Variance per strategy decision:

[AnzaExo]

Not an error. The discrepancy undoubtedly comes from whether or not the dealer has already checked his ace for a natural or not. The program that shows -.609 is allowing the dealer to have a natural; BJA3 assumes the dealer has already checked and has a playable ace.

You shouldn't find any discrepancies in the program for dealer upcards 2-9. Do you?

Don

you are absolutely right. i don't think there is a way to adjust for the A or 10 dealer card, or is there?

[DSchles]

you are absolutely right. i don't think there is a way to adjust for the A or 10 dealer card, or is there?

Yes, sure there is. Simple algebra. To get the -.609222 value, realize that it comprises two parts: the certain loss of the bet 96/309 (31.068%) of the time, when the dealer has a ten in the hole, and some other negative e.v. (which we're looking for, and that we'll call x) the remainder of the time, or 68.932%. Together, they yield the -.609222 edge that the program gives. We're trying to get the -.433 BJA3 edge. So: -1(.31068) + x(.68932) = -0.609222. Whence x = -0.433. Q.E.D.!

Don

[Norm]

I love it when a plan comes together.

[AnzaExo]

Yes, sure there is. Simple algebra. To get the -.609222 value, realize that it comprises two parts: the certain loss of the bet 96/309 (31.068%) of the time, when the dealer has a ten in the hole, and some other negative e.v. (which we're looking for, and that we'll call x) the remainder of the time, or 68.932%. Together, they yield the -.609222 edge that the program gives. We're trying to get the -.433 BJA3 edge. So: -1(.31068) + x(.68932) = -0.609222. Whence x = -0.433. Q.E.D.!

Don

How would I convert the standard deviation data though? I think it's being over-stated due to the program's treatment of a possible dealer bj.

[DSchles]

Don't have the time now, but the calculation would be similar, albeit you would need to convert all the s.d.s to variances (square them), do the math in a similar, linear way, and then convert the resulting variance back to a s.d.

Have we ever asked you what the purpose was of this exercise? What are you going to use the s.d.s of the various plays for?

Don

[AnzaExo]

Don't have the time now, but the calculation would be similar, albeit you would need to convert all the s.d.s to variances (square them), do the math in a similar, linear way, and then convert the resulting variance back to a s.d.

Have we ever asked you what the purpose was of this exercise? What are you going to use the s.d.s of the various plays for?

Don

I'm working on a small idea of mine, similar in nature to ch.13 part 2 of your book and i'd appreciate your critique. I'm comparing the EV / SD for certain plays, namely through double and split decisions that are recommended by basic strategy. I am trying to find decisions that have excessive risk relative to their reward; A-4 vs 4 comes to mind. Given 6D H17, my expectation is .065278 for doubling, and .060757 for hitting. The standard deviations are 1.952826 and .975437 respectively, effectively doubling my risk to only obtain an extra .00452 of EV. The skewed risk/reward payoff becomes even more unattractive in my view with more than a minimum bet placed. I find there are numerous plays that are given by basic strategy that ignore the additional variance associated, and solely maximize EV - no matter the margin.

I am thinking that I would rather fully utilize advantageous counts than put on additional risk for extremely slim EV opportunities like the example listed above.

I should disclaim that I study the game more than I actually play, and am trying to build a strategy for when I do have more of an adequate bankroll.

[AnzaExo]

I had initially used the combinatorial program to check the EV / SD for plays like 16 v 7 to see if I could get a different result than BS, but did not. Mainly it seems the only time this idea comes into play is for split/double decisions.

[iCountNTrack]

Thanks!

Unless I made a mistake -- quite possible -- the source code for this has disappeared from the repository. Or perhaps it was never made available.

Does anyone (e.g., iCountNTrack) know if the source code is available?

-TX

Never released the source code because it was simply too complicated to make much sense of it. I have cleaned it since then though, but never had the time to post it again. I will try to find it

[iCountNTrack]

I had initially used the combinatorial program to check the EV / SD for plays like 16 v 7 to see if I could get a different result than BS, but did not. Mainly it seems the only time this idea comes into play is for split/double decisions.

you will only see an effect if your hand warrants forfeiting a chance for getting another card to improve your hand, for example suppose you have A2 vs 5 you get the following:



The DI for hitting is higher but the EV for doubling is higher. For 16 vs 7 getting another card is not advantageous (it is actually suicidal) that is why there is no difference.

[DSchles]

Be careful here. BS rarely, if ever, gives us a positive e.v. And, ironically, the notion of risk-aversion, or the calculation of a SCORE, really doesn't apply when you have no edge.

