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  #1  
Old 09-29-2004, 10:29 AM
ChicagoTroy ChicagoTroy is offline
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Default Bankroll downswings and ROR recalculations

Suppose I have $6K (300 BB) for my $10-20 game. My first month playing I play fine but experience a 120BB downswing. Am I *still* bankrolled for this game?

Since the cards have no memory I could "reset the clock" to t=0 and I'm starting out playing a game with only 180BB. OTOH, if I look at things from the beginning of month 1, I'm experiencing a downswing but my bankroll started out adequate so I'm fine. This line of thinking seems to imply my short term luck will even out however, and that's not logical.

Thoughts?
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  #2  
Old 09-29-2004, 10:45 AM
imitation imitation is offline
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Default Re: Bankroll downswings and ROR recalculations

I can't explain the maths but i've read it before and in english, you have ~1% chance of running -300BB but using your example say you run -120BB this increases your chances that you are in the middle of your -300BB run, infact it would increase it to the same chance you have of running -180BB. So in essence yes you should probably drop back to 5/10.
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  #3  
Old 09-29-2004, 11:16 AM
ChicagoTroy ChicagoTroy is offline
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Default Re: Bankroll downswings and ROR recalculations

I know it would be a good idea, I just can't figure out mathematically why.
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  #4  
Old 09-29-2004, 02:12 PM
fnord_too fnord_too is offline
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Default Re: Bankroll downswings and ROR recalculations

It has to do with conditional probability. Say you have a 1% chance of going bust. That is over all possible outcomes. You will definitely lose 120BB at some point in the -300BB run.

So, say there is a 10% chance of losing 120BB (number pulled out of nowhere). Then, 100% of the 1% chance of going bust is in this 10% of runs that you lose 120BB. Then, the chance of losing 300BB given you have lost 120BB is in fact 10%. Use Bayes Theorem to get here.

Bayes Theorem: P(A|B) = [Prob(B|A)*Prob(A)]/Prob(B)

A = lose 300BB P(A) = .01
B = lose 120 BB P(B) = .10
P(B|A) = 1 (since lose 300BB cannot be true if lose 120BB is not true)

So P(A|B) = (1 * .01)/.10 = .01/.10 = .10

You can run your model to find out exactly what P(B) is, then just divide P(A) by this to get the conditional probability.
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  #5  
Old 09-29-2004, 03:04 PM
ChicagoTroy ChicagoTroy is offline
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Default Re: Bankroll downswings and ROR recalculations

Excellent, thanks.
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  #6  
Old 09-30-2004, 09:55 PM
cigarzfan cigarzfan is offline
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Default Re: Bankroll downswings and ROR recalculations

Seems to me the answer to this question is pretty easy. The whole point of having a high number like 300BB for a bankroll is to absorb the eventual run of bad cards. Thus, a bad run has already been incorporated into the 300BB figure. You can continue to play the same table, because probability dictates that you should pull out of your extended slump (assuming you are a positive expectation player), and regain your money. You cannot simply "reset the clock" when you lose 1/2 your bankroll. Doing this would assume that you were a negative expectation player trying to start over.

In short, the whole point of having a 300BB is to ride out the lulls. If you keep restarting the clock when you lose money, then you are admitting that you should have lost that money and are attempting a "do over".

If you keep refilling your bankroll to 300BB and keep losing half of it, then you should reevaluate your play and not your statistics.
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  #7  
Old 10-02-2004, 07:01 PM
pfkaok pfkaok is offline
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Default Re: Bankroll downswings and ROR recalculations

[ QUOTE ]
Seems to me the answer to this question is pretty easy. The whole point of having a high number like 300BB for a bankroll is to absorb the eventual run of bad cards. Thus, a bad run has already been incorporated into the 300BB figure. You can continue to play the same table, because probability dictates that you should pull out of your extended slump (assuming you are a positive expectation player), and regain your money. You cannot simply "reset the clock" when you lose 1/2 your bankroll. Doing this would assume that you were a negative expectation player trying to start over.

[/ QUOTE ]

I don't agree with this at all, and I think that this is a good way to maximize your chances to lose your BR. Bayes theorom as stated before basically explains it mathematically, but just think about it logically... Cards and hands are independant, and you're never "due" to have a good run. If you've had AA cracked 5 times in a row, they're still just as likely to get cracked next time you have them. If you have, say a 1% chance of having a horrible session and losing 100 BB, then after having one, you still have a 1% chance of having one in your next session. Probably even more if you think like this b/c you will expect to hit your hands so you likely wouldn't be playing in an optimal manner. The 300 BB figure certainly doesn't ALWAYS absorb bad runs, even for a winning player, thats why your Risk of ruin isn't ZERO.

The whole point of having a 300 BB roll is that if you go on a horrible run and lose 150 BB, then you can drop back down to half the level and still have a 300 BB roll for that game.
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