The classic payout on player Blackjack is 3 to 2. However, some casinos change the payout to increase the house edge. The payout on blackjack thus may vary from 1:1 to 6:5. As a Blackjack hand frequency is approximately 4.8% (see the table Two Card Hand Frequency), the payout of 1:1 will increase house edge by 2.3% and the payout of 6:5 - by 1.4%. The first rule (1:1) is only rarely found, while the second (6:5) can be found at some tables with a single deck blackjack game.
I don't know if this will get you any closer to a closed form solution, but maybe it helps you think about it. So the odds of not getting blackjack the first hand would be ((52.51)-((4.16).2))/(52.51) after which point you know the deck is either out one or two face cards, out one or two aces, or neither. College Probability BlackJack: What is the probability neither you nor the dealer is dealt the blackjack. RESOLVED I have the solution, but I need a few explanations. Number of times you have to bet your bonus amount to be able to cash out winnings from it. For instance, you may get a $25 no deposit bonus with a 30x wagering requirement. This means probability of getting blackjack statistics you will have to wager a total of $750 – 30 times $25 – to cashout the maximum cap winning amount.
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L8lori
Wizard,
You have an awesome sight & I have enjoyed your articles for a few years now.. thank you for giving us this web site.
I am been running different numbers thru my head trying to come up with a way to compute the probabilities of getting 3 blackjacks in a row with a 6 deck continuous shuffle machine? Or even the odds of getting 5 blackjacks in a row with the same situation?
Also.. I saw a casino promotion where you got a drawing ticket every time you were dealt a red queen and a black ace.. is there a way to calculate my odds of getting this type of hand in blackjack?
Thank you
L8lori
You have an awesome sight & I have enjoyed your articles for a few years now.. thank you for giving us this web site.
I am been running different numbers thru my head trying to come up with a way to compute the probabilities of getting 3 blackjacks in a row with a 6 deck continuous shuffle machine? Or even the odds of getting 5 blackjacks in a row with the same situation?
Also.. I saw a casino promotion where you got a drawing ticket every time you were dealt a red queen and a black ace.. is there a way to calculate my odds of getting this type of hand in blackjack?
Thank you
L8lori
HKrandom
Odds of getting a blackjack are roughly 4/13*1/13*2 so the chances of that happening 3 times in a row are about 1 in 10,000 and the chances of getting 5 in a row are about 1 in 4,200,000. The chance of getting a red queen and a black ace is 2*2/52*2/52 or 1/338.
Ayecarumba
From the Wizard of Odds Blackjack page, the probability of getting a blackjack on a six deck game is: 0.047489. The probability of this happening three times in a row is: (.047489)^3 = .00010709743 or 1 in 9,337Edit: FYI, from the Wizard's Blackjack page, the general formula for calculating the odds of a blackjack in n decks is: 2*(4/13)*(4n/(52n-1)).
Simplicity is the ultimate sophistication - Leonardo da Vinci
mkl654321
From the Wizard of Odds Blackjack page, the probability of getting a blackjack on a six deck game is: 0.047489. Poker freeroll passwords. The probability of this happening three times in a row is: (.047489)^3 = .00010709743 or 1 in 9,337
I'm glad it's that low, because I ever actually got three blackjacks in a row, the earth would open up and swallow me and the casino.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
ElectricDreams
I'm glad it's that low, because I ever actually got three blackjacks in a row, the earth would open up and swallow me and the casino.
I got two in a row a few days ago. I was all like 'yay! Blackjack again!'
I think I colored up after that, because I wanted to leave on a high point ;-)
Ayecarumba
I saw a casino promotion where you got a drawing ticket every time you were dealt a red queen and a black ace.. is there a way to calculate my odds of getting this type of hand in blackjack?
In a six deck shoe (or CSM) there are 12 red queens and 12 black aces. The odds of getting dealt one of each is (12/312)*(12/311) = .001484 or 1-in-674.
Simplicity is the ultimate sophistication - Leonardo da Vinci
Ayecarumba
I'm glad it's that low, because I ever actually got three blackjacks in a row, the earth would open up and swallow me and the casino.
1-in-9,337 is not that unusual, when, according to the Wizard, the odds of getting a royal flush in VP, if all you did was try to get a Royal every draw are 1-in-23,081.
Simplicity is the ultimate sophistication - Leonardo da Vinci
PapaChubby
In a six deck shoe (or CSM) there are 12 red queens and 12 black aces. The odds of getting dealt one of each is (12/312)*(12/311) = .001484 or 1-in-674.
You neglected that the cards may be dealt in either order, so the probability is twice as great as you calculated.
PapaChubby
I don't think I have ever been dealt three blackjacks in a row, but I'm sure I've seen the dealer get 'em on several occasions. Sometimes the math just doesn't tell the whole story.
