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Thread: ROR question

  1. #14
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    Quote Originally Posted by ZeeBabar View Post
    Eliot, by the way, I read your book, found I am “Fred” so I am using the time away from playing to strengthen up my game. As 21forme and others mentioned, my first 5 years were variance was with me and I came out ahead but 2019 and your book says the approach is all wrong. Thanks.
    Nothing wrong with being "Fred." He has a good time, mostly.

    As for your variance keeping you in the black, Max Rubin told me that this was his favorite article at AP Heat and that it was required reading by anyone involved in game protection at Barona, "Getting Lucky at Blackjack"

    https://www.888casino.com/blog/black...y-at-blackjack
    Last edited by Eliot; 06-02-2020 at 11:33 AM.
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  2. #15
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    James, if you apply the simple "Gambler's Ruin" formula, where you exit at either bust (0) or winning 10x, you get a ROR of 13.4486%. So, I also agree with that result in your list.

    Use the following, with a = 0, x = 144, b = 1440, mean = 0.094595, var = 13.57889. This will show a probability of 0.865514 of exiting at the top value 1440, hence a ROR of 13.4486%.

    BM_01.jpg

    https://www.ime.usp.br/~fmachado/MAE...chModeling.pdf

    Beginning page 508, you will get the derivation of this result, and further explanation. It may be worthwhile to start skimming at the beginning of the chapter to see the requirement for a normal distribution. We use the Central Limit Theorem to be able to use this result as an approximation for a casino game. But, when that game is highly skewed and the process can exit quickly then this approximation is not very good.
    Last edited by Eliot; 06-02-2020 at 04:04 PM.
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  3. #16


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    Quote Originally Posted by Eliot View Post
    James, if you apply the simple "Gambler's Ruin" formula, where you exit at either bust (0) or winning 10x, you get a ROR of 13.4486%. So, I also agree with that result in your list.

    Use the following, with a = 0, x = 144, b = 1440, mean = 0.094595, var = 13.57889. This will show a probability of 0.865514 of exiting at the top value 1440, hence a ROR of 13.4486%.

    BM_01.jpg

    https://www.ime.usp.br/~fmachado/MAE...chModeling.pdf

    Beginning page 508, you will get the derivation of this result, and further explanation. It may be worthwhile to start skimming at the beginning of the chapter to see the requirement for a normal distribution. We use the Central Limit Theorem to be able to use this result as an approximation for a casino game. But, when that game is highly skewed and the process can exit quickly then this approximation is not very good.
    Eliot, Thanks for your explanation.

    Another question arise, when player use 1 to 20 betting ramp in blackjack, say the ev > 0.02 and and Variance > 5, how to find more accurate ROR ?

    a) Just use new ev =0.02 and Var = 5 and key into existing ROR formula will do ? where ROR = e^(-BR * ev/Var), BR = bankroll.

    b) or is it better to set up a proper model and find ROR through simulation ?

    James
    Last edited by James989; 06-02-2020 at 06:51 PM.

  4. #17


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    Quote Originally Posted by Eliot View Post
    My quick and dirty simulation agreed with your results and gave a ROR of 12.05% m/l. I chose a different stopping point to see if I could get substantially different results than yours. I also double checked your EV/SD and agree with your analysis.

    I have encountered this sort of theory v. simulation discrepancy before, when analyzing loss rebates. The more that the higher moments in the distribution diverge from normal, the more theory diverges from simulations. Blackjack has a very tame distribution compared to this game.

    Here is my code:

    Attachment 4147

    I edited this code and ran 10M trips and got:

    10000000, 0.120483

    I think you will get same answer even for n = 13660(instead of 10M), could you please check this ? I choose n=13660 with certain reason.

  5. #18
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    Quote Originally Posted by James989 View Post
    I think you will get same answer even for n = 13660(instead of 10M), could you please check this ? I choose n=13660 with certain reason.
    Is it something to do with the periodicity of rand()? I can re-write with drand48 or a more modern RNG and run it if you wish. I doubt the result will be much different. In the mean time, here are the requested results:

    $ ./a.exe
    13660, 0.124671
    1000000, 0.120268
    2000000, 0.120455
    3000000, 0.120313
    4000000, 0.120312
    5000000, 0.120361
    6000000, 0.120417
    7000000, 0.120509
    8000000, 0.120519
    9000000, 0.120577
    10000000, 0.120642
    Last edited by Eliot; 06-02-2020 at 08:21 PM.
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  6. #19


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    Quote Originally Posted by Eliot View Post
    Is it the periodicity of rand()? No problem to re-write it with drand or any other RNG.
    No. It is related to "LONG TERM" rounds . . . Normally I use 3 Standard Deviation as "long term" play, it give accurate results most of the time, but this assumption may be wrong, I am not sure.

    n = 3^2 * Var/ev/ev = 13660

    What is drand ?

    Any comments for POST #16 ?
    Last edited by James989; 06-02-2020 at 08:13 PM.

