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Thread: the correct formula to compute the true count

  1. #1


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    the correct formula to compute the true count

    Hi, i found that actually i might have a wrong formula to compute the true count, thus I found that I am so hard to get a high true count.

    Currently by visiting the novice card counting site, it taught a theory of high-low index and they defined the computation of true count as

    true count = running count / numOfDeckOfGame

    I want to ask if it is the correct formula to compute the true count or there is other dependence that I have missed?

    thanks.

  2. #2


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    You divide by number of decks REMAINING. This is something you have to estimate (but after a little bit of practice and play, it's quite easy and quick). For the "decks remaining", it does NOT matter how many cards or decks are behind the cut card, FYI. Sometimes that throws people off for whatever reason.


    For example, you're in a 6 deck game, you're at an RC of +8 and 2 decks into the shoe. Since there are 6-2 decks remaining (4), the TC is 8/4 = +2.


    An excellent card counting guide can be found here: https://www.blackjackinfo.com/blackjack-school/ That's what I used when I first learned how to count cards, and I know several others have used it as well. If you're still serious about card counting, I recommend purchasing CVCX and CVBJ from qfit.com. The first is a simulator -- you enter the type of game you're playing, bankroll, game rules, what kinda spread you want, risk, etc. etc. etc....and it tells you proper bet ramping (when to bet what), your EV, SD, ROR, N0, etc. CVBJ is practice software with drills to speed up your counting and get you accurate, as well as a real-life BJ game (for fake money obviously) that lets you know if you've made an error, it keeps count along with you and you can check it periodically to make sure you have the correct running count, true count, etc.
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

  3. #3


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    To stir up the pot. How about getting the true count by multiplying. Multiply the running count by the following factors.
    .2---- 6 decks left
    .2---- 5 decks left
    .3---- 4 decks left
    .4---- 3 decks left
    .5---- 2 decks left
    .6---- 1.75 decks left
    .7---- 1.50 decks left
    .8---- 1.25 decks left
    1----- 1 decks left
    1.3---.75 decks left
    2-----.50 decks left

    Multiplying is easier than dividing. Suppose you are playing a 6 deck game with 2 decks left with a running count of 4.
    Multiply 4 x .5 = 2.0
    Last edited by Midwest Player; 04-23-2017 at 09:50 PM.

  4. #4
    Random number herder Norm's Avatar
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    Some people multiply, particularly in pitch games. In CV, use Custom TC Calc to support this.
    Last edited by Norm; 04-24-2017 at 06:48 AM.
    "I don't think outside the box; I think of what I can do with the box." - Henri Matisse

  5. #5
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    Quote Originally Posted by Midwest Player View Post
    "How about getting the true count by multiplying?"
    Division = Multiply by the reciprocal. That means ~
    When less than a deck remains unseen, multiplication
    is best. You take the fraction of cards unseen, turn it over
    and multiply. e.g. 1/3 = 3 ... 2/3 = 1.5 etc.

    Last edited by ZenMaster_Flash; 04-24-2017 at 07:17 AM.

  6. #6


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    Everyone's brain is wired a bit different. I'm constantly extrapolating, as in where am I at exactly between true 2 and 3 etc. The most fun exrptrapolations are with less than a deck.

    "Gee, I'm at RC 8, there's only a half deck left - wow, it's +16"

    The most practical extrapolation is After the first hand with RC 6 or 9 - am I even with the house, or do I have a clear, albeit small, advantage.

  7. #7


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    "Multiplying is easier than dividing. Suppose you are playing a 6 deck game with 2 decks left with a running count of 4.
    Multiply 4 x .5 = 2.0"

    Because, clearly, dividing 4 by 2 to get 2 is much too hard?! :-)

    Don

  8. #8


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    Quote Originally Posted by DSchles View Post
    "Multiplying is easier than dividing. Suppose you are playing a 6 deck game with 2 decks left with a running count of 4.
    Multiply 4 x .5 = 2.0"

    Because, clearly, dividing 4 by 2 to get 2 is much too hard?! :-)

    Don
    Some halves players (not me) double RC, then divide by 2 to get to a proper true count, so,

    (4x2) /(2x2) = 4/2 =2.

    There used to be old adage - there was no unemployment in the Soviet Union, however, there was a surplus of Salt.

