July 6th, 2018 at 3:18:21 PM
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Hi Everyone,

I'm less interested in a final answer and more interested in a process to answer the following question.

Moreover, is my math estimate of probability a * probability b * probability c * probability d = final probability a valid way to solve this problem or not? Thanks for reading?

I'm interested in the following situational question - In a standard Double Deck game, What is the probability of being dealt a two card 20 as a starting hand, the dealer having an Ace up, and the player not taking insurance, and the dealer not having a blackjack, and the dealer drawing to a 5 card 21? What is the probability the same occuring, but being dealt a two card combination without including A,9 combos? For some reason when I do the math of the 52,2 20 combos by hand for % P = x/y, My brain freezes up :x. I lost my math skills 10 years ago :D. Any help/contribution/jumble of formulas/estimates welcome.

The number I got was roughly around .016% of the time. This seems awful low, and doesn't account for the dealer not having it, as well as only counting 3+ card 21 draws. Solve for whatever you like, however you like. More interested in the process. :]

I'm less interested in a final answer and more interested in a process to answer the following question.

Moreover, is my math estimate of probability a * probability b * probability c * probability d = final probability a valid way to solve this problem or not? Thanks for reading?

I'm interested in the following situational question - In a standard Double Deck game, What is the probability of being dealt a two card 20 as a starting hand, the dealer having an Ace up, and the player not taking insurance, and the dealer not having a blackjack, and the dealer drawing to a 5 card 21? What is the probability the same occuring, but being dealt a two card combination without including A,9 combos? For some reason when I do the math of the 52,2 20 combos by hand for % P = x/y, My brain freezes up :x. I lost my math skills 10 years ago :D. Any help/contribution/jumble of formulas/estimates welcome.

The number I got was roughly around .016% of the time. This seems awful low, and doesn't account for the dealer not having it, as well as only counting 3+ card 21 draws. Solve for whatever you like, however you like. More interested in the process. :]

July 6th, 2018 at 3:27:19 PM
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The two-card (besides A9) 20 combos in a double-deck game is easy. The only way to get a 2-card 20 besides an A9 is with two 10-count cards.

There are 32 of these in a double deck, so there are (32)C(2) = 496 different pairs you can get.

There are (104)C(2) = 5356 total pairs of cards in two decks.

The probability is 496 / 5356 = 0.0926, or about 1 / 10.8.

There are 32 of these in a double deck, so there are (32)C(2) = 496 different pairs you can get.

There are (104)C(2) = 5356 total pairs of cards in two decks.

The probability is 496 / 5356 = 0.0926, or about 1 / 10.8.

July 6th, 2018 at 6:06:46 PM
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Thanks Don for that sequence of it. My answer was coming up differently. Any insights into the other parts of the question appreciated.

Meanwhile... And I quote another resource with the correct answer..

Probability of obtaining 20 points from the first two cards is P = 68/663 = 10.25641% in the case of a 1-deck game and P = 140/1339 = 10.45556% in the case of a 2-deck game.

[[P = 140/1339]] is the same as

[[P = 560/5356]]

We've got 496 different unique combos of face cards to make our 20, now what type of math do we use to get our apparent 64 combos of A,9?

Meanwhile... And I quote another resource with the correct answer..

Probability of obtaining 20 points from the first two cards is P = 68/663 = 10.25641% in the case of a 1-deck game and P = 140/1339 = 10.45556% in the case of a 2-deck game.

[[P = 140/1339]] is the same as

[[P = 560/5356]]

We've got 496 different unique combos of face cards to make our 20, now what type of math do we use to get our apparent 64 combos of A,9?

July 6th, 2018 at 6:34:13 PM
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I guess, visually, we just have to say _ * _, or 8 x 8, because there are 8 Aces we could draw for card1 to make 20, and 8 nines to fill card2.

When I try to choose 2 of 16 = 120 , and eliminate A,A combos and 9,9 combos, I come up with the wrong answer, still :/

When I try to choose 2 of 16 = 120 , and eliminate A,A combos and 9,9 combos, I come up with the wrong answer, still :/

July 6th, 2018 at 7:34:36 PM
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so, here's random info outlined from/for my question so far.. feel free to correct me for the sake of.. oh well..

Odds of drawing any two card 20 is 10.46%.

Odds of a Face 20 is 9.26%.

Odds of dealer having an Ace Up, disregarding our hand, is 8/104, or 7.70% of the time.

If we have an A,9 combo, the odds of one dealer Ace reduces to 7/102 or 6.86% o the time?

Moreover, the odds of the dealer having Ace Up, No blackjack in DD, disregarding our hand, should be..

