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Thread: What if the dealer has 18 every hand.

  1. #1
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    What if the dealer has 18 every hand.

    DD, DAS,DOA.,80% penetration.
    Dealer doesn't get dealt any cards.
    Players 18 is a push, Players 19 or better wins. BJ pays 3-2.
    Any idea what kind of edge this game would have? Obviously the player will bust much more frequently as 17 and below requires a hit, but all 19s and 20s are automatic winners.
    Would one split 9s or take the push?
    Let me die in my sleep like my Grandfather.
    Not screaming in agony like his passengers.

  2. #2


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    Sounds like an interesting game. I expect the answers (for strategy at least, if not for HE) will be known by hole-card players.

    How many hands per hour does this game get? I would imagine it can be dealt at a great speed.

  3. #3


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    I would think that the fact that you'd have to hit every single hard 17 and never stand on a stiff would absolutely kill you.

  4. #4
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    Hmmm . . . If you have foreknowledge that the dealer always has an 18 . . .

    The mean dealer hand is normally a bit over 18; but the casino edge is intrinsically

    linked to your having to draw to your stiffs when facing a dealer high "up card."

    That would not be the case in this hypothetical game.

    Removing the spectra of the dealer having a BJ is worth a lot,

    especially the repugnant and uninsurable "back-door" Blackjack.

    Additionally, your doubled and your split hands are very much affected !

    Hole Card strategy addresses the issue of playing one's hand's to maximize advantage.

    Any simple Level One Count will serve well here as Playing Efficiency will be 1.0%

    One thing for certain us that with any aggressive spread a Card counter would be all

    but guaranteed a magnificent opportunity enhanced I would suspect by decreased flux !

  5. #5
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    Quote Originally Posted by 21gunsalute View Post
    I would think that the fact that you'd have to hit every single hard 17 and never stand on a stiff would absolutely kill you.
    But the single most likely hand of 20 would win 100% of the time.

    Anyone want to sim a few million hands?
    Let me die in my sleep like my Grandfather.
    Not screaming in agony like his passengers.

  6. #6
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    Wouldn't the player have any advantage in a negative count as well since they are kind of taking on the rules of the dealer by hitting to 18? Knowing that you have to hit stiffs no matter what changes the whole approach. Would the proper stategy be to increase your bets as the count goes highly positive or negative and bet minimum during neutral counts? I would imagine you would have to count 9's as a big card and maybe drop the 2 to keep it balanced. Comments?
    Don't judge a man until you have walked a mile in his shoes, by then you are a mile away and have his shoes.

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    I've played a few hundred hands,flat betting, and the game so far is pretty even. It goes much faster than a regular game.

    To clarify, this game only exists in my dining room, as far as I know.
    Let me die in my sleep like my Grandfather.
    Not screaming in agony like his passengers.

  8. #8


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    Quote Originally Posted by shadroch View Post

    To clarify, this game only exists in my dining room, as far as I know.
    What's the table max?

  9. #9


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    Hmmm...this is interesting. Not quite as interesting if this actually existed, though.


    Basic strategy would be much easier, since it's one dimensional -- only the players cards, not two-dimensional by including the dealer's up card.

    There'd be no reason to side-count cards, because you will ALWAYS be hitting 17 or less. There is no need to "playing efficiency", other than knowing when to double down or split. ie: If you have 17 in a sky-rocket count, you're going to have to hit, because staying is an automatic loss.

    The only cards that would be any good for "playing efficiency" that I can think of would be 9,10,A. Maybe even an 8. Knowing there are many 9s, 10s, and ACEs remaining, you'd want to double down on 10.

    On the other hand, a big part of doubling down, at least from my understanding, is taking into consideration that the dealer busts often-enough. So, when you're doubling down your 11v10 in a normal game, part of the reason why it's +EV is because the dealer has a decent chance of busting. For this game the dealer cannot bust, making doubling down worth less. Same thing with splitting and whatnot. (Same applies for staying on stiff hands, except it's fairly obvious why you wouldn't stay with a stiff hand in this game....there's no way to win when you stay on 16.)


    The only reason to split 9,9 is if more often than not you can get at least 1 hand of 19 or higher and the other hand of 18 or higher.


    Shad, does this game offer LS?
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

  10. #10
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    You would want to split 9,9 Also you would want to hit A7 but it is a very close play according to CAA

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    There would be very few opportunities to split or double down and you'd be busting out exponentially more hands. I don't see how this could be profitable. It would probably take a very different counting system to beat it, if it can be beaten.

  12. #12
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    Obviously you would need a different counting system.
    All your 19,20s and 21 are instant winners, Every BJ is a 3-2 winner. This is a huge boon for the player.
    You would bust more frequently but is it really a killer?
    Look at 13 . In a regular game you'd stay on this about half the time and only win when the dealer busted.
    In this game you hit it and have a 3/13 chance for a win, and 1/13 chance for a push. 5/13 times you'll bust and the the other 4/13 you take another hit.
    Let me die in my sleep like my Grandfather.
    Not screaming in agony like his passengers.

  13. #13


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    So somehow I thought of this while dreaming, don't ask.

    Total 9, taking one card, there are 5 winners (T, J, Q, K, A), one pusher (9), and 7 losers (2, 3, 4, 5, 6, 7, 8).
    Total 10, taking one card, there are 6 winners (9, T, J, Q, K, A), one pusher (8), and 6 losers (2, 3, 4, 5, 6, 7). This looks like, errr, rather IS, an even proposition. Out of the thirteen ways it can turn out, you win 6, lose 6, and push one. However, when cards are removed from the deck, it means the opposite type of card are now slightly more dense.

    Take a hand of 4,6 totalling 10. There are 24 cards (four 9,T,J,Q,K,As) to help you win, 4 cards (8) to push, and 22 cards (four 2,3,4,5,6,7s....minus a 4 and a 6).


    Do the same thing with a hand totalling 11 and you get 6 winner cards (8,9,T,J,Q,K [but no ace]), 1 push card (7), and 6 loser cards (A,2,3,4,5,6).


    For doubling down on 10 or 11, instant winners are: 9,T,J,Q,K. The 8 is a push in one and a winner in the other. Opposite of the 8 (push/win) is the 7, which is a lose/push. The Ace is a winner in one but a loser in the other.

    2,3,4,5,6 = +1 each
    9,T,J,Q,K = -1 each
    7 = +0.5
    8 = -0.5

    Any time the count is positive, double 10 or 11.

    May side count Aces. Add excess Aces to the RC for doubling with 10. Subtract excess Aces from the RC for doubling with 11. If count is positive, double down (with 10 or 11). If count is negative, don't double down. True-Count calculation is not required, but 1/4 deck estimation is required for the ASC. Of course....this is all for fun.
    "Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]

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