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badoomboom
badoomboom
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December 30th, 2011 at 8:46:07 AM permalink
What's the probability of dealer getting BJ in ENHC game when 1, 2, 3, 7 boxes are open?
Wizard
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Wizard
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December 30th, 2011 at 9:50:39 AM permalink
The hole card rule and number of boxes don't matter. Just the number of decks, which you didn't specify.

The formula is for n number of decks the probability of a blackjack is (8/13)*(4*n)/(52*n-1)
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badoomboom
badoomboom
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January 1st, 2012 at 8:48:50 AM permalink
I think I'll have to disagree with you becuase with ENHC game players can bust before dealer will deal himself second card therefore dealer should have less blackjacks, considering total number of hands played, than in hole card games. And I meant 6-deck :)
NowTheSerpent
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January 6th, 2012 at 4:50:27 AM permalink
Quote: Wizard

The hole card rule and number of boxes don't matter. Just the number of decks, which you didn't specify.

The formula is for n number of decks the probability of a blackjack is (8/13)*(4*n)/(52*n-1)



What about BJ in general? for six decks, would prob(BJ) = (24*96*2)/(312*311) = 4.7489% and for eight decks, (32*128*2)/(416*415) = 4.7451%?
DJTeddyBear
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January 6th, 2012 at 6:22:10 AM permalink
Quote: badoomboom

I think I'll have to disagree with you becuase with ENHC game players can bust before dealer will deal himself second card therefore dealer should have less blackjacks, considering total number of hands played, than in hole card games. And I meant 6-deck :)

If all players bust before the dealer takes his second card, does that really affect the odds or number of BJs the dealer gets?

After all, this will happen whether the dealer has a ten or ace or something else showing, proportionally equally.
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Switch
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January 6th, 2012 at 7:10:21 AM permalink
Due to the player busting before the dealer receives a 2nd card then this would lower the amount of 'Blackjacks' that the dealer ended up with. Less players will result in fewer Blackjacks for the dealer.

For example, with heads-up play, player has 14 and dealer has an Ace. Player draws and busts so the hand is over. Next card out (which is now the player's first card of the next hand) is a 'Ten'. So the dealer would have had a 'Blackjack' but didn't need to complete his hand as the dealer won anyway.

I would guess that you could use rough calculations based on the typical dealer bust rate of 28% (as the player will draw like the dealer against a 10 and Ace with the slight exception of A,A vs 10 in ENHC). So, the player will bust roughly 28% of the time against a dealer 10 and Ace. The dealer will get a 'Blackjack' 4.75% of the time in a 6-deck game if he can complete the hand. So, if the dealer is playing against 1 player then the dealer will get 'Blackjacks' using the simple formula:-

P(Dealer Blackjack) X P(Player not busting) = 4.75 x (1 - 0.28) = 4.75 x 0.72 = 3.42%

Against 2 players it would be 4.75 x (1- (0.28)^2 = 4.75 x 0.9216 = 4.38%

For n players it would simply be 4.75 x (1 - (0.28)^n)

So, on a table with 6 players this would still give 4.75% to 2 dec.pl.
badoomboom
badoomboom
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January 6th, 2012 at 7:15:00 AM permalink
Thank you very much for that

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