Blackjack hands calculation method
July 7th, 2014 at 4:22:55 AM
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Does anyone know how to calculate the probability of dealer busting 15, with multiple hits? Not just the next card. If possible please put down the full maths equation. thanks a lot.
July 7th, 2014 at 2:20:11 PM
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Let's assume it's a hard 15...... (calculating a soft hand basically the same, but much more work).
Having 15, out of 13 cards 1 will make him an 16, and 7 will make him bust.
Having 16, out of 13 cards 8 will make him bust.
So bust(16) = 8/13 and bust(15) = 7/13 + 1/13 * bust(16) = 7/13 + 8/169 = 99/169
Having 15, out of 13 cards 1 will make him an 16, and 7 will make him bust.
Having 16, out of 13 cards 8 will make him bust.
So bust(16) = 8/13 and bust(15) = 7/13 + 1/13 * bust(16) = 7/13 + 8/169 = 99/169
July 7th, 2014 at 2:27:23 PM
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For a more comprehensive answer, you'd also need to know composition of all hands, number of decks, and compensate for the removal of those cards. Calculations work in the same general manner though.
July 7th, 2014 at 4:51:55 PM
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Assuming an "infinite deck" (where 1/13 of the remaining cards are always aces, 1/13 are 2s, 1/13 are 3s, and so on through 1/13 are kings), and assuming the dealer stands on all 17s:
Since a 7 or higher on the next card busts, that's 7/13 right there
A 2-6 gives the dealer 17-21
There is 1/13 chance of an ace, giving the dealer 16; from there, a 6 or higher busts, while an A-5 gives the dealer 17-21, so there is a further 1/13 x 6/13 = 6/169 chance.
The total chance of busting from 15 is 7/13 + 6/169 = 97/169, or about 57.4%.
Since a 7 or higher on the next card busts, that's 7/13 right there
A 2-6 gives the dealer 17-21
There is 1/13 chance of an ace, giving the dealer 16; from there, a 6 or higher busts, while an A-5 gives the dealer 17-21, so there is a further 1/13 x 6/13 = 6/169 chance.
The total chance of busting from 15 is 7/13 + 6/169 = 97/169, or about 57.4%.
July 8th, 2014 at 1:42:32 PM
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Quote: ThatDonGuya 6 or higher busts, [...] so there is a further 1/13 x 6/13 = 6/169 chance.
There are 8/13 cards 6 or higher, not 6/13.
July 8th, 2014 at 4:32:49 PM
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Starting with hard 15 (I will presume its the initial Dealer hand of 2 cards)
The chance of drawing an Ace is 1/13.
The chance of that 3-card 16 busting is 7/13.
So the chance of the final result (busting) given an initial hand of hard 15, with the draw of an Ace (for the only way a 2-hit bust can occur) is 1/13 * 7/13 when evaluating the complete situation (7/169).
The chance of drawing an Ace is 1/13.
The chance of that 3-card 16 busting is 7/13.
So the chance of the final result (busting) given an initial hand of hard 15, with the draw of an Ace (for the only way a 2-hit bust can occur) is 1/13 * 7/13 when evaluating the complete situation (7/169).
Some people need to reimagine their thinking.
July 8th, 2014 at 4:56:47 PM
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Quote: MangoJQuote: ThatDonGuya 6 or higher busts, [...] so there is a further 1/13 x 6/13 = 6/169 chance.
There are 8/13 cards 6 or higher, not 6/13.
So that's why my VP app keeps giving me problems...
In that case, the answer is 7/13 + 8/169 = 99/169, or about 58.6%.
