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Thread: Average Standard Deviation

  1. #1


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    Average Standard Deviation

    Assume 40 hands of a 6 deck, Strip rules shoe.

    If for 37 of the 40 hands I have a -$1.85 E.V. and a 2 standard deviation expected outcome of about (+$140 to -$136) due to flat betting $10 at a -.5% player advantage. (By the way, Don, thank you - page 16 of my "Blackjack Attack" is well worn due to me referencing it so much.),

    And, for the remaining 3 hands (hand #4, hand # 22, hand #26), assume I have a 5% player advantage (not as a result of card counting), and I bet $20 on each of the 3 hands (EV of $3.00, or $60 x .05), yielding a 2 standard deviation range of about (+$72 to -$66).

    To arrive at my overall expected 2 standard deviation range, can I combine (add) these two sets of 2 standard ranges and get (+$212 to -$202)? Or, do I need to compute an Average Advantage, as Don talks about in "Blackjack Attack" (3rd edition)? Or will both of these methods yield the same answer?

    Thank you in advance!!
    Last edited by Overkill; 02-26-2021 at 10:39 PM.

  2. #2


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    Quote Originally Posted by Overkill View Post
    Assume 40 hands of a 6 deck, Strip rules shoe.

    If for 37 of the 40 hands I have a -$1.85 E.V. and a 2 standard deviation expected outcome of about (+$140 to -$136) due to flat betting $10 at a -.5% player advantage. (By the way, Don, thank you - page 16 of my "Blackjack Attack" is well worn due to me referencing it so much.),
    So, two s.d.s are, in fact, $140. But, you have the range backwards, because your e.v. is negative (-$1.85). It's to that value that you have to add and subtract $140. So, you get, for -2 s.d.s, -$141.85, and for +2 s.d.s, +$138.15. Clear?

    Quote Originally Posted by Overkill View Post
    And, for the remaining 3 hands (hand #4, hand # 22, hand #26), assume I have a 5% player advantage (not as a result of card counting), and I bet $20 on each of the 3 hands (EV of $3.00, or $60 x .05), yielding a 2 standard deviation range of about (+$72 to -$66).
    So, this might depend on how you're getting that 5% edge. Because, we assume, for flat betting, s.d. of 1.15 times the bet. But, if you're locating aces and getting a disproportionate amount of blackjacks, you may have a 5% edge, but it will come at the expense of greater standard deviation than normal for those three hands. Do you understand? For now, we'll keep it at 1.15, but that may not be accurate for what you're doing.

    I get a different s.d. range from you this time. Sqrt(3) = 1.732. Multiply by $20 times three hands, and I get $39.84. Let's call it $40. So, 2 s.d.s = $80. And, given the +$3.00 e.v. for those hands, the 2-s.d. range would be: -$77 to +$83.

    Quote Originally Posted by Overkill View Post
    To arrive at my overall expected 2 standard deviation range, can I combine (add) these two sets of 2 standard ranges and get (+$212 to -$202)?
    No.

    Quote Originally Posted by Overkill View Post
    Or, do I need to compute an Average Advantage, as Don talks about in "Blackjack Attack" (3rd edition)?
    You can average (or, in this case, just add) the edges (see below), but you can't average standard deviations. You have to use variances, because s.d.s are a square-root function.

    Quote Originally Posted by Overkill View Post
    Or will both of these methods yield the same answer?
    The right way to do it is as follows: First the easy part. The global e.v. for the entire 40 hands is, obviously, (-$1.85) + (+$3.00) = +$1.15. You don't have to average anything. The variance, which is the square of one s.d., is ($70)^2 = $4,900 for the first 37 hands, and ($40)^2 = $1,600 for the other three hands. We now have have to do what is called a root-mean-square calculation to add them, giving each part the proper weighting, according to number of hands. So, the $4,900 gets multiplied by 37, the $1,600 gets multiplied by 3, the two products are added, we divide by 40, to get the average variance for all the hands, and then we take the square root of the result to get the global s.d. for all 40 hands.

