Blackjack - 5 Card Charlie Probability Question

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. :]

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.

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?

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 :/

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

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.

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.

Quote: Ace2

You 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: Ace2

You 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.

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!)

Any chance you care to show the details of your work?

Quote: Jufo81

Any chance you care to show the details of your work?

I spent about 90 minutes typing in an explanatory post with two lengthy tables and then something happened and my browser lost the post. So, I'll try again but this time as a series of shorter posts.

Probability of a dealer starting with an Ace and making a 5-card 21 -Part 1

1. First, I will assume this is Hit Soft 17 -which is the harder case.

2. Be aware that in the infinite deck approximation, the probability of any card being a Ten is 4/13, and the probability of it being any other rank A-9 is 1/13.

3. Therefore, since we are looking for every series of four cards that, when added to an ace, will make a 21 (without making a soft 18-21 along the way). Any four card series that does not contain a Ten will have a probability = 1 /13^4. Example: A 2 3 8 7; each card after the Ace had a 1/13 chance of occurring.

Any four card series that does contain a Ten will have a probability = 4/13^4. Example: A 2 T 3 5; The probability of drawing those 4 cards, after the ace in that order is 1/13*4/13*1/13*1/13.

4. Now, let's convert our "4-card problem" into a "3-card problem." Let's define the criteria for the first 3 cards drawn to the Ace as being that the 4 card total must be:
- Hard 12-16 without having made a soft 18-21 after having drawn 1 or 2 cards, or
- (Rarely) Soft 14 -17 Ex: A A 2 A

Notice if that the sum = 12 after 4 cards then the 5th card must always be a 9, if the sum =16 after 4 cards, the 5th card must always be a 5. More generally, if the first three cards meet the above criteria, then then there will only be one rank that the fifth card can be to make exactly 21, and that fifth card rank will always be in the range 4-9. So the probability of the 5th card will always be 1/13 if the 2nd, 3rd and 4th card meet the above criteria. So, we don't need to actually identify what the 5th card is, it will always have a probability of 1/13.

5. Now let's define the 3 pathways that start with an Ace and result in a 5-card 21.

Pathway A: The 2nd and 3rd card, when added to the Ace, sum up to <S18; the 4th card increases the sum to a Hard 12-16, and the fifth card increases the sum to exactly 21.

Pathway B: The 2nd , when added to the Ace, make a sum <S18; the 3rd card increases the sum to a hard 12-15, the 4th card increases the sum to a hard 16 or less and the fifth card increases the sum to exactly 21.

Pathway C: The 2nd, 3rd and 4th cards are all small and when added to the Ace sum up to <S18, and the fifth card increases the sum to exactly Soft 21.

At this point,one could write out all the sequences that that satisfy those pathways and assign probabilities to them. Pathway A has 77 sequences with a total probability of 95/13^4. Pathway B
has 47 sequences with a total probability of 68/13^4. Pathway C has 20 sequences yielding a probability of 20/13^4.

It may be tedious, but if you have never written out a list like that you might try it and check my numbers. I used a few tricks to reduce the list from 165 down to about 41. I'll explain that in my next post.

Probability of getting a hard 20, dealer up card ace, and dealer draw to 21 with exactly 5 cards:

Double deck; dealer hits soft 17

First hand (solo player) gets two tens and dealer has an ace up. Player stands. Dealer flips over the hole card (cannot be 7-10 of course), draws a third card, draws a fourth card, and draws a fifth card that is exactly 21. You lose! Darn!

I started off with all 10,000 different ways for the dealer to expose the hole card and the next three cards (4 cards having one of 10 values is 10^4=10,000).

This part I did on a spreadsheet. I started with AAAAA, then AAAA2, AAAA3, etc., all the way to ATTTT.

Then, starting at the bottom, I deleted all the ones that start with AT since that is dealer blackjack. Then I deleted more, starting from the bottom, as those would be A9xyz (soft 20 with no more cards), same for A8xyz, and A7xyz. Now my last row is A6TTT. That one is three cards ending in hard 17 (the fourth and fifth cards are irrelevant but part of the entire set of A+ 4 cards of any value). Delete A6Txy. Now my last row is A69Tx, which is bust.

