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Old 08-31-2005, 02:07 AM
PairTheBoard PairTheBoard is offline
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Join Date: Dec 2003
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Default Re: Am I stupid? I can\'t fit these two concepts into any type of harmony.

[ QUOTE ]
Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[/ QUOTE ]

If you have a July magazine article please provide the link here, and if you want to cite a paper cite it here.


It's not the "the rate of growth of the expected bankroll" either.

This is always assuming you are playing with an edge. Take the case where proportional betting is More than twice Kelly. Yes, in the long run, your Bankroll will converge in Probabilty to a negatively sloping curve that asymtopes to zero. But your Expected Bankroll will always be Greater than your Initial Bankroll. Your Expected Bankroll will Always be your Inititial Bankroll plus your Edge times your Action.

Again, playing with an edge. Consider the simple case of suicide Martingale Betting where you bet your bankroll until you lose. Clearly, your Bankroll converges in probabilty to zero. ie. Your Wong Win Rate is negative. But even as your Bankroll is converging in Probabilty to Zero, your Expected Bankroll climbs throughout.

Or are you now using the term "Expected Bankroll" to mean something other than the Expected Value of your Bankroll?

PairTheBoard
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