## 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**

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**

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**

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**

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**

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**

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.

**Ace2**

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.

**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.

**gordonm888**

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!)

**Jufo81**

**h4t**

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**

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**

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**

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**

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**

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.

**Ace2**

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.

**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.

**gordonm888**

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!)

## 5 Card Poker Games

**Jufo81**