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Keywords. Blackjack. Serendipitous hand. Probability models The question is examined here by comparing results from the two kinds of sampling.


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For how to solve the problem yourself, see my cellaz.ru site, problem If I play hands of blackjack at $5 a hand at an % house edge how.


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You will also find there other issues of probability-based blackjack strategy. See the Books section for details. Check this.


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Keywords. Blackjack. Serendipitous hand. Probability models The question is examined here by comparing results from the two kinds of sampling.


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Hence the probability of neither player getting a blackjack is: 1āˆ’+,,ā€‹ Since there are four aces and 16 cards K, the raw probability of a hand being blackjack is: Hopefully this at least helps you contextualize the problem.


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You will also find there other issues of probability-based blackjack strategy. See the Books section for details. Check this.


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cellaz.ru ā€ŗ ask-the-wizard ā€ŗ blackjack ā€ŗ probability.


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The original question was, exactly, as this: "Dealing yourself a blackjack (Ace AND Face-card or Ten) from a single deck". The calculations above are accurate for.


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I. Fundamental Probability Issue: True Odds at Blackjack, Software Calculator II. Fundamental Myth of Blackjack Gambling: Counting Cards III. Theory of Streaks:ā€‹.


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I don't know if this will get you any closer to a closed form solution, but maybe it helps you think about it. Hot Network Questions. The use of neither indicates that. Mathematics Stack Exchange works best with JavaScript enabled.{/INSERTKEYS}{/PARAGRAPH} Because the chance of one player getting blackjack is small, the enrichment is small as well, so this is not far off. {PARAGRAPH}{INSERTKEYS}Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Sign up using Facebook. We're switching to CommonMark. It only takes a minute to sign up. Blackjack Probability Ask Question. Asked 5 years, 3 months ago. Suppose that you are playing blackjack against the dealer. What posts should be escalated to staff using [status-review], and how do Iā€¦. Nolohice 5 5 silver badges 12 12 bronze badges. Sign up to join this community. Post as a guest Name. The probability of both getting blackjack is just the probability of the player getting blackjack and the subsequent probability of the dealer also getting blackjack:. Viewed 19k times. Ross Millikan Ross Millikan k 24 24 gold badges silver badges bronze badges. Sign up using Email and Password. This happens every time. Home Questions Tags Users Unanswered. Featured on Meta. So I deleted the more complicated answer I tried to give earlier. Sign up or log in Sign up using Google. Related 0. Question feed. This is not quite correct, as the fact that the first player did not get blackjack enriches the average deck with cards that could make a blackjack for the second player. MaxWell MaxWell 2 2 bronze badges. The naive assumption is that the chance of each player getting blackjack is independent of the others. With a branching factor of five I know , we have two aces, once ace, two whatevers, one 10, two 10s. Michael Cotton Michael Cotton 3 3 silver badges 6 6 bronze badges. Active Oldest Votes. Email Required, but never shown. Your calculation for the first player is correct. The outermost branches are short, but the inside of the tree is massive Hopefully this at least helps you contextualize the problem. Then for the other player we either give them two aces from the 3 that are left, give them one ace from the 3 that are left and 1 of the 32 cards, or give them two non-aces from the 47 non-aces that are left. I can't help but wonder if there perhaps is a less hideous way of acquiring the same answer though. The best answers are voted up and rise to the top. Active 1 year, 7 months ago.