    (37 x $4,900) + (3 x $1,600) = $181,300 + $4,800 = $186,100. $186,100/40 = $4,653. Sqrt($4,653) = $68. QED! Whew. Hope I didn't screw up anything. Don't think so.

    So, e.v. for the whole play is +$1.15 and two s.d.s = $136. So, the 2-s.d. range is -$134.85 to $137.85.

    Clear?

    Don

  3. #3


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    Hi, Don,

    Thanks so much for the detailed reply!

    1) Yes, sorry, negative $1.85, not +$1.85.

    2) Above you use 1.15 for s.d., but in your book you use 1.1. Please explain.

    3) If I am getting a higher number of blackjacks, that is (obviously) good. How can something so GOOD reflect POORLY on my standard deviation? (By the way, by not playing alone while sitting at first base, I am hoping to reduce the chance of the dealer getting 'my' Ace,' which should 'preserve' my edge of about 5%. Is this sound reasoning? I need to ask such a question because the .51 edge that I used in the calculation to arrive at my 5% edge has not taken into account that I may NOT receive the Ace that I claim I am certain of receiving by using .51, correct?)

    4) How do I go about calculating a more appropriate s.d. for the 3 hands, or is that question 'outside the scope' of this thread?

    5) Above, where you indicate that you arrived at a different s.d. than I, did you forget to multiply the square root of 3 by 1.1?

    6) That root-mean-square calculation is neat! I suspected something was amiss with my calculation due to not taking into account proper weighting.

    7) Ok, yes, that's interesting about how one cannot average standard deviation. In the above, you added the E.V.s together. But could we also have used ONE E.V. (not 2 initially) as a result of using one edge ("Average Advantage"?) that results from appropriate weighting and averaging of the two edges (1 edge for the 37 hands and 1 edge for the 3 hands)? But if we do that, won't that low edge not accurately represent this shoe containing a very high on 3 hands of its hands? And how does one determine bet size for each hand with the use of an average (and not a specific-to-the-hand) advantage?

    8) Using the numbers you arrived at, how can I, without computer software, calculate the ideal bet size given a specified Risk of Ruin and bankroll? Can I algebraically manipulate one of the Risk of Ruin formulas in Chapter 8 of your book "Blackjack Attack" (3rd ed.)?

    Ok, I am just going to post this reply without a 16th proofread!! I already 'lost' my draft twice while editing. Thankfully, after losing the first draft, I began copying.and pasting what I had written to the clipboard.

  4. #4


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    Quote Originally Posted by Overkill View Post
    Hi, Don,

    Thanks so much for the detailed reply!

    1) Yes, sorry, negative $1.85, not +$1.85.

    2) Above you use 1.15 for s.d., but in your book you use 1.1. Please explain.
    You haven't read very far in the book, have you. Depending on rules, one-hand s.d. can be as low as 1.12 (no DAS or resplitting of pairs). I used 1.1 just to make the math easy in the beginning of the book. All the chapter 10 charts have s.d.s, by the true count for every conceivable game. Use those. Don't use anything from the very early chapters.

    Quote Originally Posted by Overkill View Post
    3) If I am getting a higher number of blackjacks, that is (obviously) good. How can something so GOOD reflect POORLY on my standard deviation? (By the way, by not playing alone while sitting at first base, I am hoping to reduce the chance of the dealer getting 'my' Ace,' which should 'preserve' my edge of about 5%. Is this sound reasoning? I need to ask such a question because the .51 edge that I used in the calculation to arrive at my 5% edge has not taken into account that I may NOT receive the Ace that I claim I am certain of receiving by using .51, correct?)
    Getting e.v. from blackjacks doesn't excuse you from the increased s.d. that comes along with them. Variance is, basically, the average squared result of a hand. If you square the global s.d. of 1.15, you get 1.32. Meanwhile, what's the average squared result of getting a 3:2 payoff? Obviously, it's 1.5^2 = 2.25. Hence the extra variance when your e.v. is derived from naturals. Variance isn't good or bad. It's just variance!