You are probably familiar with the notion of "maximum boldness," which applies to negative-e.v. wagers. If you have, say, 100 units, playing roulette, and you would like to try to double that amount, playing a terribly negative (5.26%) expectation game, common sense tells you that your greatest chance to double your bank is to bet the entire 100 units on one spin, on an even-money wager, where you have an 18/38 = 47.37% chance to succeed. If you bet smaller amounts, trying to grind out a profit, you will surely eventually go broke.

In any event, unless I am missing something, this is why the notion of a risk-averse BS really doesn't exist. Since your overall e.v. is negative, you will always bet one unit on every hand and, to maximize the overall ratio of e.v./s.d., across all hands, we want to maximize not only e.v., but ironically, we also want to maximize s.d. (or, alternatively, variance), since the value of a negative fraction is increased if the denominator is made larger.

Refusing to double A,2 v. 5 will maximize the e.v./s.d. for that lone play, but doubling will contribute more to raising the overall s.d. across all plays, which, in turn, will have more effect on lowering (making less negative) the e.v./s.d. fraction for the entire endeavor of playing all hands than would hitting.

Does this make sense to you?

Don

[iCountNTrack]

Be careful here. BS rarely, if ever, gives us a positive e.v. And, ironically, the notion of risk-aversion, or the calculation of a SCORE, really doesn't apply when you have no edge.

You are probably familiar with the notion of "maximum boldness," which applies to negative-e.v. wagers. If you have, say, 100 units, playing roulette, and you would like to try to double that amount, playing a terribly negative (5.26%) expectation game, common sense tells you that your greatest chance to double your bank is to bet the entire 100 units on one spin, on an even-money wager, where you have an 18/38 = 47.37% chance to succeed. If you bet smaller amounts, trying to grind out a profit, you will surely eventually go broke.

In any event, unless I am missing something, this is why the notion of a risk-averse BS really doesn't exist. Since your overall e.v. is negative, you will always bet one unit on every hand and, to maximize the overall ratio of e.v./s.d., across all hands, we want to maximize not only e.v., but ironically, we also want to maximize s.d. (or, alternatively, variance), since the value of a negative fraction is increased if the denominator is made larger.

Refusing to double A,2 v. 5 will maximize the e.v./s.d. for that lone play, but doubling will contribute more to raising the overall s.d. across all plays, which, in turn, will have more effect on lowering (making less negative) the e.v./s.d. fraction for the entire endeavor of playing all hands than would hitting.

Does this make sense to you?

Don

It appears we are referring to two different things, you are talking about overall EV (pre-deal EV) while I am talking about a playing strategy given a player's hand and the dealer upcard. I do agree with you that for a risk-averse BS for a -EV game is an ill-defined notion. This is geared towards high EV game such as Ace sequencing. You knew you were landing an ace as your first card, unfortunately you ended with A,2 vs dealer's 5. You have a 50 unit bet, you probably want a risk averse play which in this case would be to hit instead of doubling.

When I first started developing my CA, i really wanted to compute the probabilities of outcomes (net win on a round) as there were many other CA that would give the EV, and then I was like hmmm given the probabilities SD can be calculated as well

[AnzaExo]

Be careful here. BS rarely, if ever, gives us a positive e.v. And, ironically, the notion of risk-aversion, or the calculation of a SCORE, really doesn't apply when you have no edge.

You are probably familiar with the notion of "maximum boldness," which applies to negative-e.v. wagers. If you have, say, 100 units, playing roulette, and you would like to try to double that amount, playing a terribly negative (5.26%) expectation game, common sense tells you that your greatest chance to double your bank is to bet the entire 100 units on one spin, on an even-money wager, where you have an 18/38 = 47.37% chance to succeed. If you bet smaller amounts, trying to grind out a profit, you will surely eventually go broke.

In any event, unless I am missing something, this is why the notion of a risk-averse BS really doesn't exist. Since your overall e.v. is negative, you will always bet one unit on every hand and, to maximize the overall ratio of e.v./s.d., across all hands, we want to maximize not only e.v., but ironically, we also want to maximize s.d. (or, alternatively, variance), since the value of a negative fraction is increased if the denominator is made larger.

Refusing to double A,2 v. 5 will maximize the e.v./s.d. for that lone play, but doubling will contribute more to raising the overall s.d. across all plays, which, in turn, will have more effect on lowering (making less negative) the e.v./s.d. fraction for the entire endeavor of playing all hands than would hitting.

Does this make sense to you?

Don

I now see that risk averse BS is a bit of a contradiction. Looking at the game from purely a BS standpoint, for some reason I didn't realize the denominator effect given the negative fraction....

Thank you for explaining this to me!

[SiMi]

Hi, iCountNTrack!!

The CA is an EXCELLENT piece of work, IMHO! I would love to see the source code in an effort to understand the combinatorial math.

Thanks!
SiMi