Therefore, we calculate the probability of getting dealt a blackjack in the following way: There is only one Queen of Spades in the deck opposed to 51 cards of.

Enjoy!

What are the odds of getting 3 blackjacks in a row with 1 deck 4 players and one using one deck, what is the probability of the dealer having a blackjack?

Enjoy!

Blackjack is one of the most advantageous casino games as in it, an efficient However, knowing the odds of winning and the probability of getting a certain card There is only one such card in a single deck and the rest of the remaining

Enjoy!

A winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards Find the probability of being dealt a blackjack hand.

Enjoy!

Sixteen out of the 51 cards left in the pack are worth ten, so the branch to the F in the top right has a probability of 16/51 - one way of getting a.

Enjoy!

Software - MORE

That's easily calculated, but it varies based on how many decks are being used. For this example, we'll use one deck. To get a blackjack, you need either an ace-β.

Enjoy!

The probability to get a blackjack (natural): 64 / = %. 1) The previous probability calculations were based on one deck of cards, at the.

Enjoy!

Sixteen out of the 51 cards left in the pack are worth ten, so the branch to the F in the top right has a probability of 16/51 - one way of getting a.

Enjoy!

dealt a blackjack? The probability depends on how many decks are played. You can get blackjack in one of two orders: 1) A ten-value card.

Enjoy!

Blackjack is one of the most advantageous casino games as in it, an efficient However, knowing the odds of winning and the probability of getting a certain card There is only one such card in a single deck and the rest of the remaining

Enjoy!

There is no sound bite answer to explain why you should hit. If I'm playing for fun then I leave the table when I'm not having fun any longer. Your question however could be rephrased as, "what is the value of the ace, given that the other card is not a ten. Determine the probability that the player will resplit to 3 hands. Blackjack is not entirely a game of independent trials like roulette, but the deck is not predisposed to run in streaks. My question though is what does that really mean? When I said the probability of losing 8 hands in a row is 1 in I meant that starting with the next hand the probability of losing 8 in a row is 1 in The chances of 8 losses in a row over a session are greater the longer the session. So the probability of winning six in a row is 0. Multiply dot product from step 11 by probability in step 9. In that case, the probability of a win, given a resolved bet, is The probability of winning n hands is a row is 0. Take another 8 out of the deck. It took me years to get the splitting pairs correct myself. Unless you are counting cards you have the free will to bet as much as you want. Putting aside some minor effects of deck composition, the dealer who pulled a 5 to a 16 the last five times in a row would be just as likely to do it the next time as the dealer who had been busting on 16 for several hours. All of this assumes flat betting, otherwise the math really gets messy. The standard deviation of one hand is 1. Cindy of Gambling Tools was very helpful. To test the most likely case to favor hitting, 8 decks and only 3 cards, I ran every possible situation through my combinatorial program. According to my blackjack appendix 9H the expected return of standing is So my hitting you will save 6. It depends whether there is a shuffle between the blackjacks. When the dealer stands on a soft 17, the dealer will bust about When the dealer hits on a soft 17, the dealer will bust about According to my blackjack appendix 4 , the probability of a net win is However, if we skip ties, the probability is So, the probability of a four wins in a row is 0. Add values from steps 4, 8, and The hardest part of all this is step 3. Determine the probability that the player will resplit to 4 hands. I know, I know, its some sort of divine intervention betting system I am talking about and no betting system affects the house edge. If there were a shuffle between hands the probability would increase substantially. For the non-card counter it may be assumed that the odds are the same in each new round. So standing is the marginally better play. Is it that when I sit down at the table, 1 out of my next playing sessions I can expect to have an 8 hand losing streak? Here is how I did it. If you were to add a card as the dealer you should add a 5, which increases the house edge by 0. Probability of Blackjack Decks Probability 1 4. Because the sum of a large number of random variables always will approach a bell curve we can use the central limit theorem to get at the answer. This is not even a marginal play. Any basic statistics book should have a standard normal table which will give the Z statistic of 0. Streaks, such as the dealer drawing a 5 to a 16, are inevitable but not predictable. I have a very ugly subroutine full of long formulas I determine using probability trees.{/INSERTKEYS}{/PARAGRAPH} {PARAGRAPH}{INSERTKEYS}This is a typical question one might encounter in an introductory statistics class. From my section on the house edge we find the standard deviation in blackjack to be 1. If the probability of a blackjack is p then the probability of not getting any blackjacks in 10 hands is 1- 1-p For example in a six deck game the answer would be 1- 0. That column seemed to put the mathematics to that "feeling" a player can get. Resplitting up to four hands is allowed. So, the best card for the player is the ace and the best for the dealer is the 5. There are cards remaining in the two decks and 32 are tens. Repeat step 3 but multiply by 4 instead of 2, and this time consider getting an 8 as a third card, corresponding to the situation where the player is forced to stop resplitting. What you have experienced is likely the result of some very bad losing streaks. Let n be the number of decks. You ask a good question for which there is no firm answer. There are 24 sevens in the shoe. However if you were going to cheat it would be much better to remove an ace, which increases the house edge by 0. Multiply dot product from step 7 by probability in step 5. However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. Since this question was submitted, a player held the dice for rolls on May 23, in Atlantic City. It is more a matter of degree, the more you play the more your results will approach the house edge. Determine the probability that the player will not get a third eight on either hand. The fewer the decks and the greater the number of cards the more this is true. Multiply this dot product by the probability from step 2. Following this rule will result in an extra unit once every hands. You are forgetting that there are two possible orders, either the ace or the ten can be first. Expected Values for 3-card 16 Vs. Steve from Phoenix, AZ. From my blackjack appendix 7 we see that each 9 removed from a single deck game increases the house edge by 0. If you want to deviate from the basic strategy here are some borderline plays: 12 against 3, 12 against 4, 13 against 2, 16 against Deviating on these hands will cost you much less. Thanks for the kind words. The probability of this is 1 in 5,,, For the probability for any number of throws from 1 to , please see my craps survival tables. The following table displays the results. For how to solve the problem yourself, see my MathProblems. I would have to do a computer simulation to consider all the other combinations. Go through all ranks, except 8, subtract that card from the deck, play out a hand with that card and an 8, determine the expected value, and multiply by 2. I recently replaced my blackjack appendix 4 with some information about the standard deviation which may help. Or does it mean that on any given loss it is a 1 in chance that it was the first of 8 losses coming my way? The best play for a billion hands is the best play for one hand. As I always say all betting systems are equally worthless so flying by the seat of your pants is just as good as flat betting over the long term. These expected values consider all the numerous ways the hand can play out. Repeat step 3 but multiply by 3 instead of 2. For each rank determine the probability of that rank, given that the probability of another 8 is zero. I have no problem with increasing your bet when you get a lucky feeling. In general the variation in the mean is inversely proportional to the square root of the number of hands you play. It depends on the number of decks. What is important is that you play your cards right. Here is the exact answer for various numbers of decks. It may also be the result of progressive betting or mistakes in strategy. Take the dot product of the probability and expected value over each rank. Besides every once in awhile throwing down a bigger bet just adds to the excitement and for some reason it seems logical that if you have lost a string of hands you are "due" for a win. I hope this answers your question. Thanks for your kind words. According to my blackjack appendix 4 , the probability of an overall win in blackjack is I'm going to assume you wish to ignore ties for purposes of the streak. It would take about 5 years playing blackjack 40 hours a week before this piece of advice saved the player one unit.