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Therefore, we calculate the probability of getting dealt a blackjack in the following way: P (Ace) * P (Ten-Value Card) * 2 = (4/52) * (16/51) * 2 = *

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Therefore, we calculate the probability of getting dealt a blackjack in the following way: P (Ace) * P (Ten-Value Card) * 2 = (4/52) * (16/51) * 2 = *

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The probability of drawing those 2 cards in that order is 1/52 X 1/51, or 1/ Example 2: You want to know the probability that you'll get a blackjack.

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Therefore, we calculate the probability of getting dealt a blackjack in the following way: P (Ace) * P (Ten-Value Card) * 2 = (4/52) * (16/51) * 2 = *

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rating.bonusmoneygames.site āŗ Casino āŗ Blackjack.

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The probability of drawing those 2 cards in that order is 1/52 X 1/51, or 1/ Example 2: You want to know the probability that you'll get a blackjack.

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