A blackjack game has a dealer and one or more players. Let's call p the total probability of winning a pass line bet (so p is the number we are trying to that a whole deck will be used to choose among when you are dealt your next card?

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(i.e., choose two cards for the first player and then two for the second.) Then, the # of ways to deal a blackjack to both players would be: (41)()(31)()=

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Now we need to find probability of getting a black jack. Overall there are 2 black What is the probability of picking an ace and a king from 52 card? 4, Views.

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list.jabkino.ru āŗ ask-the-wizard āŗ blackjack āŗ probability.

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In fact, it's easier for computer programs to calculate blackjack probability by running billions of Your odds of picking the correct number are therefore 37 to 1.

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list.jabkino.ru āŗ ask-the-wizard āŗ blackjack āŗ probability.

Enjoy!

In fact, it's easier for computer programs to calculate blackjack probability by running billions of Your odds of picking the correct number are therefore 37 to 1.

Enjoy!

Now we need to find probability of getting a black jack. Overall there are 2 black What is the probability of picking an ace and a king from 52 card? 4, Views.

Enjoy!

(i.e., choose two cards for the first player and then two for the second.) Then, the # of ways to deal a blackjack to both players would be: (41)()(31)()=

Enjoy!

In fact, it's easier for computer programs to calculate blackjack probability by running billions of Your odds of picking the correct number are therefore 37 to 1.

Enjoy!

Hot Network Questions. Asked 5 years, 3 months ago. The outermost branches are short, but the inside of the tree is massive Hopefully this at least helps you contextualize the problem. Sign up to join this community. MaxWell MaxWell 2 2 bronze badges. I can't help but wonder if there perhaps is a less hideous way of acquiring the same answer though. Your calculation for the first player is correct. It only takes a minute to sign up. Blackjack Probability Ask Question. So I deleted the more complicated answer I tried to give earlier. Post as a guest Name. Mathematics Stack Exchange works best with JavaScript enabled.{/INSERTKEYS}{/PARAGRAPH} {PARAGRAPH}{INSERTKEYS}Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. With a branching factor of five I know , we have two aces, once ace, two whatevers, one 10, two 10s. Sign up using Facebook. Question feed. 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 or log in Sign up using Google. The naive assumption is that the chance of each player getting blackjack is independent of the others. Active Oldest Votes. Active 1 year, 7 months ago. What posts should be escalated to staff using [status-review], and how do Iā¦. Home Questions Tags Users Unanswered. I don't know if this will get you any closer to a closed form solution, but maybe it helps you think about it. The best answers are voted up and rise to the top. Viewed 19k times. Sign up using Email and Password. 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. 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. Featured on Meta. We're switching to CommonMark. Because the chance of one player getting blackjack is small, the enrichment is small as well, so this is not far off. Related 0. Ross Millikan Ross Millikan k 25 25 gold badges silver badges bronze badges. Email Required, but never shown. Michael Cotton Michael Cotton 3 3 silver badges 6 6 bronze badges. Suppose that you are playing blackjack against the dealer. The use of neither indicates that. This happens every time. Nolohice 5 5 silver badges 12 12 bronze badges.