# Blackjack - Probability

Kevin

This is a typical question one might encounter in an introductory statistics class. 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.

From my section on the house edge we find the standard deviation in blackjack to be 1.17. You won't understand this if you haven't studied statistics, but the probability of being at a loss in your example will be the Z statistic of 45000*0.005/(45000^{1/2}*1.17) =~ 0.91.

Any basic statistics book should have a standard normal table which will give the Z statistic of 0.8186. So the probability of being ahead in your example is about 18%.

John from Westminster, USA

When the dealer stands on a soft 17, the dealer will bust about 29.1% of the time. When the dealer hits on a soft 17, the dealer will bust about 29.6% of the time. According to my blackjack appendix 4, the probability of a net win is 42.42%. However, if we skip ties, the probability is 46.36%. So, the probability of a four wins in a row is 0.4636^{4} = 4.62%.

Anyway, on to my question. Well, more of an observation: when the dealer pulls a 5 on a 16 for their sixth consecutive win, there's always someone who gets up and leaves the table, muttering that the dealer is a mean cruel heartless soul, and goes in search of a "hotter" table. But is there any truth in this? Obviously the dealer is inconsequential to the cards dealt (I like to say the dealer is "simply a messenger of the cards") but are streaks in an 8-deck shoe inevitable, and even predictable? Or is it more like your roulette example, where the odds of each new round are exactly the same? Thanks once again for your web site.

Dave K. from Beverly Hills, California

Thanks for your kind words. Streaks, such as the dealer drawing a 5 to a 16, are inevitable but not predictable. Blackjack is not entirely a game of independent trials like roulette, but the deck is not predisposed to run in streaks. For the non-card counter it may be assumed that the odds are the same in each new round. 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.

James from Palo Alto, California

According to my blackjack appendix 4, the probability of an overall win in blackjack is 42.22%, a tie is 8.48%, and a loss is 49.10%. I'm going to assume you wish to ignore ties for purposes of the streak. In that case, the probability of a win, given a resolved bet, is 46.36%. The probability of winning n hands is a row is 0.4636^{n}. So the probability of winning six in a row is 0.99% and seven in a row is 0.46%.

Chris from Gaithersburg, Maryland

What you have experienced is likely the result of some very bad losing streaks. It may also be the result of progressive betting or mistakes in strategy. The basic strategy flat bettor should have a roughly symmetrical expectation in terms of steep ups and downs, slightly favoring steep downs due to the house edge and a 48% chance of a losing hand compared to 43% chance of winning. If I'm playing for fun then I leave the table when I'm not having fun any longer.

Ed from Lynnwood, USA

Let n be the number of decks. The probability of a blackjack is 2*(4/13)*(4n/(52*n-1)). If n=6 the probability is 192/4043 = 4.75%.

Kim from Helsinki, Finland

Thanks for the kind words. You ask a good question for which there is no firm answer. It is more a matter of degree, the more you play the more your results will approach the house edge. I recently replaced my blackjack appendix 4 with some information about the standard deviation which may help. For example this table shows that if you play 10,000 hands of blackjack the probability is 90% of finishing within 192 units where you started after subtracting the expected loss due to the house edge. So in 10,000 hands you are likely to win or lose less than 2% of total money bet due to random variation. However if we go up to one million hands the probability is 90% of an 0.2% variation due to luck. In general the variation in the mean is inversely proportional to the square root of the number of hands you play. All of this assumes flat betting, otherwise the math really gets messy.

Mike from Bossier City, USA

The probability that the first card is an ace is 4/52. The probability that the second card is a 10 point card is 16/51. So the probability of an ace first blackjack is (4/52)*(16/51). Multiply this by 2 because the ten could just as easily be the first card and the answer is 2*(4/52)*(16/51) = 128/2652 = 0.0482655, or about 1 in 20.7 .

Mike

Since this question was submitted, a player held the dice for 154 rolls on May 23, 2009 in Atlantic City. The probability of this is 1 in 5,590,264,072. For the probability for any number of throws from 1 to 200, please see my craps survival tables. For how to solve the problem yourself, see my MathProblems.info site, problem 204.