Ayecarumba
You neglected that the cards may be dealt in either order, so the probability is twice as great as you calculated.
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Thank you for the correction PapaChubby. This is my calculation: (24 (12 red Q's and 12 black A's) / 312 total cards in six decks) * (12 red Q's or black A's, depending on what was dealt first/311) = 1-in-337
Simplicity is the ultimate sophistication - Leonardo da Vinci
mcavanaugh8
Hey all! I'm not all that math-inclined, so I was just curious as to what the odds are (if this is even possible to calculate) of NOT being dealt a blackjack in a shoe? And how that extrapolates over multiple shoes? I ask because this past weekend I played roughly 10 shoes in 6D blackjack and was not dealt a single blackjack. The guy three seats over from me was dealt, I kid you not, 8 blackjacks in a single shoe. Should have sat there instead!
weezrDASvegas
I think you askin the complementary probability for getting a blackjack
q = 1 – p
Probability to get blackjack p ~ 5%
Probability to NOT get blackjack: q=1-5%=95% (95% of all hands you deal to yourself).
One site analyzes probably all possible situations – number of decks, number of players, cards remaining in the deck. It goes like this (urls not permitted here)
forums . / blackjack-natural-odds-probability . html
(remove spaced surrounding / .)
You can google also odds to get a blackjack.
Player with 8 blackjacks in a 6D shoe seems unreal. If it was the dealer I would have suspected the automatic shuffler.
q = 1 – p
Probability to get blackjack p ~ 5%
Probability to NOT get blackjack: q=1-5%=95% (95% of all hands you deal to yourself).
One site analyzes probably all possible situations – number of decks, number of players, cards remaining in the deck. It goes like this (urls not permitted here)
forums . / blackjack-natural-odds-probability . html
(remove spaced surrounding / .)
You can google also odds to get a blackjack.
Player with 8 blackjacks in a 6D shoe seems unreal. If it was the dealer I would have suspected the automatic shuffler.
mcavanaugh8
I think you askin the complementary probability for getting a blackjack
q = 1 – p
Probability to get blackjack p ~ 5%
Probability to NOT get blackjack: q=1-5%=95% (95% of all hands you deal to yourself).
One site analyzes probably all possible situations – number of decks, number of players, cards remaining in the deck. It goes like this (urls not permitted here)
forums . / blackjack-natural-odds-probability . html
(remove spaced surrounding / .)
You can google also odds to get a blackjack.
Player with 8 blackjacks in a 6D shoe seems unreal. If it was the dealer I would have suspected the automatic shuffler.
q = 1 – p
Probability to get blackjack p ~ 5%
Probability to NOT get blackjack: q=1-5%=95% (95% of all hands you deal to yourself).
One site analyzes probably all possible situations – number of decks, number of players, cards remaining in the deck. It goes like this (urls not permitted here)
forums . / blackjack-natural-odds-probability . html
(remove spaced surrounding / .)
You can google also odds to get a blackjack.
Player with 8 blackjacks in a 6D shoe seems unreal. If it was the dealer I would have suspected the automatic shuffler.
Hmm, so do you mean it's a 95% chance when being dealt a hand that it will not be a natural blackjack? That seems to make sense. But if it's 95% per hand that you won't get one, my real question is that if you play through 10 shoes and you have a 5% chance of getting blackjack on a hand, what are the chances that never occurs? I don't think I'd ever gone ~3 shoes without one before this weekend! Haha.
And yeah, it was crazy! He had four in a row at one point.. the dealer was flabbergasted.
Unfortunately on the next shoe the dealer had three in a row, sandwiched between a couple of 20s and 21s. Was not a good day for me!
![Probability Of Getting A Blackjack Probability Of Getting A Blackjack](/uploads/1/2/5/2/125215835/140226122.png)
gordonm888
the answer to your question depends upon the number of hands you can expect to be dealt in 6 shoes, which in turn depends upon the number of players at the table (you have already indicated there was at least one other player.)So: How many players were at the table?
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
weezrDASvegas
Hmm, so do you mean it's a 95% chance when being dealt a hand that it will not be a natural blackjack? That seems to make sense. But if it's 95% per hand that you won't get one, my real question is that if you play through 10 shoes and you have a 5% chance of getting blackjack on a hand, what are the chances that never occurs? I don't think I'd ever gone ~3 shoes without one before this weekend! Haha.
And yeah, it was crazy! He had four in a row at one point.. the dealer was flabbergasted.
Unfortunately on the next shoe the dealer had three in a row, sandwiched between a couple of 20s and 21s. Was not a good day for me! forums . / saliu . com / blackjack-natural-odds-probability . html
And yeah, it was crazy! He had four in a row at one point.. the dealer was flabbergasted.
