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Thread: Those not so neutral 789s at a special RC 0

  1. #1


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    Those not so neutral 789s at a special RC 0

    Hello all and especially the math heads as well as those who pay attention to the 789s, the so called “neutral” cards.

    Imagine a 6D shoe at the 144 cards mark (12 cards past the 156 cards half point)
    The HiLo count is RC 0 with 12 aces left, 48 tens and 60 low cards. The 12 “neutral” cards missing are four sevens, four eights and four nines. So there are 24 x 789s left in the 144-card shoe.

    It’s time to bet and the RC 0 suggests a minimal bet BUT…

    Would you consider a 2-unit bet (or more)?

    I would like feedbacks from the experts.
    Personnaly I think this subset justifies a 2-unit bet because…

    Despite the RC 0, the Ten ratio is at 0,333 (48/144) right at the insurance threshold.
    The blackjack odds are up at 0,56 instead of 0,48 so that’s one blackjack every 17 hands instead of one every 21 hands.
    The odds to get a hard 20 (TT) are up at 0,112 instead of 0,095.
    The odds of a two-card stiff (dealer’s and ours) involving a Ten are 0,140 instead of 0,119. Since the dealer has to hit to 17 (and we don’t) that’s a player advantage.
    One downside is that double down opportunities will be fewer but not sure how much.

    There are other implications that the math experts like Don could mention.

    As a final note, the CV deck edit function suggests and improvement of 0,43 over a perfectly balanced count which seems to justify a 2-unit bet.

    Your thoughts, anyone!

  2. #2


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    So your doing an Ace side count AND a Neutral side count?

  3. #3


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    Quote Originally Posted by tunicadave View Post
    So your doing an Ace side count AND a Neutral side count?
    Thats what he’s doing. I will eventually give a more detailed response, but I’m really kinda busy with a lot of this almost retired stuff.

  4. #4


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    Quote Originally Posted by Secretariat View Post
    The 12 “neutral” cards missing are four sevens, four eights and four nines. So there are 24 x 789s left in the 144-card shoe.
    What am I missing? A six-deck shoe has 24 7s, 24 8s, and 24 9s. You wrote, "The 12 'neutral' cards missing are four sevens, four eights and four nines." My higher math tells me that that leaves 60 7s, 60 8s, and 60 9s. And yet, you wrote, "So there are 24 x 789s left in the 144-card shoe." What sense does that make??

    Don

  5. #5


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    Quote Originally Posted by Secretariat View Post
    Hello all and especially the math heads as well as those who pay attention to the 789s, the so called “neutral” cards.

    Imagine a 6D shoe at the 144 cards mark (12 cards past the 156 cards half point)
    The HiLo count is RC 0 with 12 aces left, 48 tens and 60 low cards. The 12 “neutral” cards missing are four sevens, four eights and four nines. So there are 24 x 789s left in the 144-card shoe.

    It’s time to bet and the RC 0 suggests a minimal bet BUT…

    Would you consider a 2-unit bet (or more)?

    I would like feedbacks from the experts.
    Personnaly I think this subset justifies a 2-unit bet because…

    Despite the RC 0, the Ten ratio is at 0,333 (48/144) right at the insurance threshold.
    The blackjack odds are up at 0,56 instead of 0,48 so that’s one blackjack every 17 hands instead of one every 21 hands.
    The odds to get a hard 20 (TT) are up at 0,112 instead of 0,095.
    The odds of a two-card stiff (dealer’s and ours) involving a Ten are 0,140 instead of 0,119. Since the dealer has to hit to 17 (and we don’t) that’s a player advantage.
    One downside is that double down opportunities will be fewer but not sure how much.

    There are other implications that the math experts like Don could mention.

    As a final note, the CV deck edit function suggests and improvement of 0,43 over a perfectly balanced count which seems to justify a 2-unit bet.

    Your thoughts, anyone!
    I have been using three counts for a very long time. I use Zen, ace side count and 8/9 side count, so I have covered all ranks of cards. This is similar to Tarzan Count. For people who are not familiar with Tarzan Count: In short, basic version of Tarzan count is three columns (2-5, 6-9, 10's) with ASC.

  6. #6


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    Quote Originally Posted by DSchles View Post
    What am I missing? A six-deck shoe has 24 7s, 24 8s, and 24 9s. You wrote, "The 12 'neutral' cards missing are four sevens, four eights and four nines." My higher math tells me that that leaves 60 7s, 60 8s, and 60 9s. And yet, you wrote, "So there are 24 x 789s left in the 144-card shoe." What sense does that make??

    Don
    12 aces plus 48 face cards plus 60 low cards plus 24 neutral cards equals the 144 cards OP stated. His further statement of 12 neutral cards was simply a brain fart. He clearly meant to say 24.

    Further, it’s pretty tough to identify 180 remaining cards, all neutral as you’ve stated when OP has mentioned 144 in categories listed identifying low neutral and high cards.

  7. #7


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    Quote Originally Posted by DSchles View Post
    What am I missing? A six-deck shoe has 24 7s, 24 8s, and 24 9s. You wrote, "The 12 'neutral' cards missing are four sevens, four eights and four nines." My higher math tells me that that leaves 60 7s, 60 8s, and 60 9s. And yet, you wrote, "So there are 24 x 789s left in the 144-card shoe." What sense does that make??

