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Thread: actual result vs expected result

  1. #1


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    actual result vs expected result

    How do I figure out the SD if my actual result is -$3300 and my expected reult is +$7900?

  2. #2


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    Quote Originally Posted by roliin View Post
    How do I figure out the SD if my actual result is -$3300 and my expected reult is +$7900?
    You can't. You need more information.

    Don

  3. #3


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    Quote Originally Posted by DSchles View Post
    You can't. You need more information.

    Don
    15500 hands, %W/L 1.475%. Is that enough information?

  4. #4


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    Quote Originally Posted by roliin View Post
    15500 hands, %W/L 1.475%. Is that enough information?
    No. 1.475% of WHAT?? Ideally, at the very least, we need the average bet size, along with the win rate, and the per-hand s.d.

    Don

  5. #5


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    %W/L 1.475% is the %win/loss (EV or IBA) written on the bottom line of my version of CVCX.

    Average bet is $34.54, win per hour is $50.96 ($0.51 per round), s.d. per round is $72.44 ($724.40 per hr). ROR is 2%, N0 is 20,208.

  6. #6


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    The difference between +$7,900 and -$3,300 is $11,200. Total hours is EV/Hourly_EV = 155 hours. Your per hour SD is $724.40, so your SD over 155 hours is (I hope I'm doing this right) $9,018*. So, you are at -$11,200/$9,018 SDs = -1.24 SDs.


    * ($724.40^2 * 155)^0.5 = $9,018.
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

  7. #7


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    Quote Originally Posted by RS View Post
    The difference between +$7,900 and -$3,300 is $11,200. Total hours is EV/Hourly_EV = 155 hours. Your per hour SD is $724.40, so your SD over 155 hours is (I hope I'm doing this right) $9,018*. So, you are at -$11,200/$9,018 SDs = -1.24 SDs.


    * ($724.40^2 * 155)^0.5 = $9,018.

    Thank you. Is the formula to reach $9018 always the same, (hourly s.d.^2 * total hours)^0.5?

    Also, in CVCX under expected results:

    prob.

    66.7% -$832 to $16,630
    90% -$7002 to $22,800
    99.7% -$18,902 to $34,700

    So, any results that fall within the first row are within the mean, anything within the second row is 1 s.d. from the mean, anything withinin the 3rd row is 2 s.d, correct?

  8. #8


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    In a Normal distribution :

    68.2% falls within the mean +/- 1 s.d.
    95.4% falls within the mean +/- 2 s.d.
    99.7% falls within the mean +/- 3 s.d.

    Your results would technically fall within the mean +/- 2 s.d. range

  9. #9


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    Quote Originally Posted by RS View Post
    ($724.40^2 * 155)^0.5 = $9,018.
    Why would you square the $724 only to then take the square root? Just take the square root of the number of hours (155) and multiply that result by $724.

    Don

  10. #10


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    1.24 SDs happens with probability = 10.75%, so surely not a rare event!

    Don

  11. #11


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    Quote Originally Posted by DSchles View Post
    1.24 SDs happens with probability = 10.75%, so surely not a rare event!

    Don
    Care to share that math ( how to get to 10.75%)?

    If results fall within 2 SD 95% of the time, that's the same as saying a 2 SD event happens 5% of the time, correct? Have you experienced a 2 SD event in your blackjack life, CVCX or otherwise?

  12. #12


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    Quote Originally Posted by roliin View Post
    Care to share that math ( how to get to 10.75%)?
    You look it up in a normal distribution table. See, for example, BJA3, p. 147.

    Quote Originally Posted by roliin View Post
    If results fall within 2 SD 95% of the time, that's the same as saying a 2 SD event happens 5% of the time, correct?
    Two-sided, yes. Actually, 4.56%--2.28% for each side.

    Quote Originally Posted by roliin View Post
    Have you experienced a 2 SD event in your blackjack life, CVCX or otherwise?
    EVERYONE has!

    Don

  13. #13


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    roliin, why do you consider 5% to be rare? Being dealt a natural is only a ~1/21 = 4.8% event.

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