  7. #20
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    Quote Originally Posted by James989 View Post
    Any comments for POST #16 ?
    In my opinion, choice (b). A simulation will give a more accurate estimate of the ROR for the situation you described.

    Surely there is commercial software that does this. Hint. Hint.

    By the way, regarding this formula: ROR = e^(-BR * ev/Var). What happens in the gambler's ruin formula above if you set a = 0, x = BR and let b get large?
    Last edited by Eliot; 06-02-2020 at 09:47 PM.
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  8. #21


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    Quote Originally Posted by Eliot View Post
    Is it something to do with the periodicity of rand()? I can re-write with drand48 or a more modern RNG and run it if you wish. I doubt the result will be much different. In the mean time, here are the requested results:

    $ ./a.exe
    13660, 0.124671
    1000000, 0.120268
    2000000, 0.120455
    3000000, 0.120313
    4000000, 0.120312
    5000000, 0.120361
    6000000, 0.120417
    7000000, 0.120509
    8000000, 0.120519
    9000000, 0.120577
    10000000, 0.120642
    I get the same results ! It's a bit weird, why ROR for n=13660 more than ROR of 10M ? I thought more rounds will give results closer to one Kelly ROR = 13.53% ? No ?

  9. #22
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    Quote Originally Posted by James989 View Post
    I get the same results ! It's a bit weird, why ROR for n=13660 more than ROR of 10M ? I thought more rounds will give results closer to one Kelly ROR = 13.53% ? No ?
    The left-hand column is the total number of trials, the value "tripno" in your code. It has nothing to do with the number of hands played in any given trip.

    Please study my code
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  10. #23


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    Quote Originally Posted by Eliot View Post
    The left-hand column is the total number of trials, the value "tripno" in your code. It has nothing to do with the number of hands played in any given trip.

    Please study my code

    13660 is rounds played(n) in a trip, NOT the tripno. I always set tripno = 500,000. Sorry for the confusion.

    n = rounds/hands played in a trip.

    My simulation results :-
    1) n= 2,000, tripno = 500000, ROR = 8.91
    2) n= 5,000, tripno = 500000, ROR = 11.56
    3) n= 13660, tripno = 500000, ROR = 12.044
    4) n= 1000,000, tripno = 500000, ROR = 12.010
    Last edited by James989; 06-02-2020 at 10:45 PM.

  11. #24
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    The correct formula for kelly betting on a bet like this with two possible results is:

    bet size = bankroll * edge / payoff_odds

    The approximation for games with more complicated payoffs is bet size = bankroll * edge / var, but you've got a binary game, so the first formula is more exact. Rearranging the equation, the kelly bankroll for your game is about 132 bets. Your 144 bet bankroll is slightly oversized, so it's not surprising that your risk numbers are down in the 12% range. Though as Eliot mentioned, you shouldn't expect exactly 13.53% on a high edge game like this where the correct bet is a decent chunk of your bankroll and the payoff distribution is skewed. In the trials where you lose your money, you probably aren't getting enough plays for a normal distribution to be a good approximation of your range of outcomes.

    If you've got some choice in bet sizing, bet approximately (bankroll * kelly_fraction / 132), resizing frequently, and don't worry about the ROR calculations for fixed betting. If there is any uncertainty regarding some of those 5 balls occasionally getting mixed in, adjust the calculations accordingly. If you're stuck with a fixed available bet size, or you're up against a table minimum or maximum, that's where the simulations with fixed betting are more useful.
    Last edited by Nyne; 06-07-2020 at 09:01 PM.

  12. #25


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    Quote Originally Posted by Nyne View Post
    The correct formula for kelly betting on a bet like this with two possible results is:

    bet size = bankroll * edge / payoff_odds

    The approximation for games with more complicated payoffs is bet size = bankroll * edge / var, but you've got a binary game, so the first formula is more exact. Rearranging the equation, the kelly bankroll for your game is about 132 bets. Your 144 bet bankroll is slightly oversized, so it's not surprising that your risk numbers are down in the 12% range. Though as Eliot mentioned, you shouldn't expect exactly 13.53% on a high edge game like this where the correct bet is a decent chunk of your bankroll and the payoff distribution is skewed. In the trials where you lose your money, you probably aren't getting enough plays for a normal distribution to be a good approximation of your range of outcomes.

    If you've got some choice in bet sizing, bet approximately (bankroll * kelly_fraction / 132), resizing frequently, and don't worry about the ROR calculations for fixed betting. If there is any uncertainty regarding some of those 5 balls occasionally getting mixed in, adjust the calculations accordingly. If you're stuck with a fixed available bet size, or you're up against a table minimum or maximum, that's where the simulations with fixed betting are more useful.

    I thought the formula should be, bet size = kelly ratio * bankroll * edge / variance ? where kelly ratio = 1.0

  13. #26


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    Welcome back, Nyne. Been a long time!

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