A combo hand of one of [8] aces, plus any card A(7) thru 9, or [71] cards out of 5356 2 card combos for the deelah?!?!

8 x 71 / 5356

or 10.6% of the time

Odds of drawing any two card 20 is 10.46%.

Odds of a Face 20 is 9.26%.

Odds of dealer having an Ace Up, disregarding our hand, is 8/104, or 7.70% of the time.

If we have an A,9 combo, the odds of one dealer Ace reduces to 7/102 or 6.86% o the time?

Moreover, the odds of the dealer having Ace Up, No blackjack in DD, disregarding our hand, should be..

A combo hand of one of [8] aces, plus any card A(7) thru 9, or [71] cards out of 5356 2 card combos for the deelah?!?!

8 x 71 / 5356

or 10.6% of the time

July 7th, 2018 at 7:38:00 AM
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The only part of OP's question that is hard is, given that dealer starts with an ace, what is the probability of dealer making a 5-card 21.

Does anyone know any way of calculating that probability other than essentially listing all the 4 card draws to an Ace that total 21 without making a soft 18-20 or a hard 17-20?

Assuming dealer hits a soft 17, basically there are two pathways for dealer to make a 5 card 21 starting from an Ace:

1. Two small cards totally 2-6 to make a hand between S13-S17, followed by a card larger than a 4 to make a hard 12-16, followed by a final card to make 21.

2. One small card, A-6, followed, by a card (larger than a 4) to make a hard 12-15, followed by a small card to keep the total under 16, followed by a final card to make 21.

But, as you look at the details, I don't think that is easy to write as an analytic algorithm.

Does anyone know any way of calculating that probability other than essentially listing all the 4 card draws to an Ace that total 21 without making a soft 18-20 or a hard 17-20?

Assuming dealer hits a soft 17, basically there are two pathways for dealer to make a 5 card 21 starting from an Ace:

1. Two small cards totally 2-6 to make a hand between S13-S17, followed by a card larger than a 4 to make a hard 12-16, followed by a final card to make 21.

2. One small card, A-6, followed, by a card (larger than a 4) to make a hard 12-15, followed by a small card to keep the total under 16, followed by a final card to make 21.

But, as you look at the details, I don't think that is easy to write as an analytic algorithm.

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

July 7th, 2018 at 5:35:43 PM
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You can easily do it with a Markov chain. You should only need 65 cells - 5 rows and 13 columns.

That’s assuming infinite deck which will be very close.

You could also list out all 5 card permutations, there are only 100,000.

That’s assuming infinite deck which will be very close.

You could also list out all 5 card permutations, there are only 100,000.

July 8th, 2018 at 7:47:57 AM
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Quote:Ace2You can easily do it with a Markov chain. You should only need 65 cells - 5 rows and 13 columns.

I'm an excellent Excel programmer but I'm weak on Markov chains. Still trying to figure out how to do that.

Quote:Ace2You could also list out all 5 card permutations, there are only 100,000.

No, the first card is always an Ace, which reduces it from 100,000 to 10,000. And then, you can eliminate whenever the 2nd card is a 7-T, which reduces it to 6,000. Then, also eliminate whenever the last card is a Ten or A-4, which reduces it to 3,000. Then, also eliminate these combinations:

- whenever the 2nd and 3rd card add to 7,8,9 or 10.

- whenever the 2nd and 3rd cards add up to 15 or more

- whenever the 2nd, 3rd and 4th cards add to 7,8,9 or 10.

- whenever the 2nd, 3rd and 4th cards add up to 16 or more.

In reality, the number of permutations to write out is reasonable.

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

July 8th, 2018 at 10:31:23 AM
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I have figured out a straightforward way to list all the permutations for dealer making a 5-card 21 starting with an ace and to calculate the probability - assuming an infinite deck approximation. By defining some new terminology (or at least it was new to me) I needed only 41 groups of permutations; i.e. my list need only 40 equations in it for the probability.

Assuming, the H17 rule and infinite decks, I got that the probability of dealer starting with an Ace and making a 5 card 21 to be 183/13^4 = 0.006407.

If you assume the S17 rule, I calculate the probability for a 5 card 21 to be 138/13^4 =0.004832

Edit: corrected numbers (whoops!)

Assuming, the H17 rule and infinite decks, I got that the probability of dealer starting with an Ace and making a 5 card 21 to be 183/13^4 = 0.006407.

If you assume the S17 rule, I calculate the probability for a 5 card 21 to be 138/13^4 =0.004832

Edit: corrected numbers (whoops!)

Last edited by: gordonm888 on Jul 8, 2018

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.