Baccarat question

Hi all,

I'm new to the forum and if I'm posting in the wrong place please accept my apology. I've been looking for an answer all over the Internet, and I hope you guys could help me.

In a traditional baccarat game, 8 decks, how do I calculate the house edge of either banker or player hand will get all 3 face cards? And 3 face cards I mean just any combination of J, Q, and K. And also what's the appropriate math to see the probability of getting any 3 face cards suites? As well as the straight flush of J,Q,K same suit?

I found this math off of this site base on a single deck for 3 cards baccarat and the house edge is 83% of the pay out is 16 to 1

Using this math

(12/52)*(11/51)*(10/50)=0.009954751. The return to the player is easily calculated as 0.009954751 × 16 - (1-0.009954751) = -0.830769231

so to simulate this math for 8 decks I did this:

(96/416)*(88/408)*(80/400)=0.00995475113
same as the above

However, I calculate it with the payout of 30 to 1 for an example
0.00995475113*30-(1-0.00995475)=0.691402719

So 69.1% house edge is what I got if the payout is 30 to one for 8 decks.

Am I right here? I'm just copying and pasting with the amount of cards for 8 decks but I have no idea if this is the right formula. Ultimately, I'd really love to know how to calculate the probability for any 3 face cards value in 8 decks, any 3 face cards same suit and the probability of getting a 3 face cards straight flush in 8 decks too.


Thank you so much in advance.

also for my education and understanding...

(12/52)*(11/51)*(10/50)=0.009954751

in this formula I get the 12/52 because we have 4 J, 4 Q, 4 K=12 value and we have total 52 cards in a single deck

but what are these number represented 11/51? 10/50?

Thanks again =D
Tommy

You're close, but you're misapplying the single-deck math to the 8-deck game.

There are 52 x 8 = 416 cards in an 8-deck shoe. Of these, 4 x 8 = 32 are Jacks, 32 are Queens, and 32 are Kings (there are 8 of each particular rank and suit - e.g. 8 Kings of Spades), so there are 96 face cards.

The probability that the first three cards are face cards = 96/416 x 95/415 x 94/414 = about 1/83.37178.

For three face cards of the same suit, 96 cards of the 416 available for the first card are face cards; for the second card, there are 23 face cards remaining of that suit (8 Jacks, 8 Queens, and 8 Kings, minus the one that was the first card) out of the 415 remaining; for the third card, there are 22 face cards remaining of that suit out of 414, so it's 96/416 x 23/415 x 22/414 = about 1/1471.3636.

For J,Q,K of the same suit, again any of the 96 face cards out of the original 416 can be the first card; however, there are only 16 cards of the first card's suit that don't match the first card's rank (e.g. if the first card is the King of spades, the second can be the Jack or Queen of spades, and there are 8 of each) out of the 415 remaining, and there are only 8 cards (whichever of J, Q, K is not in the first two cards) out of 414 for the third card, so it's 96/416 x 16/415 x 8/414 = about 1/5816.4844.

^^^^^^ At least it appears as though our math checks one another =P.

Well, the odds of a dealer hand getting 1 face card (K Q J Only), from an 8D shoe is... 3x4x8 / 416 cards = 96/416 = 0.2318, about 1 in 4.25

P(3 face cards) = P(1st face card) * P(2nd face card) * P(3rd face card) = (96/416) * (95/415) * (94/414) = 0.012, about 1 in 84.


Now on to problem 2: P(3 face cards same suit) = P(1st face card) * P(2nd face card SAME SUIT) * P(3rd face card SAME SUIT) = (96/416) * (23/415) * (22/414) = 0.0006786, about 1 in 1500.

*For problem 2 I took it as 3 suited face cards... Thus Kh-Kh-Kh is valid.


Lastly, problem 3: P(KQJ Straight Flush) = P(1st face card) * P(2nd face same suit diff rank) * P(3rd face same suit last rank) = (96/416) * (16/415) * (8/414) = 0.0001719, about 1 in 5825.

