š Blackjack Odds and Probability
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Option 2: First event (getting a 10 card): There are 16 cards worth 10, as before, but this time we've got a full pack, so the probability is 16/
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If the player's first two cards total 21, this is a blackjack and she wins times her the dealer's up card is 10 or A, and double with 9 only against a dealer's 2 to 6. but in this case blackjack rules allow you to get only one card on each hand, Let's call p the total probability of winning a pass line bet (so p is the number.
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2) An ace followed by a ten-value card. If we add the probabilities of these two events, this will give us the odds of getting blackjack. The odds of.
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Obviously the dealer is inconsequential to the cards dealt (I like to say the dealer is The probability of a suited blackjack in a six-deck game is 2*(4/13)*(6/) What are the odds of a dealer getting 3 blackjacks in a row on a single deck.
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2) An ace followed by a ten-value card. If we add the probabilities of these two events, this will give us the odds of getting blackjack. The odds of.
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(a) If a player is dealt 2 cards from a standard deck of 52 well-shuffled cards, what is the probability that the player will receive a blackjack?
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Option 2: First event (getting a 10 card): There are 16 cards worth 10, as before, but this time we've got a full pack, so the probability is 16/
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Since there are four aces and 16 cards K, the raw probability of a hand being blackjack is: Pblackjack=4Ć16(). The probability of both getting blackjack is.
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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:. Sign up using Email and Password. Michael Cotton Michael Cotton 3 3 silver badges 6 6 bronze badges. Blackjack Probability Ask Question. Sign up to join this community. 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. Suppose that you are playing blackjack against the dealer. Email Required, but never shown. I don't know if this will get you any closer to a closed form solution, but maybe it helps you think about it. Featured on Meta. Question feed. Ross Millikan Ross Millikan k 24 24 gold badges silver badges bronze badges. Sign up or log in Sign up using Google. Related 0. The naive assumption is that the chance of each player getting blackjack is independent of the others. Post as a guest Name. We're switching to CommonMark. Nolohice 5 5 silver badges 12 12 bronze badges. Home Questions Tags Users Unanswered. The use of neither indicates that. Viewed 19k times. Hot Network Questions. Asked 5 years, 3 months ago.
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MaxWell MaxWell 2 2 bronze badges. It only takes a minute to sign up. Active 1 year, 7 months ago. What posts should be escalated to staff using [status-review], and how do Iā¦. Active Oldest Votes. So I deleted the more complicated answer I probability of getting blackjack with 2 cards to give earlier.
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Your calculation for the first player is correct. Because the chance of one player getting blackjack is small, the enrichment is small as well, so this is not far off. I can't help but wonder if there perhaps is a less hideous way of acquiring the same answer though. This happens every time. Sign up using Facebook. 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. The outermost branches are short, but the inside of the tree is massive Hopefully this at least helps you contextualize the problem. With a branching factor of five I know , we have two aces, once ace, two whatevers, one 10, two 10s.