulo

5. ## Double21: Re: Playing bankroll

Thanks for the response Karel.

In a sense my question is very simple---at some point most people spend or otherwise use part of their winnings. I have often wondered what impact this has on your starting ROR, which of couse assumes never taking any winnings out. I thought that at a certain level you should be able to withdraw without a significant impact on this starting ROR.

After all, the cards don't know you are starting over with the same bankroll. Teams routinely play until they reach a certain point (time or money), disband, distribute the winnings (or remaining proceeds) then start over thinking the ROR is still the same. You are saying this is a bad assumption because the ROR increases with this strategy. If this is true, then after breaking a bank many times the ROR would get quite high. Am I understanding you correctly?

6. ## Karel: Re: Playing bankroll

> After all, the cards don't know you are
> starting over with the same bankroll. Teams
> routinely play until they reach a certain
> point (time or money), disband, distribute
> the winnings (or remaining proceeds) then
> start over thinking the ROR is still the
> same. You are saying this is a bad
> assumption because the ROR increases with
> this strategy. If this is true, then after
> breaking a bank many times the ROR would get
> quite high. Am I understanding you
> correctly?

Well, I am not sure if you understand me correctly. Whenever you start again you ROR is still the same. There is no influence of the past on the future (so-called Markov process).

On the other hand, if you start now and ask "What is the chance of busting at least once if I consume my winnings after doubling the bankroll, and then try to double again", the chance of busting at least once is certainly higher, just because you have two trials.

Another example: Throw a coin twice in a row. What is the probability of getting heads AT LEAST once? The probability is 3/4. However, if you through once and have tails, the probability of getting heads for the second round is 1/2, regardless of the first result.

Regards,

Karel

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