Blair from Christchurch, New Zealand

Your expected loss would be 100*$5*.005=$2.50. The standard deviation of one hand is 1.17, which can be found in my blackjack appendix 7. So, one standard deviation in your example is $5*1.17*sqr(100)=$58.5. So, the probability of losing $295 or more due to bad luck is .00135 (the Z statistic for -3).

T from Las Vegas, U.S.

There are 103 cards remaining in the two decks and 32 are tens. So the probability of a blackjack is 32/103=31.07%.

RWR from Tuscon, USA

The probability of a suited blackjack in a six-deck game is 2*(4/13)*(6/311) = 0.0118723.

Rodrigo from Costa Rica

I attempt to work this out in my blackjack appendix 8 but I’ll work through it more slowly here. We’ll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin(312,3)=5,013,320. There are 24 sevens in the shoe. The number of ways to arrange 3 sevens out of 24 is combin(24,3)=2024. The probability is the number of winning combinations divided by total combinations, or 2024/5013320=0.0004, or about 1 in 2477.

Jeffrey from Loveland

You are forgetting that there are two possible orders, either the ace or the ten can be first. Multiply by 2 and you’ll have your answer.

Victor from Yakima, Washington

According to Stanford Wong’s ’Basic Blackjack’ he says the player’s edge given the first card is an ace is 50.5% (page 124). Your question however could be rephrased as, "what is the value of the ace, given that the other card is not a ten." Using an infinite deck for the sake of simplicity we can breakdown Wong’s number as follows: 0.505 = (4/13)*1.5 + (9/13)*x, where x is what you want to know. Doing some simple algebra we get x=28.5%.

Rodney from Clarence, New York

Yes! Good question, even I didn’t know this. The fewer the decks and the greater the number of cards the more this is true. To test the most likely case to favor hitting, 8 decks and only 3 cards, I ran every possible situation through my combinatorial program. The following table displays the results.

### Expected Values for 3-card 16 Vs. 10 in 8-deck game

Hand | EV Hit | EV Stand | Best Play |
Probability | Return Hit |
Return Stand |

1/5/10 | -0.540978 | -0.539872 | Stand | 0.132024 | -0.071422 | -0.071276 |

1/6/9 | -0.536558 | -0.540151 | Hit | 0.059837 | -0.032106 | -0.032321 |

1/7/8 | -0.537115 | -0.537003 | Stand | 0.059837 | -0.032139 | -0.032133 |

2/4/10 | -0.540947 | -0.541 | Hit | 0.237478 | -0.128463 | -0.128475 |

2/5/9 | -0.542105 | -0.540534 | Stand | 0.039891 | -0.021625 | -0.021563 |

2/6/8 | -0.537701 | -0.540773 | Hit | 0.059837 | -0.032174 | -0.032358 |

2/7/7 | -0.538271 | -0.537584 | Stand | 0.028983 | -0.015601 | -0.015581 |

3/3/10 | -0.540385 | -0.540995 | Hit | 0.115028 | -0.06216 | -0.06223 |

3/4/9 | -0.541769 | -0.540536 | Stand | 0.059837 | -0.032418 | -0.032344 |

3/5/8 | -0.54295 | -0.540022 | Stand | 0.039891 | -0.021659 | -0.021542 |

3/6/7 | -0.538575 | -0.540228 | Hit | 0.059837 | -0.032227 | -0.032326 |

4/4/8 | -0.543188 | -0.54003 | Stand | 0.028983 | -0.015743 | -0.015652 |

4/5/7 | -0.544396 | -0.539483 | Stand | 0.039891 | -0.021717 | -0.021521 |

4/6/6 | -0.539446 | -0.542878 | Hit | 0.028983 | -0.015635 | -0.015735 |

5/5/6 | -0.545033 | -0.542137 | Stand | 0.009661 | -0.005266 | -0.005238 |

Total | 1 | -0.540355 | -0.540293 |

The two right numbers in the bottom row show that the overall expected value for hitting is -0.540355 and for standing is -0.540293. So standing is the marginally better play. Following this rule will result in an extra unit once every 1117910 hands. It would take about 5 years playing blackjack 40 hours a week before this piece of advice saved the player one unit.

