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.

Quote:weezrDASvegasI 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.

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!

So: How many players were at the table?

Quote:mcavanaugh8Hmm, 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.

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.

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.

Quote:gordonm888the 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.

Quote:BlackjackGuy123Depends, 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.

Quote:weezrDASvegasThe 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.