    Can't help with the uncertainty of getting the ace. That's up to you to determine. I don't know what you're doing, so I can't tell you how likely or not you are to get an ace. But I can tell you with certainty that your overall variance, and hence s.d., is larger than what I calculated, as I warned you it might be.

    Quote Originally Posted by Overkill View Post
    4) How do I go about calculating a more appropriate s.d. for the 3 hands, or is that question 'outside the scope' of this thread?
    Yes, it's somewhat outside. You may have read the recent reference to James Grosjean's 42.08% article. It refers to the optimal percent of one's bank to wager given one card in the hand will be an ace. But it requires playing a different BS than normal, to be more risk-averse, given such a large wager.

    Quote Originally Posted by Overkill View Post
    5) Above, where you indicate that you arrived at a different s.d. than I, did you forget to multiply the square root of 3 by 1.1?
    Actually, no, I didn't; I just forgot to write that I did! Sorry about that. But the answer is right.

    Quote Originally Posted by Overkill View Post
    7) OK, yes, that's interesting about how one cannot average standard deviation. In the above, you added the E.V.s together. But could we also have used ONE E.V. (not 2 initially) as a result of using one edge ("Average Advantage"?) that results from appropriate weighting and averaging of the two edges (1 edge for the 37 hands and 1 edge for the 3 hands)? But if we do that, won't that low edge not accurately represent this shoe containing a very high on 3 hands of its hands? And how does one determine bet size for each hand with the use of an average (and not a specific-to-the-hand) advantage?
    I don't see the point in one e.v. But, you have it. You play 40 hands and you win $1.15. So the per-hand global edge is $1.15/40 = $0.02875. But, it isn't really pertinent or useful for anything.

    Quote Originally Posted by Overkill View Post
    8) Using the numbers you arrived at, how can I, without computer software, calculate the ideal bet size given a specified Risk of Ruin and bankroll? Can I algebraically manipulate one of the Risk of Ruin formulas in Chapter 8 of your book "Blackjack Attack" (3rd ed.)?
    You can use Norm's calculators, at the top, all based on my formulas. But there's no point. You really shouldn't be betting $10 on all the losing hands and then $20 for the aces. That makes no sense at all. You should be betting table minimum for the bad hands and whatever you can afford and get away with for the good ones. You have -0.5% edge on 37 hands and 5% edge (not sure where that comes from) for the other three, and all you want to bet is twice as much? Not very logical.

    Don

  5. #5
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    Quote Originally Posted by DSchles View Post
    You have -0.5% edge on 37 hands and 5% edge (not sure where that comes from) for the other three, and all you want to bet is twice as much? Not very logical.

    Don
    It's hilarious how people overcomplicate things that are shocking simple. At least we have Don to point out the obvious.

    Dalmatian

  6. #6


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    What is the difference between the strip rules and the station rules? My impression is that all station casinos often do not allow double down after splitting. Is this correct?

  7. #7


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    Quote Originally Posted by aceside View Post
    What is the difference between the strip rules and the station rules? My impression is that all station casinos often do not allow double down after splitting. Is this correct?

    I know a guy that will be there this summer. I’ll let you know what he finds out.

  8. #8


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    Quote Originally Posted by aceside View Post
    What is the difference between the strip rules and the station rules? My impression is that all station casinos often do not allow double down after splitting. Is this correct?
    The term "Strip rules" refers to a specific set of rules that used to be the standard for Blackjack games on Vegas Strip. They are:
    S17, DAS and BJ pays 3:2
    G Man

  9. #9


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    Quote Originally Posted by G Man View Post
    The term "Strip rules" refers to a specific set of rules that used to be the standard for Blackjack games on Vegas Strip. They are:
    S17, DAS and BJ pays 3:2
    But the strip rules usually means the more liberal rules, and the station rules means the more conservative rules. Conservatives often do not allow double down after splitting and surrender.

  10. #10


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    Quote Originally Posted by G Man View Post
    The term "Strip rules" refers to a specific set of rules that used to be the standard for Blackjack games on Vegas Strip. They are:
    S17, DAS and BJ pays 3:2
    Change " They are" to "They were."

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