Richard S. from Memphis, USA

According to my blackjack appendix 9H the expected return of standing is -0.476476 and of hitting is -0.408624, assuming the 16 is composed of a 10 and 6. So my hitting you will save 6.79 cents for each dollar bet. This is not even a marginal play. There is no sound bite answer to explain why you should hit. 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. 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 10. Deviating on these hands will cost you much less.

Beau from Toronto, Canada

(1) It depends what happens if the dealer does have a blackjack. If the player is guaranteed to lose no more than the original wager then it doesn’t matter if the dealer takes a second card or not. If the player stands to lose the total amount bet after doubling or splitting and the dealer gets a blackjack then that works to the dealer’s advantage. (2) I don’t have to simulate this because the number of players makes no difference.

"Anonymous" .

It depends on the number of decks. If the number of decks is n then the probability is 2*pr(ace)*pr(10) = 2*(1/13)*(16*n/(52*n-1)), which is conveniently about 1 in 21.
Here is the exact answer for various numbers of decks.

### Probability of Blackjack

Decks | Probability |
---|---|

1 | 4.827% |

2 | 4.780% |

3 | 4.764% |

4 | 4.757% |

5 | 4.752% |

6 | 4.749% |

7 | 4.747% |

8 | 4.745% |

Matt from Radford, USA

If the probability of a blackjack is p then the probability of not getting any blackjacks in 10 hands is 1-(1-p)^{10}. For example in a six deck game the answer would be 1- 0.952511^{10}= 0.385251.

Joe P from Parma Heights, USA

I’m going to assume there is never a shuffle between hands. The three other players don’t matter. The answer would be 2^{3}*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47)= 0.00004401, or about 1 in 22722. If there were a shuffle between hands the probability would increase substantially.

Steve from Solva, United Kingdom

The number of hands doesn’t matter. The probability is 2*(4/13)*(8/103) = 0.0478.

J.A.S. from Las Vegas, USA

It depends whether there is a shuffle between the blackjacks. Assuming there isn’t the probability would be 8*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47) = 0.000044011058. The number of other players doesn’t matter, except if they cause a shuffle.

Matt

From my blackjack appendix 7 we see that each 9 removed from a single deck game increases the house edge by 0.20%. However if you were going to cheat it would be much better to remove an ace, which increases the house edge by 0.58%. If you were to add a card as the dealer you should add a 5, which increases the house edge by 0.80%. So, the best card for the player is the ace and the best for the dealer is the 5.

That column seemed to put the mathematics to that "feeling" a player can get. In that columns’ example of a player losing 8 consecutive hands of blackjack the odds were (.5251^8 or about 1 in 173). My question though is what does that really mean? Is it that when I sit down at the table, 1 out of my next 173 playing sessions I can expect to have an 8 hand losing streak? Or does it mean that on any given loss it is a 1 in 173 chance that it was the first of 8 losses coming my way?

I know, I know, its some sort of divine intervention betting system I am talking about and no betting system affects the house edge. I’m still curious though. 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.

Steve from Phoenix, AZ

I have no problem with increasing your bet when you get a lucky feeling. What is important is that you play your cards right. Unless you are counting cards you have the free will to bet as much as you want. 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. When I said the probability of losing 8 hands in a row is 1 in 173 I meant that starting with the next hand the probability of losing 8 in a row is 1 in 173. The chances of 8 losses in a row over a session are greater the longer the session. I hope this answers your question.

Jason

The probability of this occurring in which your other two cards are any two 10-point cards is 4*COMBIN(6,4)*COMBIN(6*16,2)*(4/6)*(3/5)*(1/2)/combin(312,6) 1 in 22,307,231. However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. I would have to do a computer simulation to consider all the other combinations. However to make a rough guess I’d say the 7 million looks about right.

