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Thread: It says odds of busting with 14 is 56% but 6/13 = 46%, why?

  1. #14


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    Quote Originally Posted by PinkChip View Post
    6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
    1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
    1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
    1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)

    = 6/13 + 7/169 + 8/169 + 8/2197
    = 1014/2197 + 91/2197 + 104/2197 + 8/2197
    = 1217/2197
    = 55.4 %
    Thank you, mate I was thinking of it the same, then made the edit because I thought 6/13 stays the same if we start on 14 but the thing is, we've already counted that probability in by writing 6/13 and your calculations make sense to me because we can't assume the probability of bust remains static no matter what. In that case, the resources on the internet were correct about the 56% (excluding the second screenshot of mine).

    Thank you very much PinkChip If the other people agree with the calculations, please let me know

  2. #15


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    I'm sorry if that's not the right thread to ask but PinkChip, are you a card counter?

  3. #16


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    Quote Originally Posted by LukizKing View Post
    Thank you, mate I was thinking of it the same, then made the edit because I thought 6/13 stays the same if we start on 14 but the thing is, we've already counted that probability in by writing 6/13 and your calculations make sense to me because we can't assume the probability of bust remains static no matter what. In that case, the resources on the internet were correct about the 56% (excluding the second screenshot of mine).

    Thank you very much PinkChip If the other people agree with the calculations, please let me know
    You are welcome. I also had to go through my calculations and correct them several times. Should there still be any error, Don probably will tell us :-)

  4. #17


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    Quote Originally Posted by LukizKing View Post
    I'm sorry if that's not the right thread to ask but PinkChip, are you a card counter?
    You could ask that to any forum member. I have not yet counted in real casino and money conditions, since I'm living in CSM dreamland Europe ;-)
    Anyway I'm quite mathematically educated and in the process of learning counting. Still working on my strategies (tayloring a system which I can use best under real conditions) and have to practice, improving speed and accuracy. There are lots of things to take into consideration, especially if it's not possible to easily spot the casino conditions just a few miles away.

  5. #18


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    Quote Originally Posted by Midwest Player View Post
    I have hit soft 21 in error already, and have seen other folks do it quite often in error.
    Years of living in cold weather causes occasional brain freeze.

  6. #19


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    Most likely amount of cards in the discard and depending on the running count if your a +5 TC the odds to bust on a 14 are increased as there's more 10s left in the deck - Mathmatics differ but can be calculated if keeping perfect running count and can calculate True count accurately

  7. #20


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    % to bust will correspond with your True count if your at a +5 or greater TC then you have a higher chance to bust on a 14 than you would if the TC was a -5 count

  8. #21


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    Yes, I know about the Illustrious 18 table where you should split 10s at the +5 TC or higher and so on. However, I'll never play games with the TC that low so don't have to worry about those little adjustments (not sure if the probability from ~ 56% can go straight to the less than 50% with the true count of -5 though).

  9. #22


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    Quote Originally Posted by PinkChip View Post
    But in the second case (dealer gets an Ace, with probability of 1/13):
    when dealer has 15, his probability of busting by the next card is not 6/13 anymore.
    If he has Hard 15, there are now 7 cards which would him bust: 7,8,9,10,J,Q,K.
    So you must compute 1/13 * 7/13 in the second case.

    Analogously, 1/13 * 8/13 in the third case, since if the dealer has 16,
    he busts with 8 cards: 6,7,8,9,10,J,Q,K.

    It gets further complicated, since in the second case, if the dealer has 15,
    he could draw another Ace (1/13) and then get into the third case.
    In my opinion, the result of these overall four cases is

    6/13 (start on Hard 14 and draw 8,9,10,J,Q,K) +
    1/13 * 7/13 (start on Hard 14, then draw an Ace to Hard 15, then draw 7,8,9,10,J,Q,K) +
    1/13 * 8/13 (start on Hard 14, then draw a Two to Hard 16, then draw 6,7,8,9,10,J,Q,K) +
    1/13 * 1/13 * 8/13 (start on Hard 14, then draw two Aces to Hard 16, then draw 6,7,8,9,10,J,Q,K)

    = 6/13 + 7/169 + 8/169 + 8/2197
    = 1014/2197 + 91/2197 + 104/2197 + 8/2197
    = 1217/2197
    = 55.4 %

    Nicely done. Obviously, this is an infinite-deck calculation, where you don't account for the two cards that comprise the original 14, nor the fact, in the last part, that, since you've already drawn one ace to get to 15, the probability of drawing the second ace to get to 16 is now less than 1/13.

    Don

  10. #23


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    Quote Originally Posted by DSchles View Post
    Nicely done. Obviously, this is an infinite-deck calculation, where you don't account for the two cards that comprise the original 14, nor the fact, in the last part, that, since you've already drawn one ace to get to 15, the probability of drawing the second ace to get to 16 is now less than 1/13.

    Don
    Thank you Don. Yes, I referred to the simplification of infinite decks from the OP. With changing deck composition, the combinatorial analysis gets much more complicated.

    I once wrote a program which computes some probabilities like dealer bust rate, depending on number of decks (see the other thread on number of cards per hand). It doesn't take into account player hands, just the dealer hand. But even this requires a tree with a depth of 12 or 13 levels, since the dealer might draw that many low cards (Aces and Twos) before busting.

    Of course these dealer hands are very rare. Mostly the dealer hand will comprise 2, 3 or 4 cards, so the path in these cases can be aborted because its subtree needs not be traversed. The program counts the number of pathes it had to traverse, and luckily it was not 10^12 or 10^13 but only about 50000 (for S17 rule) or 70000 (for H17 rule).

    It guess it would be impossible to compute the play through a full 6 deck shoe with 312 cards in this manner, which could produce exact results, in contrast to a Monte Carlo simulation.
    Last edited by PinkChip; 11-17-2019 at 03:47 PM.

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