Unfortunately on the next shoe the dealer had three in a row, sandwiched between a couple of 20s and 21s. Was not a good day for me! forums . / saliu . com / blackjack-natural-odds-probability . html
You described a phishy session in that casino. Automatic or continuous shufflers in “beast mode” maybe?
NOT a blackjack in 6D is the opposite of 1 bj in 6D. The site I pointed you to was not printed correctly here because of no urls allowed policy. It has the formulas for many situations. Number of players is very important also. Number of bjs per deck should be distributed equally amongst all players. Like every player should get 2 bjs on average per deck if plying some 10 decks.
I don’t trust those automatic shufflers for the life of me. But they are part of life now.
weezrDASvegas
The probability of getting a blackjack is a hypothetical situation looks like. You shuffle a deck and deal 2 cards. If you do it 100 times, the 2-card hand is a bj in 5 situations; in other 95 shuffles the hand is not a blackjack.
But there is the dealer also. So you get the bj every 21 cards and so does the dealer. That should be also the case with dealer plus 5 players in a 6D shoe. Of course the cards are not perfectly distributed. In one 6D shuffle you might get 8 bjs while the dealer gets one or none—and vice versa. That is very rare if the game is fair. It’s hard to believe you got NO bj in 10 6D shuffles even considering penetration. I never remember something similar happening to me.
But there is the dealer also. So you get the bj every 21 cards and so does the dealer. That should be also the case with dealer plus 5 players in a 6D shoe. Of course the cards are not perfectly distributed. In one 6D shuffle you might get 8 bjs while the dealer gets one or none—and vice versa. That is very rare if the game is fair. It’s hard to believe you got NO bj in 10 6D shuffles even considering penetration. I never remember something similar happening to me.
BlackjackGuy123
Depends, how many players are playing?
If it is just heads up and we are talking about a shoe where there is 1 decks cut off out of six, then we can expect to get 52 rounds in playing heads up. So the odds of not getting blackjack in one trial is 95.17%.
.9517 to the 52nd power is 7.6%
But at a crowded table the probability would be much higher.
If it is just heads up and we are talking about a shoe where there is 1 decks cut off out of six, then we can expect to get 52 rounds in playing heads up. So the odds of not getting blackjack in one trial is 95.17%.
.9517 to the 52nd power is 7.6%
But at a crowded table the probability would be much higher.
mcavanaugh8
the answer to your question depends upon the number of hands you can expect to be dealt in 6 shoes, which in turn depends upon the number of players at the table (you have already indicated there was at least one other player.)
So: How many players were at the table?
So: How many players were at the table?
Calculating Blackjack Probabilities
It was a full table--maybe a hand or two lag between a player getting up and another sitting down, but it was almost always full for the shoes I was in.
mcavanaugh8
Depends, how many players are playing?
If it is just heads up and we are talking about a shoe where there is 1 decks cut off out of six, then we can expect to get 52 rounds in playing heads up. So the odds of not getting blackjack in one trial is 95.17%.
.9517 to the 52nd power is 7.6%
But at a crowded table the probability would be much higher.
If it is just heads up and we are talking about a shoe where there is 1 decks cut off out of six, then we can expect to get 52 rounds in playing heads up. So the odds of not getting blackjack in one trial is 95.17%.
.9517 to the 52nd power is 7.6%
But at a crowded table the probability would be much higher.
Full table in 6D BJ with about 2 decks penetration on average.
Hmm interesting. Maybe not as uncommon as I thought, then.
mcavanaugh8
The probability of getting a blackjack is a hypothetical situation looks like. You shuffle a deck and deal 2 cards. If you do it 100 times, the 2-card hand is a bj in 5 situations; in other 95 shuffles the hand is not a blackjack.
But there is the dealer also. So you get the bj every 21 cards and so does the dealer. That should be also the case with dealer plus 5 players in a 6D shoe. Of course the cards are not perfectly distributed. In one 6D shuffle you might get 8 bjs while the dealer gets one or none—and vice versa. That is very rare if the game is fair. It’s hard to believe you got NO bj in 10 6D shuffles even considering penetration. I never remember something similar happening to me.
But there is the dealer also. So you get the bj every 21 cards and so does the dealer. That should be also the case with dealer plus 5 players in a 6D shoe. Of course the cards are not perfectly distributed. In one 6D shuffle you might get 8 bjs while the dealer gets one or none—and vice versa. That is very rare if the game is fair. It’s hard to believe you got NO bj in 10 6D shuffles even considering penetration. I never remember something similar happening to me.
Variance was not on my side, it seems. I was honestly baffled--I was dealt PLENTY of tens and aces. I can't even remember how many soft hands I doubled on against dealer low cards. I just never got the ten and ace together lol.
Blackjack Probability Chart
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