"Anonymous" .

It took me years to get the splitting pairs correct myself. Cindy of Gambling Tools was very helpful. Peter Griffin also addresses this topic in chapter 11 of the The Theory of Blackjack Let’s say I want to determine the expected value of splitting eights against a dealer 2. Resplitting up to four hands is allowed. Here is how I did it.

- Take a 2 and two 8’s out of the shoe.
- Determine the probability that the player will not get a third eight on either hand.
- 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. For each rank determine the probability of that rank, given that the probability of another 8 is zero. Take the dot product of the probability and expected value over each rank.
- Multiply this dot product by the probability from step 2.
- Determine the probability that the player will resplit to 3 hands.
- Take another 8 out of the deck.
- Repeat step 3 but multiply by 3 instead of 2.
- Multiply dot product from step 7 by probability in step 5.
- Determine the probability that the player will resplit to 4 hands.
- Take two more 8’s out of the shoe.
- 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.
- Multiply dot product from step 11 by probability in step 9.
- Add values from steps 4, 8, and 12.

The hardest part of all this is step 3. I have a very ugly subroutine full of long formulas I determine using probability trees. It gets especially ugly when the dealer has a 10 or ace up.

Geoff

Let’s assume six decks of cards and the player always takes a third card (whether by hitting or splitting). The number of ways to draw 3 suited sevens is the number of suits (4) times the number of ways to choose 3 out of 6 sevens of that suit in the shoe. In other words 4×combin(6,3)=4×20=80. The number of ways to draw 3 colored sevens, including 3 suited sevens, is the number of colors times the number of ways to pick 3 out of the 12 sevens in the shoe of that color, or 2×combin(12,3)=2×220=440. The number of ways to draw any 3 sevens, including 3 colored and suited sevens, is the number of ways to pick 3 cards out of the 24 sevens in the shoe, or combin(24,3)=2024. The total number of combinations for any 3 cards out of 312 is combin(312,3)=5013320. So the probability of 3 suited sevens is 80/5013320=0.000015957. The probability of three colored, but not suited, sevens is (440-80)/5013320=0.0000718. The probability of three sevens of mixed colors is (2024-440)/5013320= 0.00031596.

"Anonymous" .

Yes, I calculate blackjack odds using a combinatorial approach, analyzing every possible ways the player and dealer cards can come out, taking the greatest expected value at every decision point. This is harder to program than a simulation but I feel is more elegant and a nice challenge in recursive programming. However I still respect my peers to do simulations. With today’s computers it doesn’t take long to run off a billion bets, which gets very close to the optimal strategy return.

"Anonymous" .

Not too many places allow resplitting aces, so be glad you were playing somewhere that did. Your seat position does not matter. The probability of this is the probability that the first four cards out of the shoe are aces, and the next four are tens, or (combin(24,4)/combin(312,4))*(combin(96,4)/combin(308,4)) = 1 in 4,034,213.

"Anonymous" .

I seem to get a variation of this question at least once a month. Let’s assume for now the deck is shuffled after every hand, to make the math easier. If the probability of something happening is p then the probability of it happening n times in a row is p^{n}. The probability of a blackjack in a single deck game is 4*16/combin(52,2) = 64/1326. So the probability of four in a row is (64/1326)^{4} = 16777216/ 3091534492176 = 1 in 184270. However the actual probability is much less, because as the player gets each blackjack the ratio of aces to cards left in the deck decreases. Without knowing what cards the dealer got I can’t tell you the exact answer.

"Anonymous" .

You’re welcome, thanks for you compliments. The probability of being dealt 3 jokers in a row from a six deck shoe (plus the 3 jokers) is 1/combin(315,3) = 1 in 5,159,805. Another solution is (3/315)*(2/314)*(1/313).

"Anonymous" .

