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Thread: Odds of Getting Three Blackjacks in a Row on Single Deck vs 6 Deck

  1. #1


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    Odds of Getting Three Blackjacks in a Row on Single Deck vs 6 Deck

    Can anybody tell me what are the odds of getting 3 blackjacks in a row on a single deck blackjack game versus the odds of three blackjacks in a row on a six deck blackjack game? I assume the odds are better on a 6 deck game.

  2. #2


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    It's a bit tedious to compute and write down. Here is my computation. Every bracket is for one natural and contains the probabilities for drawing an Ace and a Ten. This must be multiplied by 2 because there are two possible orders for a natural: Ace-Ten and Ten-Ace. The decreasing numerators and denominators reflect the number of still available aces, tens (numerator) and overall cards (denominator) in the pack. In the end, all these probabilities must be multiplied together.

    Single Deck:

    P(three naturals in a row)
    = (2 * 4/52 * 16/51) * (2 * 3/50 * 15/49) * (2 * 2/48 * 14/47)
    = 2*2*2 * (4*3*2 * 16*15*14) / (52*51*50*49*48*47)
    = 240 / 5453175 (after simplifying)
    = 1 / 22722 (approximately)
    = 4.401105778 * 10^(-5)

    Which means this happens in one out of 22722 three-round-sequences.

    Six decks:

    = (2 * 24/312 * 96/311) * (2 * 23/310 * 95/309) * (2 * 22/308 * 94/307)
    = 1 / 10552 *approximately)
    = 9.47654849 * 10^(-5)

    So this happens in one out of 10552 cases and indeed more than twice as often as in Single Deck.

    By the way, for infinite decks and cards, the probability would simply compute to

    = (2 * 1/13 * 4/13)^3
    = (8 / 169)^3
    = 8^3 / 169^3
    = 512 / 4826809
    = 1 / 9427 (approximately)
    = 10.6074 * 10^(-5)

    and thus even larger than for six decks, since the Aces and Tens are not depleted anymore.
    Last edited by PinkChip; 10-16-2019 at 03:38 AM.

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    PinkChip
    You need to factor in the dealers cards as well, which you haven’t done. Also, not sure, though sounds intuitive to say that the order of getting an ace ir king, will ever so slightly alter the odds.

    Fun problem to work out, which I just might do at breakfast in a couple of hours.

  4. #4


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    Quote Originally Posted by Freightman View Post
    PinkChip
    You need to factor in the dealers cards as well, which you haven’t done. Also, not sure, though sounds intuitive to say that the order of getting an ace or king, will ever so slightly alter the odds.

    Fun problem to work out, which I just might do at breakfast in a couple of hours.
    Hi Freightman,

    I have done and read the computation for one natural often and am sure the factor of 2 is correct, because the order of multiplication is irrelevant:

    P(Ace,Ten) = 4/52 * 16/51 = (4*16)/(52*51) = 16/52 * 4/51 = P(Ten, Ace)

    So it makes no difference if Ace or Ten comes first, and since you have to add the two equal probabilities (in the combinatorial analysis tree), it is simpler to multiply by 2.

    But you are right in that my computation does not yet take into account whether or not the player blackjacks are matched by the dealer. MWP has not explicitly told if this matters for him but supposedly it does.
    Last edited by PinkChip; 10-16-2019 at 05:32 AM.

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    PinkChip
    What you’re missing is the dealer has to get a card before you get your second card. Further, assuming US style hole card play, dealer always gets a second card if his upcard is an ace or king, and may or may not get a second card if his upcard is less than 10 value, depending on how quick you flip the cards. also assumes heads up play.

    So, to make this an easier calculation (we can calculate whatever we want as long we have the time, and I have things to do after breakfast , so let’s just say that dealer never gets an ace or king, and always deals himself a hole card.

    Just multiply the odds of getting a blackjack on your first hand, by the odds of getting blackjack on your second hand, multiplied by the odds of getting blackjack on your third hand.

    (4/52 x 16/50) x (3/47 x 15/45) x (2/43 x 14/41) = .000831791517 - single deck calculation - same principle for 6 decks

    If this is wrong, Don will wake up soon

  6. #6


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    Quote Originally Posted by Freightman View Post
    PinkChip
    What you’re missing is the dealer has to get a card before you get your second card. Further, assuming US style hole card play, dealer always gets a second card if his upcard is an ace or king, and may or may not get a second card if his upcard is less than 10 value, depending on how quick you flip the cards. also assumes heads up play.

