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#1
April 28th, 2010, 12:53 PM
 rabbit215 Member Join Date: Jan 2010 Posts: 25
Numbers

Im looking for some info about the game.

Whats the probability of a dealer to win

1 time

2 times in a row

3 times in a row

4 5 6 7 8 9 10 ?

Whats the probabilty that i can get a 20 on my first two cards ?
#2
April 28th, 2010, 01:07 PM
 Sucker Executive Member Join Date: Feb 2009 Location: U.S.A. Posts: 1,504

It depends upon how well you play, and to a lesser extent the rules that are offered by the particular house; but assuming perfect basic strategy, the dealer will win approximately 53% of the time. The chance of the dealer winning twice in a row is 53/100 squared, or about 28% of the time. For three in a row, four in a row, and more; it's 53/100 to the 3rd power, 53/100 to the fourth power; and so on respectively.

The chance of your first two cards totaling 20 is about 11.1%, or one in nine.
#3
April 28th, 2010, 01:30 PM
 Sucker Executive Member Join Date: Feb 2009 Location: U.S.A. Posts: 1,504

Quote:
 Originally Posted by Sucker It depends upon how well you play, and to a lesser extent the rules that are offered by the particular house; but assuming perfect basic strategy, the dealer will win approximately 53% of the time. The chance of the dealer winning twice in a row is 53/100 squared, or about 28% of the time. For three in a row, four in a row, and more; it's 53/100 to the 3rd power, 53/100 to the fourth power; and so on respectively.
I misunderstood the question, and gave you incorrect information here; so you can safely ignore this part of my answer.

But my second statement is correct for 8 decks; and for a single deck game the chance of being dealt a 20 is ((16/52 X 15/51) + (8/52 X 4/51)) or 10.1%. (21forme forgot to include in the chance of being dealt A-9).
#4
April 28th, 2010, 01:48 PM
 Sucker Executive Member Join Date: Feb 2009 Location: U.S.A. Posts: 1,504

Quote:
 Originally Posted by FLASH1296 The following is REQUIRED for your calculations; irrespective of voodoo nonsense like “streaks” is an exercise in self-destructive futility. On average your chances of winning a hand in BJ is 43% On average your chances of losing a hand in BJ is 48% On average your chances of pushing a hand in BJ is 9%
If this is correct (and I have no reason to believe it's NOT), then the answer to the first part of the question is:

1 time - 48%
2 times - .48 x .48 = 23.04%
3 times - .48 ^ 3 = 11.06%
4 times - .48 ^ 4 = 5.31%
etc.; etc.
#5
April 28th, 2010, 01:08 PM
 21forme Executive Member Join Date: Oct 2006 Posts: 3,120

1 time = 43%
2 times = 0.43 x 0.43 = 18.5 %
and so on.

This is at a neutral count. At +3, it's about 50%, and the basis for playing the Streak side bet, when available.

Odds of getting a 20 depends on the number of decks. If single deck, it's 16/52 x 15/51.

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