I hate to link to another board, but the poster poses an interesting question. Of the two bet schedules, which one is superior, optimal or custom?
The optimal has a higher SCORE but lower CE whereas the custom has a lower SCORE but higher CE.
Also, shouldn't CE be about 50% of WR/Hr? In both schedules the CE > 50% of WR/Hr.
I used the equation below and plugged in the variables given by the software but did not come up with the same CE for either optimal or custom.
CE = EV - Var/(2 * BR * Kf)
> I hate to link to another board, but the poster poses
> an interesting question. Of the two bet schedules,
> which one is superior, optimal or custom?
That's a very silly question. "Optimal" means optimal. If there were something better, then it wouldn't be optimal, now, would it?
> The optimal has a higher SCORE but lower CE whereas
> the custom has a lower SCORE but higher CE.
Apples and oranges. The RORs are completely different, so what's the sense of comparing?
> Also, shouldn't CE be about 50% of WR/Hr?
It is, when you bet optimally.
> In both
> schedules the CE > 50% of WR/Hr.
The CE is much greater than 50%, because, for "optimal," the ROR isn't anywhere near the 13.5% that it ought to be. You chose unit sizes, in dollars, that were much too small for your $10,000 bank. Garbage in, garbage out.
> I used the equation below and plugged in the variables
> given by the software but did not come up with the
> same CE for either optimal or custom.
> CE = EV - Var/(2 * BR * Kf)
See above.
Don
> The CE is much greater than 50%, because, for
> "optimal," the ROR isn't anywhere near the
> 13.5% that it ought to be.
Given the formula for CE, wouldn't a smaller ROR/Kf result in a smaller CE ( less than 50% WR)? You seem to be implying the smaller the ROR/kf, the larger the CE.
> You chose unit sizes, in
> dollars, that were much too small for your $10,000
> bank.
Perhaps the poster selected a small unit size b/c he was risk averse. Is CE only 50% of WR when the ROR is 13.5%?
> Given the formula for CE, wouldn't a smaller ROR/Kf
> result in a smaller CE ( less than 50% WR)? You seem
> to be implying the smaller the ROR/kf, the larger the
> CE.
The smaller the Kelly fraction, and, thus, the smaller the variance, the LARGER the CE. Nothing else would make sense. if you take practically no risk at all, why would you want to accept anything less than your expectation, with "certainty"?
> Perhaps the poster selected a small unit size b/c he
> was risk averse. Is CE only 50% of WR when the ROR is
> 13.5%?
CE is 50% of e.v. when you bet full Kelly optimally.
Don
Posting Permissions
-*-
Bookmarks