    I used that exact expression while fighting g a traffic ticket 40 years or so ago. In an empty parking lot, I parked in a handicapped stall. I didn't see the sign as it was perpendicular to the stall on a ridiculously high pole, about 10 feet high. After testifying g to that fact, the prosecutor actually asked me, if, get this - Mr. Freightman, where the handicapped sign should have been, was there another sign placed directing you to the location of the handicapped sign. And lawyers are supposed to be intelligent.
    Last edited by Freightman; 04-24-2017 at 10:25 AM. Reason: Sorry Frank, on that last line - just some lawyers :)

  9. #9
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    Quote Originally Posted by Freightman View Post
    " ... double RC, then divide by 2."
    IF you double a number and then halve
    it you are left with the original number.

    n'est pas ?


  10. #10


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    Quote Originally Posted by ZenMaster_Flash View Post
    IF you double a number and then halve
    it you are left with the original number.

    n'est pas ?

    Flash, note the responded to quote, and the last paragraph of my post. I think I made it clear that doubling the values to divide by 2 to come to the same answer is essentially the same as mining salt.

  11. #11


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    "n'est pas ?"

    Make that n'est-ce pas! :-)

    Don


  12. #12
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    Since you all are discussing different methods of calculating TC I will put my method in the mix. An source of possible error I wanted to eliminate was thinking about numbers besides my 2 RC's causing me to lose the count. I don't think it was happening but I had to recognize the possibility. The solution I came up with is math free and very easy once you retrain your brain.

    The quick overview is you convert everything to RC. You think of the multiples of the number of decks remaining. These were engrained into everyone's memory in elementary school and are automatic for most people. If they are not for you, you simply need to learn the multiples of your remaining deck estimations that you use. There should be a handful of deck estimations used. Here is an example for a 6 deck game:
    Decks remaining: multiple list
    6: 0, 6, 12, 18, 24, 30, 36, 52, 48, 54, 60, 66...
    5: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55...
    4: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44...
    3: 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33...
    2.5: 0, 2.5, 5, 7.5, 10, 12.5, 15, 17.5, 20, 22.5, 25, 27.5, 30...
    2: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22...
    1.5: 0, 1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12, 13.5, 15, 16.5...
    1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11...
    0.75: 0, 0.75, 1.5, 2.25, 3, 3.75, 4.5, 5.25, 6, 6.75, 7.5, 8.25, 9, 9.75, 10.5, 11.25, 12...
    0.5: 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5...

    Almost every one of the above should be automatic since elementary school. If they aren't it takes little study effort to make them automatic. I just use multiplication by 4/3 for 3/4 of a deck unseen. Anyway you just know the multiple that is the same or just less than the RC for the multiplier being the TC. Like you have a RC of 16 with 2.5 decks left you know the multiple is 15 and that is 6 times 2.5 (I think of it as double the 5 multiple's 3 multiplier, 3*2. for the x.5 deck estimates I see the double the deck estimate list of multiples with midpoint hashes in my head. So I would see 0, *, 5, *, 10, *... which eliminates half the lists needed. I double the next lower multiplier and add one if a hash is the mark just below the RC). For playing decisions you find the multiple of decks unseen that corresponds to the index as a RC index for that deck estimate. Like index of +7 the RC index with 4 decks left is 28. The key is to eliminate thinking about multiplying and just know the answers without thinking the math problem. It is much faster and you only think about you RC the entire time.

    With practice you just know the right play or bet without any thought. No math. It actually becomes a graphics problem rather than a math problem. Since you have a small set of deck estimates the number of lists to have automatic graphical recall is quite small and we all should already be wired for this. Do you need to think that 35 with 4 decks unseen is 35/4 = 8.xxx or that 32, the next lower multiple of 4 from 35, is the 8th multiple of 4? You might have to divide if you think about the RC but if you think about the multiples of 4 you just know 32 is the eighth multiple of 4 since the appropriate part of the list is 32, 36 (the 8th and 9th multiples of 4). Do you really have to do any math to know that the bookend multiples are 32 and 36 and that 8 corresponds to 32? With practice you just know the TC without thinking about the calculation and just know the RC index for the current deck estimate without thinking about it. All you need to do is decide if you are at or above the current deck estimate RC index. No math there. All you think about the entire time is your count's RC. All calculations are eliminated so things are instantaneous as long as you know the deck estimate.

    So with basic elementary school multiples no math is needed you just know the right plays and TC.
    Last edited by Three; 04-25-2017 at 09:14 AM.

  13. #13


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    Quote Originally Posted by DSchles View Post
    "Multiplying is easier than dividing. Suppose you are playing a 6 deck game with 2 decks left with a running count of 4.
    Multiply 4 x .5 = 2.0"

    Because, clearly, dividing 4 by 2 to get 2 is much too hard?! :-)

    Don
    How about a double deck game with a running count of 4 and 1.25 decks remaining. What is easier, multiply 4 x .8 = 3.2 or divide 4 / 1.25?

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