Probability Of Getting A Blackjack With One Deck

| 24-05-2021
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The probability of winning for the player sum 12-16 should ideally be equal to the probability of dealer going burst.Your chance of getting a blackjack is now 16.9%.Casino regina kenny rogers - Probability of getting a blackjack with one deck - Winner casino uk By admin on May 19, 2016 888 casino refer a friend its private longer The the less. Adding the probabilities of the two different orders in which you can get blackjack (16/663 + 16/663) yields the total probability of getting blackjack. 16/663 + 16/663 = 32/663 = 4.827%. For a 6-deck shoe, the same principles apply but the number of cards changes.How To Calculate The Probability Of Blackjack. How to calculate the probability. 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/ 176 = 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.

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
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 9,337
Edit: 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. 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?

Probability Of Blackjack Single Deck


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.


Probability blackjack questions1-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.


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.
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.


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.

Blackjack Probability Chart

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!
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


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.
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.
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?


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.


Full table in 6D BJ with about 2 decks penetration on average.
Hmm interesting. Maybe not as uncommon as I thought, then.
mcavanaugh8

Probabilities In Blackjack

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.


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.

What Is The Probability Of Getting Blackjack With One Deck

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