At this point I decided to write a VBA (Visual Basic for Access) program to analyze the 590 remaining possibilities. I don't like Excel Visual Basic. Access can handle millions of rows in a table (limit 2GB of data in a table, something that would bog down Excel for an eternity).

Anyway, I wrote the program, acting like the dealer, where I start with A (score of 11), flip over the hole card, add its score (if A, it must be 1 since I already used 11), and so on. If a soft score>21, switch to hard total and subtract 10 (e.g. A3T4x is hard 18, no 5th card needed).

For each row, the program output the result, which is: bust, 3 card 21 (e.g. A64), 4 card 21, 3 cards soft 18-20, 3 cards hard 17-20, 4 cards soft 18-20, and 4 cards hard 17-20. If none of those, what is left is 4 cards that add up to soft 17, or 16 or less. Now I can easily calculate what the 5th card must be (e.g. soft 17 on 4 cards, draw a 4 to get to 21 in 5 cards).

I ended up with 135 possibilities:

up hole
card card 3rd 4th 5th
A A A A 7
A A A 2 6
A A A 3 5
A A A 4 4
A A A 9 9
A A A 10 8
A A 2 A 6
A A 2 2 5
A A 2 3 4
A A 2 8 9
A A 2 9 8
A A 2 10 7
A A 3 A 5
A A 3 2 4
A A 3 7 9
A A 3 8 8
A A 3 9 7
A A 3 10 6
A A 4 A 4
A A 4 6 9
A A 4 7 8
A A 4 8 7
A A 4 9 6
A A 4 10 5
A A 5 5 9
A A 5 6 8
A A 5 7 7
A A 5 8 6
A A 5 9 5
A A 10 A 8
A A 10 2 7
A A 10 3 6
A A 10 4 5
A 2 A A 6
A 2 A 2 5
A 2 A 3 4
A 2 A 8 9
A 2 A 9 8
A 2 A 10 7
A 2 2 A 5
A 2 2 2 4
A 2 2 7 9
A 2 2 8 8
A 2 2 9 7
A 2 2 10 6
A 2 3 A 4
A 2 3 6 9
A 2 3 7 8
A 2 3 8 7
A 2 3 9 6
A 2 3 10 5
A 2 4 5 9
A 2 4 6 8
A 2 4 7 7
A 2 4 8 6
A 2 4 9 5
A 2 9 A 8
A 2 9 2 7
A 2 9 3 6
A 2 9 4 5
A 2 10 A 7
A 2 10 2 6
A 2 10 3 5
A 3 A A 5
A 3 A 2 4
A 3 A 7 9
A 3 A 8 8
A 3 A 9 7
A 3 A 10 6
A 3 2 A 4
A 3 2 6 9
A 3 2 7 8
A 3 2 8 7
A 3 2 9 6
A 3 2 10 5
A 3 3 5 9
A 3 3 6 8
A 3 3 7 7
A 3 3 8 6
A 3 3 9 5
A 3 8 A 8
A 3 8 2 7
A 3 8 3 6
A 3 8 4 5
A 3 9 A 7
A 3 9 2 6
A 3 9 3 5
A 3 10 A 6
A 3 10 2 5
A 4 A A 4
A 4 A 6 9
A 4 A 7 8
A 4 A 8 7
A 4 A 9 6
A 4 A 10 5
A 4 2 5 9
A 4 2 6 8
A 4 2 7 7
A 4 2 8 6
A 4 2 9 5
A 4 7 A 8
A 4 7 2 7
A 4 7 3 6
A 4 7 4 5
A 4 8 A 7
A 4 8 2 6
A 4 8 3 5
A 4 9 A 6
A 4 9 2 5
A 4 10 A 5
A 5 A 5 9
A 5 A 6 8
A 5 A 7 7
A 5 A 8 6
A 5 A 9 5
A 5 6 A 8
A 5 6 2 7
A 5 6 3 6
A 5 6 4 5
A 5 7 A 7
A 5 7 2 6
A 5 7 3 5
A 5 8 A 6
A 5 8 2 5
A 5 9 A 5
A 6 5 A 8
A 6 5 2 7
A 6 5 3 6
A 6 5 4 5
A 6 6 A 7
A 6 6 2 6
A 6 6 3 5
A 6 7 A 6
A 6 7 2 5
A 6 8 A 5

Back to Excel, I calculated the probability of each occurrence. I have 4 tables of deck compositions, starting with the same for everyone (30 tens, 7 aces, and 8 of everything else). The 2nd table has the deck composition after the hole card, and so on, until the last table with the deck composition after the 4th card has been drawn (to calculate the probability of the 5th needed card being drawn). I calculated the probability of that particular hand happening (e.g. A23T5 as 1.13851693510761E-06).

