5 Card Blackjack
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FrostieFingerz
Guide to Blackjack Card Counting (updated 2021) - Learn how to count cards with our FREE & Easy Card Counting Training Game. Start practicing here today! A five-card Charlie in blackjack is when you have a total of 5 cards and you do not exceed a point total of 21. How many such hands are there? Of course, the natural next question concerns six-card Charlies, etc. When using the Wong Halves counting method, the 3, 4, and 6 cards are valued at +1, the 2 and 7 cards are valued as +0.5, and the 5 is worth +1.5. All 8s are 0, 9 is valued at -0.5, and all Ace and face cards are valued as -1. If you're having a hard time counting with fractions, double each value for a simplified strategy.
Hello
Regardless of other rules or variations, let's say there is a game of Blackjack that pays 1,5 times your bet when you have a winning hand consisting of 5 cards. I think a hand like this occurs every 50 hands or so, but you'll only get paid if it's a winning hand. Moreover, a winning 6 card charlie pays twice your bet, and a winning 7 card charlie pays 5 times your bet.
Anyone has an idea which effect this has on the house edge?
Would that influence my basic strategy play? Would I be hitting some hands where I would normally stand?
Thanks!
Regardless of other rules or variations, let's say there is a game of Blackjack that pays 1,5 times your bet when you have a winning hand consisting of 5 cards. I think a hand like this occurs every 50 hands or so, but you'll only get paid if it's a winning hand. Moreover, a winning 6 card charlie pays twice your bet, and a winning 7 card charlie pays 5 times your bet.
Anyone has an idea which effect this has on the house edge?
Would that influence my basic strategy play? Would I be hitting some hands where I would normally stand?
Thanks!
Romes
It would definitely effect basic strategy. To the exact extent, I'm not sure.
I believe you can find the % difference in rules from the Rules Variations:
http://wizardofodds.com/games/blackjack/rule-variations/
I hope this doesn't come off rude, but more so encouraging... Try Google first =)...
http://lmgtfy.com/?q=wizard+of+odds+5+card+charlie
Any of those top 3 search results should point you at information you're interested in.
I believe you can find the % difference in rules from the Rules Variations:
http://wizardofodds.com/games/blackjack/rule-variations/
I hope this doesn't come off rude, but more so encouraging... Try Google first =)...
http://lmgtfy.com/?q=wizard+of+odds+5+card+charlie
Any of those top 3 search results should point you at information you're interested in.
FrostieFingerz
Thank you, Romes :)
The first link is indeed a good link to see the % difference, I should have found that one by myself, but I guess I've read so many info I looked over it. It says there a 5 card win paying 3:2 adds 0,33 % to the player's edge, which is not bad at all I think.
However, I find very little info on how this would effect basic strategy, most of the search results from Google treat the topic of automatic win if you don't bust with 5 cards, no matter what the dealer's cards are.
With this variation, you'll only win if you can also beat the dealer, and that info is harder to find! :-)
The first link is indeed a good link to see the % difference, I should have found that one by myself, but I guess I've read so many info I looked over it. It says there a 5 card win paying 3:2 adds 0,33 % to the player's edge, which is not bad at all I think.
However, I find very little info on how this would effect basic strategy, most of the search results from Google treat the topic of automatic win if you don't bust with 5 cards, no matter what the dealer's cards are.
With this variation, you'll only win if you can also beat the dealer, and that info is harder to find! :-)
Romes
It will definitely be tougher to find exact basic strategy deviations for these rules, probably because they're so unique and rare. However, while waiting for SIM people to respond (or someone with similar prior experience), don't let the google stop! Try different searches, etc.For example: 'blackjack 5 card charlie if you beat the dealer' resulted in:
http://blackjack-authority.com/articles/charlies.html
on the 3rd hit. This has some strategy deviations, but be ware of the exact EV's of each play. There's some strategy from this link that I would and would not follow:
'If you are holding a hand with a total of 4 - 11 and the Dealer is holding 2, 3, 4, 5, 6, 7, 8, 9, 10 or an Ace you should always Hit that hand.
You'll definitely be hitting more to draw to these special pay hands, but personally I wouldn't sacrifice doubling on 10 and 11, unless maybe it was a super negative count (that I would have wong'd out of) or something.
Hopefully someone else who's ran sims for this before can chime in with exact plays, but I wanted to show you that with more digging you can slowly start to uncover more information to at least point you in the right direction. Both of the links from both of my posts in this thread took under 30 seconds of Googling.
EDIT: Another 60 seconds of Googling ('blackjack 5 card charlie strategy') again came up with a ton of information, including threads on this site about similar topics:
http://wizardofodds.com/games/blackjack/strategy/charlie/
http://wizardofvegas.com/forum/gambling/blackjack/5578-charlie-strategy/
http://www.blackjackrules.org/Five-Card-Charlie.htmlYes, they might not be exact 'and you have to win the hand' but I would be fairly confident these are pretty close to what you're looking for.
Playing it correctly means you've already won.
AxiomOfChoice
This has nothing to do with 5- or 6-card charlie. The numbers for 5-card charlies are not useful here, I don't think.
A charlie is an automatic winner on an unbusted hand. The original post describes a bonus for winning the hand. Those are very different things.
A charlie is an automatic winner on an unbusted hand. The original post describes a bonus for winning the hand. Those are very different things.
Romes
This has nothing to do with 5- or 6-card charlie. The numbers for 5-card charlies are not useful here, I don't think.
A charlie is an automatic winner on an unbusted hand. The original post describes a bonus for winning the hand. Those are very different things.
Are you referring to a bonus because it's not an automatic winner? Thus, the hand is normal, the payout is just a little extra?
Playing it correctly means you've already won.
5 Card Trick Blackjack Rules
AxiomOfChoice
I don't see anything in the OP referring to a bonus? Seems like regular charlie with the caveat that it must still beat the dealers hand (i.e. not automatic winner).
The bonus is the 1.5 to 1 payout.
'5 card charlie' refers to the rule where if you have a 5-card unbusted hand, it wins automatically.
The rule that he is referring to is that if your 5-card hand wins, it pays 3:2.
There is a big difference between a 5-card 16 winning every time, and it paying 3:2 if the dealer busts.
Dieter
It also becomes very tempting to hit a 5 card soft 21, if I'm allowed. More tempting to hit the 6 card.
FrostieFingerz
Thank you guys for your replies. It is indeed tempting to hit in some circumstances, although my fear is that would have a huge negative effect on the player's edge.
I'm still hoping someone who can run a simulation could help us out here :-)
I'm still hoping someone who can run a simulation could help us out here :-)
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. :]
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.
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?
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 :/
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
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.
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
5 Cards Blackjack
- whenever the 2nd, 3rd and 4th cards add to 7,8,9 or 10.5 Card Blackjack Charlie
- 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?