Every legitimate blackjack expert agrees the house edge decreases as the number of decks goes down, all other rules being equal. However it is hard to explain why. First, it is true that you are more likely to get one small card and one big card in single-deck than multiple-deck. For example if we define a small card as 2 through 6, and a large card as any 10-point card or ace then the probability of getting one of each in single deck is 2*(20/52)*(20/51) = 30.17%. The probability in 8-decks is 2*(160/416)*(160/415) = 29.66%. Although stiffs can cut both ways the player has the free will to stand, the dealer must always hit them.

"Anonymous" .

Following are the probabilities:

Player 1 0.048265

Player 2 0.036735

Player 3 0.024823

Dealer 0.012560

The product is 1 in 1,808,986.

"Anonymous" .

Thanks for the compliment. I’m afraid I know of no source, including myself, that shows code for game analysis. It took me years to get my blackjack engine to work perfectly (splits when the dealer had a 10 or ace showing was very tricky). An easier way to get the house edge for blackjack is to write a random simulation. One of these days I would like to write a book on how I analyzed the games, but I’m afraid only you would buy it.

"Anonymous" .

2*(4/13)*(4n/(52n-1))

"Anonymous" .

The probability is (4/52)*(3/51)*(2/50)*(1/49)*(4/48) = 1 in 3,248,700.

"Anonymous" .

Thanks for the compliment.

Assuming a six-deck game, where the dealer stands on soft 17, and the player plays basic strategy here are the rounded results based on a 100-million hand simulation.

### Player Hand Probabilities

Event | Probability |
---|---|

Dealer has only blackjack | 1 in 22 |

Player doubles or splits | 1 in 7.7 |

2 cards | 1 in 2.3 |

3 cards | 1 in 3.8 |

4 cards | 1 in 10 |

5 cards | 1 in 50 |

6 cards | 1 in 400 |

7 cards | 1 in 4,600 |

8 cards | 1 in 79,000 |

9 cards | 1 in 2,200,000 |

10 cards | 1 in 100,000,000 |

### Dealer Hand Probabilities

Event | Probability |
---|---|

Player has only blackjack | 1 in 22 |

2 cards | 1 in 3.0 |

3 cards | 1 in 2.4 |

4 cards | 1 in 6.1 |

5 cards | 1 in 31 |

6 cards | 1 in 270 |

7 cards | 1 in 3,700 |

8 cards | 1 in 79,000 |

9 cards | 1 in 2,200,000 |

10 cards | 1 in 100,000,000 |

"Anonymous" .

If you had a game with no house edge the probability of winning $200 with $5000 to risk, using any system, would be 5000/(5000+200) = 96.15%. The general formula for winning w with a bankroll of b is b/(b+w). So the larger the bankroll the better your chances. The house edge will lower the probability of success by an amount that is hard to quantify. For a low house edge game like blackjack, the reduction in the probability of success will be small. It would take a random simulation to know for sure. Forgive me if I don't bother with that. VegasClick did a small simulation about the probability of success with the Martingale.

Frank from Michigan

This is not true. The remaining deck needs to be exhibit more than a certain degree of skewness for the odds to swing to the player's favor. Consider a hypothetical side that pays 3 to 1 for any suited pair in a one-deck game. Of the top of the deck the probability of winning is 4*combin(13,2)/combin(52,2) = 23.53%. However if you burn two cards of different suits the probability of winning goes down to 2*(combin(13,2)+combin(12,2))/combin(50,2) = 23.51%. If you burn two cards of the same suit the probability of winning increases to (3*combin(13,2)+combin(11,2))/combin(52,2) = 23.59%. If one card of each rank were removed the probability of winning would go down to 4*combin(12,2)/combin(48,2) = 23.40%. What all this shows is that if cards are removed at a uniform distribution the odds of winning go down, however at a very skewed distribution the odds go up. As the deck is played down sometimes your odds get better, and sometimes worse, but in the long run they average out and stay at a 23.53% chance of winning.

Mark from Las Vegas

Under typical Vegas rules (6-deck, dealer hits soft 17) the house edge by always standing is 15.7%. In the short-run one could still overcome that, but in the long run you’ll lose badly.

