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

  1. #14


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    Yes, all exactly right. Was going to provide a similar example, but had to step out for a while. Glad to see this when I got back.

    Don

  2. #15


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    Quote Originally Posted by Gronbog View Post
    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.
    Hi Gronbog,

    this is exactly the same I wanted to write (but was on airplane on my flight back from vacation during the last hours).
    I thought of a similar example: if you don't see the burn cards, they are treated the same way as if they were behind the cut card.
    Cards you don't know and see do not change anything to the computations.

  3. #16


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    Quote Originally Posted by Dog Hand View Post
    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
    Very nice explanation, Dog Hand! It reminds me of some similar explanations in Fred Renzey's Blackjack Bluebook II,
    where he makes plausible that other players (e.g. third base "taking the dealer's bust card") don't influence your outcome,
    because it all averages out to the same as if you were ignoring other players.

    Besides, nice trick to simplify the computation using the distributive law regarding 4, and subsequently cancelling down the 51 off the fraction.
    (I'm a currently a math teacher so I really liked it).

  4. #17


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    I saw a person get four blackjacks in a row in an 8 deck shoe about a month ago. Crazy.

  5. #18


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    There's a dealer I know who told me she got 6 BJs in a row on a SD game. I asked her how that was possible with only 4 aces in the deck. She said she got 4 in a row, shuffled, then got two more.

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