    Don
    Half of the shoe is gone, Don. So let’s say there are three decks left which should be 156 cards
    At the 156 cards mark there should be 12 aces, 48 tens, 60 low cards and 36 neutral/intermediate cards for a perfectly balanced shoe.
    However, assume a shortage of 12 x 789s (4x7, 4x8, 4x9) at the 144-card mark.
    That’s 12 neutral cards below normal at slightly below the 3-deck point. That’s what I meant.
    Do you think that eliminates the house edge at 6D, S17,DAS?
    If you had that information at the table would you bet 2 units?

  8. #8


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    Quote Originally Posted by tunicadave View Post
    So your doing an Ace side count AND a Neutral side count?
    Basically yes!

  9. #9


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    Quote Originally Posted by Freightman View Post
    12 aces plus 48 face cards plus 60 low cards plus 24 neutral cards equals the 144 cards OP stated. His further statement of 12 neutral cards was simply a brain fart. He clearly meant to say 24.

    Further, it’s pretty tough to identify 180 remaining cards, all neutral as you’ve stated when OP has mentioned 144 in categories listed identifying low neutral and high cards.
    I had Freightman's dual ramping in mind when asking this. Would the OP statement be a starting point for the higher ramp, Freightman?

  10. #10


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    Quote Originally Posted by BJGenius007 View Post
    I have been using three counts for a very long time. I use Zen, ace side count and 8/9 side count, so I have covered all ranks of cards. This is similar to Tarzan Count. For people who are not familiar with Tarzan Count: In short, basic version of Tarzan count is three columns (2-5, 6-9, 10's) with ASC.
    Seems perfect fit for the Zen count since 89s are the only cards valued at zero. How do you use that info? Do you know how much more EV do you get.

  11. #11
    Senior Member Tarzan's Avatar
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    Sure, but what's that got to do with Nicki Minaj's cousin's friend's balls? Yes, displacement by volume. Surplus {6-9} in the remainder in an even distribution lower the RC, deficit {6-9} in an even distribution in the remainder raise the RC. An uneven distribution within the grouping sends it off in one direction or another from there. 0-10-0 is way different than 10-0-10, generally speaking. I'll be perhaps under or over other counts in comparison, where they get RC+2 across the board, I may get RC+1, match theirs at RC+2, or get RC+3, for example, depending upon what is going on with these middle cards and whether it is an even or uneven distribution of cards within the grouping. Verify that what you are doing checks out is all. When I make an assessment of the deck composition to come up with the RC/TC, I have a sort of "shorthand" that I use for this that I won't go into to avoid being long-winded on the post, but whatever you do or come up with has to be executable at the table.

    It's been a couple of years since I've played in a casino, and although I've enjoyed the hiatus, I look forward to being able to walk into a casino again someday. In getting back to the grind, maybe I'll hear less about Nicki Minaj's cousin's friend's balls.

  12. #12


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    Quote Originally Posted by Tarzan View Post
    Sure, but what's that got to do with Nicki Minaj's cousin's friend's balls? Yes, displacement by volume. Surplus {6-9} in the remainder in an even distribution lower the RC, deficit {6-9} in an even distribution in the remainder raise the RC. An uneven distribution within the grouping sends it off in one direction or another from there. 0-10-0 is way different than 10-0-10, generally speaking. I'll be perhaps under or over other counts in comparison, where they get RC+2 across the board, I may get RC+1, match theirs at RC+2, or get RC+3, for example, depending upon what is going on with these middle cards and whether it is an even or uneven distribution of cards within the grouping. Verify that what you are doing checks out is all. When I make an assessment of the deck composition to come up with the RC/TC, I have a sort of "shorthand" that I use for this that I won't go into to avoid being long-winded on the post, but whatever you do or come up with has to be executable at the table.

    It's been a couple of years since I've played in a casino, and although I've enjoyed the hiatus, I look forward to being able to walk into a casino again someday. In getting back to the grind, maybe I'll hear less about Nicki Minaj's cousin's friend's balls.
    Even if I knew that there were 60 low cards left, I would not be able to say exactly how many cards of each denomination are left but let’s assume equal distribution so that would be 12 of each (2, 3, 4, 5, 6). Since the twelve 6s would be in the middle group using the Tarzan count, they would join the twenty-four 789s (8x7, 8x8, 8x) mentioned in the OP for a total of 36 cards in the middle group would. With the T-count we would thus have 48 lows, 36 intermediates, 48 tens and 12 aces for a total of 144 cards.
    It’s my understanding that the Tarzan basic count would be 12-0-12, plus 12 aces, and would not suggest a 2-unit bet.

    However I am not familiar with the advanced T-count. So Tarzan how would you bet this situation with the top version of your T-count considering even distribution of 23456s? In the mean time, I am educating myself on Nicki Minaj.

  13. #13


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    Quote Originally Posted by Secretariat View Post
    Half of the shoe is gone, Don. So let’s say there are three decks left which should be 156 cards
    At the 156 cards mark there should be 12 aces, 48 tens, 60 low cards and 36 neutral/intermediate cards for a perfectly balanced shoe.
    However, assume a shortage of 12 x 789s (4x7, 4x8, 4x9) at the 144-card mark.
    That’s 12 neutral cards below normal at slightly below the 3-deck point. That’s what I meant.
    Do you think that eliminates the house edge at 6D, S17,DAS?
    If you had that information at the table would you bet 2 units?
    The precise combinatorial analytic answer is no, because the house edge is 0.1369 for your remaining shoe composition.

    Don

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