Adam from Toronto

At Cryptologic they use 8 decks and the dealer stands on a soft 17. According to my blackjack appendix 2, the probability of the dealer busting with a 6 up is 0.422922. So the probability of not busing is 1 - 0.422922 = 0.577078. The probability of not busing 7 times out of 7 is (0.577078)^{7} = 2.13%.

Timothy Rowland from Orilila

Wow! The probability of this is (combin(24,6)/combin(312,6)) * (24/306) * (combin(18,5)/combin(305,5)) = 1 in 287,209,346,813,617.

Jesse from Scottsdale

That probability would be 52/combin(260,5) = 5/9525431552 = 1 in 1,905,086,310.

Cameron from Melbourne, Australia

Assuming liberal Vegas Strip rules (six decks, dealer stands on soft 17, double after split allowed, late surrender allowed, resplitting aces allowed) the following are the probabilities of each possible outcome when doubling on the initial two cards. This does not include doubling after splitting.

Ken from Auckland, New Zealand

From my blackjack appendix 4 we see the following probabilities for each initial hand.

- Win 42.43%
- Lose 49.09%
- Draw 8.48%

So the probability of going exactly 19 losses in a row is 0.4909^19*(1-0.4909) = 1 in 1,459,921. By way of comparison, the probability of being dealt a royal flush in video poker is 1 in 649,740, or 2.25 times as likely. Avid video poker players have been known to receive several dealt royals, so if you play a lot of blackjack you’ll likely hit such a losing streak eventually.

Ernie from Toledo

The reason the strategy changes, according to the number of cards in your hand, as shown in appendix 18, is that every card that leaves the deck changes the probabilities of every card left to be played. A good example is the single-deck basic strategy says to surrender 7,7 against a 10; but for any other 14 you should hit. The reason you should surrender is half the sevens have already been removed from the deck. You need another seven to make 21, the only hand that will beat a dealer 20. So the shortage of sevens lowers the expected value of hitting to under half a bet, making surrender the better play.

In an eight-deck shoe there are 416 cards. That may seem like a lot, but 16 against a 10 is such a borderline hand that removal of just one card can making standing a better play. The rule is that for eight or fewer decks if your 16 is composed of three or more cards, and the dealer has a 10, then you should stand. In a two-card 16 the average points per card is 8, with a 3-card 16 the average is 5.33. With more small cards out of the deck in the 3-card hand the remaining deck becomes more large card rich, making hitting more dangerous, swaying the odds in favor of standing.

Stan from The Netherlands, Europe

I show that rule is worth 0.026% to the player. Despite the incentive to hit 7,7 against a dealer 2-7, the player should still follow basic strategy and split.

Scott from Long Beach

My blackjack appendix 14 shows that if your first card is a six your expected value is already about −21%. For example, if he bet $100, a fair price to sell the hand and bet would be about $79. Maybe you can take advantage of his complaining by offering to buy his hand for less than the fair 79 cents on the dollar. I’d suggest 75 cents on the dollar, to give you an edge, without taking too much advantage.

Michael L. from West Mifflin, PA

For the benefit of other readers, the full set of rules is:

- Single deck.
- Dealer stands on soft 17.
- Winning blackjack pays even money.
- Player may double any first two cards.
- No double after split.
- Player may resplit to four hands, including aces.
- No draw to split aces.
- No surrender.
- Six-card Charlie (player unbusted six cards automatically wins).
- Cards shuffled after every hand.
- If game runs out of cards, all unbusted player hands automatically win.

The house edge using total-dependent basic strategy is 2.13%. I ran a 7-player simulation, using total-dependent basic strategy, and the average number of cards used per round was 21.65, with a standard deviation of 2.72. In almost 190 million rounds played, the most cards ever used was 42, which happened 7 times.

It is my educated opinion that even with computer perfect composition-dependent strategy the player would still realistically never see the last card. You could cut down the house edge much more using composition-dependent strategy, according to all the cards seen as you go along. However bucking 2.13% house edge to start with, you’ll never get anywhere near break-even, regardless of how hard you try.

