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Thread: How do you calculate this?

  1. #1


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    How do you calculate this?

    How does one calculate the odds of losing x units after y amount of hands?

    Game is:
    payback 100.15%
    Variance 1.1

    My buddy lost $12,000 betting unit of $50 after 4500 hands. A very rare result. Want to know if this was an honest report from him or if there is a real chance, yet very rare, that it can happen.
    May the Variance be with you.

  2. #2


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    Quote Originally Posted by blueman View Post
    How does one calculate the odds of losing x units after y amount of hands?

    Game is:
    payback 100.15%
    Variance 1.1

    My buddy lost $12,000 betting unit of $50 after 4500 hands. A very rare result. Want to know if this was an honest report from him or if there is a real chance, yet very rare, that it can happen.
    So, he lost $12,000/$50 = 240 units. Betting 4,500 hands of $50, with a 0.15% edge, his e.v. was to have won 6.75 units, so his result was 246.75 units below expectation. You write that variance is 1.1, so I'll have to go with that. Hope you didn't mean that s.d. was 1.1. So s.d. is 1.049 units per round. After 4,500 rounds, s.d. is 70.37 units. Therefore, the loss is 246.75/70.37 = 3.50 s.d.s, which, while not theoretically impossible is, nonetheless, very improbable, happening about once every 4,444 attempts.

    You find the last number in a cumulative normal probability chart. And so, when you see a claim like that, you have two choices: 1) believe it and say that the person was extraordinarily unlucky, or 2) don't believe it and say he was exaggerating or didn't remember correctly. Your choice.

    Don

  3. #3


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    If this was video poker, maybe he didn't have the strategy down pat.

  4. #4


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    Quote Originally Posted by DSchles View Post
    So, he lost $12,000/$50 = 240 units. Betting 4,500 hands of $50, with a 0.15% edge, his e.v. was to have won 6.75 units, so his result was 246.75 units below expectation. You write that variance is 1.1, so I'll have to go with that. Hope you didn't mean that s.d. was 1.1. So s.d. is 1.049 units per round. After 4,500 rounds, s.d. is 70.37 units. Therefore, the loss is 246.75/70.37 = 3.50 s.d.s, which, while not theoretically impossible is, nonetheless, very improbable, happening about once every 4,444 attempts.

    You find the last number in a cumulative normal probability chart. And so, when you see a claim like that, you have two choices: 1) believe it and say that the person was extraordinarily unlucky, or 2) don't believe it and say he was exaggerating or didn't remember correctly. Your choice.

    Don
    How do you get the number 70.37 after 4500 hands?

    And how do you get 1 out of 4444?

    I'd like to know the formula to be able to calculate in the future

  5. #5


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    Quote Originally Posted by blueman View Post
    How do you get the number 70.37 after 4500 hands?

    And how do you get 1 out of 4444?

    I'd like to know the formula to be able to calculate in the future
    Ok nvm i figured it out, sqrt(hands × variance) = SD units

    How do you get the 1 in 4444?

  6. #6


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    Quote Originally Posted by blueman View Post
    Ok nvm i figured it out, sqrt(hands × variance) = SD units

    How do you get the 1 in 4444?
    Like this:

    Quote Originally Posted by DonS
    You find the last number in a cumulative normal probability chart.
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

  7. #7


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    Ok i now can understand the math. One thing i noticed:

    Probability of being within 1 SD after x amount of hands is 84% according to the chart.

    But then the probability on the bell curve is only 68% to be within 1 SD

    (And 2 SD on chart is 97.7% but 95% on bell curve)

    Why are these different?

  8. #8


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    Being within 1 SD is not the same as being down 1 SD or better.

    IE: One is "being within -$300 and +$315" while the other is "Being at -$300 [or better]".
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

  9. #9
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    Quote Originally Posted by blueman View Post
    How do you get the number 70.37 after 4500 hands?
    He showed the math. SD grows by square root of number of rounds played. SD is the square root of variance, SD = 1.0488/round. Multiply SD by the square root of 4500 rounds, which is 67.0820, and you get a SD of 70.3562 units at 4500 rounds. Multiply by $50/unit and you get a 4500 round SD as $3517.8116. Divide the $12000 lose by the latter and you get 3.41 SDs. Now either I made a mistake on the calculator or the discrepancy is from rounding error from Don not taking things to as many decimal places or the less likely but possible, Don made a mistake. My bet is on rounding error.
    Quote Originally Posted by blueman View Post
    Probability of being within 1 SD after x amount of hands is 84% according to the chart.

    But then the probability on the bell curve is only 68% to be within 1 SD

    (And 2 SD on chart is 97.7% but 95% on bell curve)

    Why are these different?
    84% is the probability of everything to the bell curve peak side of one SD . 68% is the probability of being within one SD of expectation.

  10. #10


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    "He showed the math. SD grows by square root of number of rounds played. SD is the square root of variance, SD = 1.0488/round. Multiply SD by the square root of 4500 rounds, which is 67.0820, and you get a SD of 70.3562 units at 4500 rounds. Multiply by $50/unit and you get a 4500 round SD as $3517.8116. Divide the $12000 lose by the latter and you get 3.41 SDs. Now either I made a mistake on the calculator or the discrepancy is from rounding error from Don not taking things to as many decimal places or the less likely but possible, Don made a mistake. My bet is on rounding error."

    Neither. YOU made a mistake. You didn't take into account that the game has a positive e.v. (0.15%) and that, therefore, the loss alone is not the numerator. The positive e.v. minus the loss (slightly larger number) is the numerator.

    Don

  11. #11
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    Quote Originally Posted by DSchles View Post
    Neither. YOU made a mistake. You didn't take into account that the game has a positive e.v. (0.15%) and that, therefore, the loss alone is not the numerator. The positive e.v. minus the loss (slightly larger number) is the numerator.
    Thank you Don. I figured you didn't make a mistake. It seemed like a large difference for rounding error but I didn't see my mistake, until now. But combining my post with what Don pointed out I missed illustrates how the problem should be worked.

  12. #12


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    Thanks for these replies, i learned a lot about SD and the graphs/charts and calculations.

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