    So, to make this an easier calculation (we can calculate whatever we want as long we have the time, and I have things to do after breakfast , so let’s just say that dealer never gets an ace or king, and always deals himself a hole card.

    Just multiply the odds of getting a blackjack on your first hand, by the odds of getting blackjack on your second hand, multiplied by the odds of getting blackjack on your third hand.

    (4/52 x 16/50) x (3/47 x 15/45) x (2/43 x 14/41) = .000831791517 - single deck calculation - same principle for 6 decks

    If this is wrong, Don will wake up soon
    Really pisses you off when this happens at min or close to min bet.

  7. #7


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    Quote Originally Posted by Freightman View Post
    assuming US style hole card play, dealer always gets a second card if his upcard is an ace or king, and may or may not get a second card if his upcard is less than 10 value, depending on how quick you flip the cards. also assumes heads up play.
    I've never seen a dealer abort taking a second card if the player has BJ. I believe regulations require complete hands to be dealt.

  8. #8


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    Quote Originally Posted by 21forme View Post
    I've never seen a dealer abort taking a second card if the player has BJ. I believe regulations require complete hands to be dealt.
    The dealer has to deal themselves a hand. A hand consists of two cards.

  9. #9


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    Quote Originally Posted by Freightman View Post
    If this is wrong, Don will wake up soon
    Yup. Right you are! I'd forget the dealer. His hand is really irrelevant to the calculation. I think Pink Chip's calculations are the way to do it.

    Of course, all of this is based on just shuffling and dealing once. If the real question is how rare is it to get three naturals in a row, you have to then state how many hands in total you're experiencing. And, of course, in the context of playing thousands of hands, getting three blackjacks in a row is somewhat commonplace.

    Don

  10. #10


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    The taking of cards by the dealer and by the other players is not relevant to this calculation. The mathematical concept is the same one which explains why a NHC OBO game is mathematically identical to a game in which the dealer takes a hole card and peeks. That is, the order in which the cards are dealt and to whom does not change anything.

    The executive summary is that the shuffles which deal you the three blackjacks off the top with no dealer, with a dealer, and with a dealer and other players are all equally likely.

    Another way to think of it, which may make more sense to card counters, is to imagine that the cards which go to the dealer and to the other players are all dealt face down. They are simply unseen cards. Flipping them over so you can see them doesn't change anything and doesn't affect the calculation which was done before any cards were dealt.

    Having said that, I have not taken the time to check the calculations posted above.

  11. #11


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    Didn't mean to step on Don's toes. Looks like we were typing at the same time.

    I have reviewed the calculations and I believe that PinkChip's calculations are correct.

  12. #12


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    Quote Originally Posted by Gronbog View Post
    Didn't mean to step on Don's toes. Looks like we were typing at the same time.

    I have reviewed the calculations and I believe that PinkChip's calculations are correct.
    As long as it’s you guys, nary a problem.

  13. #13


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    Freightman,

    Here's an example to illustrate Gronbog's assertion: say you are playing BJ with one player to your right. Off the top of a fresh shuffle, what is the chance that your first card is an Ace?

    Now Gronbog (and I, for that matter) would say that, without knowing first base's first card, that your chance is 4/52 = 1/13.

    You would say, "But wait! FB's first card may be an Ace, so we cannot calculate my first-card-Ace chance without knowing his first card."

    Let's work it out together. Either his 1st is an Ace, or it isn't.

    If it is an Ace, then your chance is 3/51. This will be true with probability 4/52.

    If it isn't an Ace, then your chance is 4/51. This will occur with probability 48/52.

    So, your first-card-Ace probability is

    (3/51)*(4/52) + (4/51)*(48/52) =

    (3*4 + 4*48)/(51*52) =

    (4*(3 + 48))/(51*52) =

    (4*51)/(51*52) =

    4/52 = 1/13.

    Thus, without knowing beforehand what first base's first card is, we see that your first-card-Ace probability is 1/13.

    If you think about it, this is the same as asking, "What is the probability that the 6th card from the top of the deck is a 4?"

    Hope this helps!

    Dog Hand

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