Add them all together, and I get 5.00105E-05, or 1 in 19995.78.

Could be a super high stakes side bet. $17k for a $1 bet, or $15k and a progressive meter.

I'll resurface soon in this thread, hopefully, wow, got some responses, thanks all! Been quite busy, and unfortunately nothing with gambling and math :(. I thought this was a great question, first of all, because it's kind of like a very polar situation to having a near perfect hand, thinking it's good, and getting 110% eradicated, especially for new blackjack players.

Fun fact to employ to play with the mind, enjoy the game, and learn the math aspect, ect. Wanted to get my math fresh, branch off into other probability scenarios, or other situational scenarios if the opportunity arose.

More recently I was just thinking, in my final answer, I need to make sure what I am asking for, specifically SAMPLE SIZE wise. If i have the probability of ALL possible dub deck combos, that's what i'm solving for. I think a neat question to ask is, how many average blackjack hands does a user play in the past, present, or future day/month/year/session/lifetime, and refine the question that way, too.

Quote: KevinAA

Probability of getting a hard 20, dealer up card ace, and dealer draw to 21 with exactly 5 cards:

Double deck; dealer hits soft 17

First hand (solo player) gets two tens and dealer has an ace up. Player stands. Dealer flips over the hole card (cannot be 7-10 of course), draws a third card, draws a fourth card, and draws a fifth card that is exactly 21. You lose! Darn!

I haven't had a chance to review everything you did, but there is no reason to assume that player has two tens (and that there are only 30 tens in the double deck shoe.) You can calculate the general chance of a dealer making a five card 21 if you make no assumptions about what the player has -its as if you deal the dealer's cards first and assume the player will be randomly dealt get his cards afterwards. Thta would be the correct way to do this calculation, I believe.

And by assuming player has two tens you've made a bounding assumption that provides the highest possible probability for a dealer 5-card 21.

Quote: gordonm888

I haven't had a chance to review everything you did, but there is no reason to assume that player has two tens (and that there are only 30 tens in the double deck shoe.) You can calculate the general chance of a dealer making a five card 21 if you make no assumptions about what the player has -its as if you deal the dealer's cards first and assume the player will be randomly dealt get his cards afterwards. Thta would be the correct way to do this calculation, I believe.

And by assuming player has two tens you've made a bounding assumption that provides the highest possible probability for a dealer 5-card 21.

It's not an assumption; it's part of the probability question.

i.e. freshly shuffled double deck (32 tens and 8 aces in it), the procedure is:
Dealer hands the player a card
Dealer draws the up card
Dealer hands the player a second card (if you did player card 1, player card 2, then dealer up card it would make no difference)

What is the probability this particular event will happen:
Player's first card is a ten
Up card is an ace
Player's second card is a ten

That would be:
32/104 * 8/103 * 31/102

At this point, the rest of the deck (including the hole card, which is the next card) has 30 tens and 7 aces.

I determined that there are 135 ways the hole card requires drawing 3 more cards (i.e. no 21 and no soft 18-20 and no hard 17-20 along the way) to total exactly 21 on the 5th card.

For example, hole card 2, 3rd card 4, 4th card 8, 5th card 6 (i.e. start with soft 13, draw a 4, hit soft 17, draw an 8, hit on 15 and draw a 6 = 21). The probability of this event (starting from the beginning) is:

32/104 * 8/103 * 31/102 * 8/101 * 8/100 * 8/99 * 8/98

It's less likely if there are repeating cards, such as A2279 which is:

32/104 * 8/103 * 31/102 * 8/101 * 7/100 * 8/99 * 8/98

Notice the denominators are always the same, but in this one, one numerator is a 7 instead of 8 (the 2nd 2 is drawn from a deck that contains only 7 2's since the 1st one was the hole card).