I just wanted to express my disappointment in this change, if it is true. I never had a chance to take advantage of the promotion and doubt I will be able to now. The amount of time necessary to receive 30 blackjacks (I’m told about 8 hours of continuous play) seems unreasonable at $15/hand when the promotion still offers only $100.

Here is the reply I received:

In response to your e-mail on the blackjack blackout promotion, I’m not sure where you received your information on how long it takes to complete the blackout card. We’ve seen players complete the card in less than four hours. Also, you have thirty days in which to complete the card. I hope you understand this is not a task that is unreachable with that much time. I THANK YOU for your letter. It’s good to hear feedback from our customers. Hope you can give it a try and win some money!

What is the probability of getting 30 blackjacks in four hours?

nyuhoosier

According to my game comparison, blackjack players play about 70 hands per hour. The probability of a blackjack in a six-deck game is 24*96/combin(312,2)=4.75%. I assume a blackjack tie still gets a stamp. So it should take about 30/0.0475=632 hands to fill the card, or 9.02 hours.

The probability of filling the card in 4 hours, assuming 280 hands, is 1 in 30,000 playing one hand at a time. I suspect any player achieving the goal in four hours was playing at least two hands at a time.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

aceofspades

The probability of the dealer getting exactly a 9-card 21 under those rules is 1 in 32,178,035. Here is the probability for various numbers of decks and whether dealer hits or stands on soft 17.

### Probability of Dealer 9-Card 21

Decks | Stand Soft 17 | Hit Soft 17 |
---|---|---|

1 | 1 in 278,315,855 | 1 in 214,136,889 |

2 | 1 in 67,291,581 | 1 in 41,838,903 |

4 | 1 in 38,218,703 | 1 in 22,756,701 |

6 | 1 in 32,178,035 | 1 in 18,980,158 |

8 | 1 in 29,749,421 | 1 in 17,394,420 |

Assuming six decks and the dealer stands on soft 17, here is the probability of the dealer getting a 21 (or a blackjack in the case of two cards), according to the total number of cards.

### Probability of Dealer 21/BJ

by Number of Cards

Cards | Probability |
---|---|

2 | 1 in 21 |

3 | 1 in 19 |

4 | 1 in 56 |

5 | 1 in 323 |

6 | 1 in 3,034 |

7 | 1 in 42,947 |

8 | 1 in 929,766 |

9 | 1 in 32,178,035 |

10 | 1 in 1,986,902,340 |

11 | 1 in 270,757,634,011 |

12 | 1 in 167,538,705,629,468 |

Not that you asked, but the next table shows the probability of the dealer making any non-busted hand under the same rules by the number of cards.

### Probability of Dealer 17-21/BJ

by Number of Cards

Cards | Probability |
---|---|

2 | 1 in 3 |

3 | 1 in 4 |

4 | 1 in 12 |

5 | 1 in 67 |

6 | 1 in 622 |

7 | 1 in 8,835 |

8 | 1 in 193,508 |

9 | 1 in 6,782,912 |

10 | 1 in 424,460,108 |

11 | 1 in 58,597,858,717 |

12 | 1 in 36,553,902,750,535 |

For more discussion about this question, please visit my forum at Wizard of Vegas.

At Resorts World, they allow "even money" in blackjack, including tables that pay 6 to 5 on a winning blackjack. How much does this lower the house edge?

Jay

For my answer, I shall assume six decks of cards.

If the dealer pays 6 to 5 only on a winning blackjack, then "even money" is usually not allowed. However, I'll take your word for it that they do offer it at Resorts World.

At a win of 6 to 5, a player blackjack against an ace is worth 83% of the bet amount. So, to get 100% is a great deal. This situation will happen with probability 0.352%. Overall, the is worth 0.00352 × (1 - 0.83) = 0.0006 to the player. In other words, it decreases the house edge by 0.06%.

I must remind my readers that if blackjack pays 3 to 2, the player should decline it. In that case, a blackjack against an ace is worth 1.037 times the bet amount, so to accept only one unit would be a bad decision.