you will win about 43% of your hands, push 9% and lose 48%. (Even “winning” blackjack players lose more hands than th If so can win most of the times?

Enjoy!

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So, the probability of a four wins in a row is = %. First of What are the odds against winning seven hands of blackjack in a row? To test the most likely case to favor hitting, 8 decks and only 3 cards, I ran every possible situation.

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I won 23 hands in a row once and ended my session there. That was enough of good variance for one day.

Enjoy!

Software - MORE

So, the probability of a four wins in a row is = %. First of What are the odds against winning seven hands of blackjack in a row? To test the most likely case to favor hitting, 8 decks and only 3 cards, I ran every possible situation.

Enjoy!

nodownload.amarant-ural.ru › ask-the-wizard › blackjack › probability.

Enjoy!

I won 23 hands in a row once and ended my session there. That was enough of good variance for one day.

Enjoy!

I won 23 hands in a row once and ended my session there. That was enough of good variance for one day.

Enjoy!

you will win about 43% of your hands, push 9% and lose 48%. (Even “winning” blackjack players lose more hands than th If so can win most of the times?

Enjoy!

Enjoy!

you will win about 43% of your hands, push 9% and lose 48%. (Even “winning” blackjack players lose more hands than th If so can win most of the times?

Enjoy!

However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. If there were a shuffle between hands the probability would increase substantially. I know, I know, its some sort of divine intervention betting system I am talking about and no betting system affects the house edge. Determine the probability that the player will resplit to 3 hands. Add values from steps 4, 8, and The hardest part of all this is step 3. If you were to add a card as the dealer you should add a 5, which increases the house edge by 0. Determine the probability that the player will resplit to 4 hands. I hope this answers your question. To test the most likely case to favor hitting, 8 decks and only 3 cards, I ran every possible situation through my combinatorial program. There are 24 sevens in the shoe. Take the dot product of the probability and expected value over each rank. Cindy of Gambling Tools was very helpful. So, the best card for the player is the ace and the best for the dealer is the 5. 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? Blackjack is not entirely a game of independent trials like roulette, but the deck is not predisposed to run in streaks. I would have to do a computer simulation to consider all the other combinations. I recently replaced my blackjack appendix 4 with some information about the standard deviation which may help. Steve from Phoenix, AZ. The fewer the decks and the greater the number of cards the more this is true. Any basic statistics book should have a standard normal table which will give the Z statistic of 0. There is no sound bite answer to explain why you should hit. 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. According to my blackjack appendix 9H the expected return of standing is So my hitting you will save 6. What is important is that you play your cards right. 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. That column seemed to put the mathematics to that "feeling" a player can get. For each rank determine the probability of that rank, given that the probability of another 8 is zero. Determine the probability that the player will not get a third eight on either hand. These expected values consider all the numerous ways the hand can play out. The best play for a billion hands is the best play for one hand. The standard deviation of one hand is 1. There are cards remaining in the two decks and 32 are tens. In that case, the probability of a win, given a resolved bet, is The probability of winning n hands is a row is 0. My question though is what does that really mean? Let n be the number of decks. For the non-card counter it may be assumed that the odds are the same in each new round. Multiply this dot product by the probability from step 2. The following table displays the results. If I'm playing for fun then I leave the table when I'm not having fun any longer. 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. What you have experienced is likely the result of some very bad losing streaks. All of this assumes flat betting, otherwise the math really gets messy. For how to solve the problem yourself, see my MathProblems. Your question however could be rephrased as, "what is the value of the ace, given that the other card is not a ten. This is not even a marginal play. Multiply dot product from step 7 by probability in step 5. You ask a good question for which there is no firm answer. Following this rule will result in an extra unit once every hands. Expected Values for 3-card 16 Vs. Probability of Blackjack Decks Probability 1 4. Resplitting up to four hands is allowed. Since this question was submitted, a player held the dice for rolls on May 23, in Atlantic City. Here is the exact answer for various numbers of decks. I have a very ugly subroutine full of long formulas I determine using probability trees.{/INSERTKEYS}{/PARAGRAPH} Take another 8 out of the deck. From my blackjack appendix 7 we see that each 9 removed from a single deck game increases the house edge by 0. Thanks for the kind words. Unless you are counting cards you have the free will to bet as much as you want. 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. It depends whether there is a shuffle between the blackjacks. However if you were going to cheat it would be much better to remove an ace, which increases the house edge by 0. You are forgetting that there are two possible orders, either the ace or the ten can be first. Streaks, such as the dealer drawing a 5 to a 16, are inevitable but not predictable. So standing is the marginally better play. 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. So the probability of winning six in a row is 0. It depends on the number of decks. 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? It took me years to get the splitting pairs correct myself. 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. It may also be the result of progressive betting or mistakes in strategy. 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 have no problem with increasing your bet when you get a lucky feeling. From my section on the house edge we find the standard deviation in blackjack to be 1. It is more a matter of degree, the more you play the more your results will approach the house edge. 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. Multiply dot product from step 11 by probability in step 9. {PARAGRAPH}{INSERTKEYS}This is a typical question one might encounter in an introductory statistics class. Here is how I did it. 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. In general the variation in the mean is inversely proportional to the square root of the number of hands you play. 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. 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. 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. It would take about 5 years playing blackjack 40 hours a week before this piece of advice saved the player one unit. Repeat step 3 but multiply by 3 instead of 2. Thanks for your kind words.