## 5 Card Blackjack

5-Card Charlie is a special blackjack rule according to which a player automatically wins if they draw 5 cards without reaching 21. There are some variations, mind you, but that’s the gist. What is 5-Card Charlie? According to this rule, drawing 5 cards. The automatic win upon receiving five cards in blackjack is called “5 Card Charlie”. However, the rule doesn’t apply to most blackjack games online or in land-based casinos. You’ve probably heard this from someone who plays blackjack at home. That’s where the five card rule is the most prominent. Remember, you don't win because you are closer to the value of 21 - you win because your combined value of the cards is greater than that of dealer. Blackjack Strategy. When the value of dealer's revealed card is 4,5 or 6, it may be fruitful to double your bet with an Ace and 4 in hand. 5-Card Charlie is a special blackjack rule according to which a player automatically wins if they draw 5 cards without reaching 21. There are some variations, mind you, but that’s the gist. What is 5-Card Charlie? According to this rule, drawing 5 cards results in a win regardless of what the rest of the hand shows. In the game of Blackjack, the term '5 card charlie' can be an option in the favor of the blackjack player instead of the casino. In this case, the player would win if they have 5 cards on the table for a single hand and yet have not busted. If the dealer is playing 5 card charlie, then the player would automatically win.

h4t
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. :]
ThatDonGuy
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.
h4t
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?
h4t
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 :/
h4t
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
gordonm888
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.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Ace2
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.
gordonm888

Wink bingo games bonus. 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.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
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!)
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Jufo81
Any chance you care to show the details of your work?
h4t
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. :]
ThatDonGuy
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.
h4t
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?
h4t
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 :/
h4t
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

## 5 Card Blackjack Strategy

gordonm888
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.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Ace2
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.
gordonm888

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.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
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!)
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

## 5 Card Poker Games

Jufo81
Any chance you care to show the details of your work?