#1




4 Blackjacks in a row!
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 Does anyone know the exact odds of this happening? I know they must be astronomical. 
#2




Blue
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You are on a roll, a Royal now this! Take some advice from a neophyte, employ a side of aces in the DD. CP 
#3




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I believe it's something like 192,000 to 1 for double deck. at any rate it's a huge! . congrats.
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There Is No Second Place Winner ...even if your not brusque and merely informative. 
#4




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". 
#5




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!
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Only those who will risk going too far can possibly find out just how far one can go. We cannot direct the wind, we can only adjust our sails. 
#6




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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*n1 cards left. So for n decks the probability is 2*(4/13)*(4*n/(52*n1)), which is conveniently about 1 in 21. For example for 6 decks the answer is 2*(4/13)*((4*6)/(52*61)) = 192/4043 = 0.047489. http://wizardofodds.com/askthewizard/blackjackfaq.html 
#7




I figured about one in 150,000 or something like that.
Thanks! 
#8




Done and done. It really isn't that much extra work to keep track of 8 aces.

#9




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.

#10




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