Today I got 4 blackjacks in a row. On a double deck game no less! Too bad I was betting table minimum for the last 3 :laugh: Does anyone know the exact odds of this happening? I know they must be astronomical.

Blue You are on a roll, a Royal now this! Take some advice from a neophyte, employ a side of aces in the DD. CP

I believe it's something like 192,000 to 1 for double deck. at any rate it's a huge! .:laugh: congrats.

consecutive BJ's No, it is nothing remotely like 192,000 to 1 -- UNLESS the precise 4 hand sequence is predicted in advance. That is to say, that with every hand that you play you create yet another 4 hand sequence starting point. Ergo, you will have hundreds, even thousands of chances to pull 4 consecutive B J' s in the course of just a year. On an interesting note, I was playing single deck in No. Nevada once when the dealer gave herself 4 consecutive 'snappers'. It turned out that black chip players with skill were not barred at J.A.'s Nugget in Sparks. They were simply cheated until they "got the message".

Very nice. I've had three on a bunch of occasions, and even have gotten the Ten or Ace on the next hand, but have yet to seal the deal. Now get to work on getting 5!

odds for 1 blackjack is about 1 in 20 (closer to 1 in 21 but this keeps the math simpler) odds for 2 in a row is one in 400. odds for 3 in a row is one in 8,000 odds for 4 in a row is one in 160,000 How did you calculate those numbers? The general formula is 2*(probability of 10)*(probability of ace). The reason for the 2 is that there are two ways to order the two cards; either the ace or 10 can come first. The probability that the first card will be a 10 is always 4/13, regardless of the number of decks, because there are 16 out of 52 tens in the deck and 16/52 = 4/13. The probability of an ace, given that a 10 has already been removed from the deck is the number of aces divided by the number of cards left. Let n be the number of decks. There are 4*n aces and 52*n-1 cards left. So for n decks the probability is 2*(4/13)*(4*n/(52*n-1)), which is conveniently about 1 in 21. For example for 6 decks the answer is 2*(4/13)*((4*6)/(52*6-1)) = 192/4043 = 0.047489. http://wizardofodds.com/askthewizard/blackjack-faq.html

Inplay, is this assuming we have an infinite number of decks present - see if we are playing a 6 deck games, we only have 24 aces present and if one ace is depleted, the chances of another successive BJ further decreases. I wont worry about the tens because there are 96 tens compared to 24 aces.

Nice! Not sure where you were with the 4 in a row...I got 4 in six hands at Mystic two weeks ago on the DD. Of course those were about the only hands I won. rock

i also got four in a row, two weeks ago. and then i got another one two hands later-so 5 out of 6 hands...and i was betting my minimum because the count was neutral..too bad..it's funny because the whole table was like "wow you're so lucky" but it really didn't do anything for me with my ten dollar bet...too bad it never happens at those high counts..

Just to upstage everyone I recently drew 4oak at a VP game with a probability of approx 1 in 400 3 hands in a row. I make that approximately a 1 in 64000000 chance RJT

In a row 4 in a row! WOW! I have banged 3 in a row numerous times. I once had three hands on the table and banged a blackjack on all three. Both of these sorts of events are incredibly rare and I don't think I have ever hit 4 BJ's in a row and I play 20-60 hours a week! I have done 4 or 5 blackjacks in 6-7 hands but not consecutive.

Not to one-up but i have received 5 in a row, with a winning double-down to follow. Single deck, obviously with a shuffle (after the first two, with a dealer change). Was fun, but didn't have big bets out. Before counting, so had a slight progression after the first two BJ's. SSSSSSSSSSStretchhhh