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Thread: Card Counting Spanish 21

  1. #27
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    Quote Originally Posted by Bushie View Post
    "is only 8. Can you see what I mean?"
    If you sum the positive and the negative integers and get the
    same amount on each side of the equation, the count is BALANCED.

  2. #28


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    Quote Originally Posted by ZenMaster_Flash View Post
    If you sum the positive and the negative integers and get the
    same amount on each side of the equation, the count is BALANCED.
    Ugh. I'm fully aware of what a balanced count is and how it works. But that's not the issue.

    The tags of
    1 2 3 3 0 0 -2 -3
    1 2 3 3 0 0 -2 -3
    v v v v v v v v
    2 3 4 5 7 8 10 A

    Can you see what is wrong? You've missed listing two tags, that being (I'd imagine) 6 and 9, with 7 and 8 being neutral at 0. Is it fair to assume that 6 and 9 are respectively either 1 or 2 and -1 or -2? Given that I know how a balanced count works, it would balance out either side at either +10 and -10 or +11 and -11 with 7 and 8 being neutral.

    Do you understand this?

  3. #29


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    Beginner card counter and rank amateur mathematician so I need help from someone to tell me if this would be a viable counting system for Sp21.

    Its mostly copied from Arnold Snyder's Red Seven blackjack count with the exception that ALL 7s are neutral/obviously the 10s are gone/ and the 4 aces are all counted as +1.5

    Each deck of 48 cards would still have a count of +2 so in an 8 deck shoe, you would start count at negative -16 just like in Arnold Snyder's Red 7.

    I'm thinking I would start increasing my bets at the same indices used in the Red Seven count but I may be mistaken and that's why I posted here?

    Thanks

  4. #30


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    Quote Originally Posted by rooneys9 View Post
    Beginner card counter and rank amateur mathematician so I need help from someone to tell me if this would be a viable counting system for Sp21.

    Its mostly copied from Arnold Snyder's Red Seven blackjack count with the exception that ALL 7s are neutral/obviously the 10s are gone/ and the 4 aces are all counted as +1.5

    Each deck of 48 cards would still have a count of +2 so in an 8 deck shoe, you would start count at negative -16 just like in Arnold Snyder's Red 7.

    I'm thinking I would start increasing my bets at the same indices used in the Red Seven count but I may be mistaken and that's why I posted here?

    Thanks
    As a beginner, you shouldn't be trying to reinvent the wheel. There are established systems available. I suggest you learn one of them, and if/when you gain sufficient experience, then consider trying to improve them.

  5. #31


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    The reason why I'm posting here is b/c I've read Katrina Walker's book and her counting system involves constantly having to convert the running count back to a true count and I don't want to have to do something like that.

    I'm more than happy to get the same .8% edge the Arnold Snyder's red seven count gets of the HILO count. The other reason I'm posting is b/c like I said I'm unsure of the indices/math when making an ACE equal to +1.5.

    Thanks

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