*For problem 3 this assumes any order... Thus, Kh-Jh-Qh is valid.

...I'm not sure what the payouts are in respect to so I can't do the house edge. Is the 16-1 payout for any 3 face cards, for 3 suited face cards, or ?? Is the 30-1 payout for straight flush 3 face cards?

Romes,

Thank you for your reply. I really appreciate that. At the moment, I'm just trying to get the House Edge % down for these 3 scenarios, and from this number perhaps I will have a different Tiers for the appropriate payouts.

For example:
Any 3 face cards= 30 to 1 (69.1% H/A)
Any 3 same suit face cards= 50 to 1 (? H/A)
And 3 face cards Straight Flush any suit= 100 to 1 (?H/A)

Would you please show me how to calculate the H/A for all three payouts above? and From there I can use your math try to figure out different payout as necessary for other scenarios..

Quote: ThatDonGuy

You're close, but you're misapplying the single-deck math to the 8-deck game.

There are 52 x 8 = 416 cards in an 8-deck shoe. Of these, 4 x 8 = 32 are Jacks, 32 are Queens, and 32 are Kings (there are 8 of each particular rank and suit - e.g. 8 Kings of Spades), so there are 96 face cards.

The probability that the first three cards are face cards = 96/416 x 95/415 x 94/414 = about 1/83.37178.

For three face cards of the same suit, 96 cards of the 416 available for the first card are face cards; for the second card, there are 23 face cards remaining of that suit (8 Jacks, 8 Queens, and 8 Kings, minus the one that was the first card) out of the 415 remaining; for the third card, there are 22 face cards remaining of that suit out of 414, so it's 96/416 x 23/415 x 22/414 = about 1/1471.3636.

For J,Q,K of the same suit, again any of the 96 face cards out of the original 416 can be the first card; however, there are only 16 cards of the first card's suit that don't match the first card's rank (e.g. if the first card is the King of spades, the second can be the Jack or Queen of spades, and there are 8 of each) out of the 415 remaining, and there are only 8 cards (whichever of J, Q, K is not in the first two cards) out of 414 for the third card, so it's 96/416 x 16/415 x 8/414 = about 1/5816.4844.

ahhh thank you Don!!! Now I understand the math behind it now.

i'm so sorry if I repeat myself again but let's say the pay out is following:

any 3 face cards=30 to 1
any 3 same suit face cards= 50 to 1
any 3 face card straight flush= 100 to 1

How do I calculate the house edge for this particular scenario?

Thanks again you guys rock.

Quote: tommyngo215

Romes,

Thank you for your reply. I really appreciate that. At the moment, I'm just trying to get the House Edge % down for these 3 scenarios, and from this number perhaps I will have a different Tiers for the appropriate payouts.

For example:
Any 3 face cards= 30 to 1 (69.1% H/A)
Any 3 same suit face cards= 50 to 1 (? H/A)
And 3 face cards Straight Flush any suit= 100 to 1 (?H/A)

Would you please show me how to calculate the H/A for all three payouts above? and From there I can use your math try to figure out different payout as necessary for other scenarios..

So, here's what you have to do (I think; I am not a math guy).

First, figure out the total number of possible 3 card hands in baccarat.

That's the total number of possible deals (You will need this number for your calculation), less the total number of hands that won't reach 3 cards (like a natural 9 in either hand).

Then subtract all possible 3 card hands that have zero faces. Then subtract those with exactly 1 face. Then exactly 2 faces.

That leaves you with all 3 card with faces hands. Save this number, but use it for the next 2 calcs.

Subtract from that number, all suited 3OAK. Save the number of 3OAKsuited possible.

Also subtract from that new number, all 3 card hands that make a straight flush. (both subsets overlap in all 3 card, but not with each other). Save the number of 3 faces SF. And save your new result (3 face card hands less the 3OAKsuited and SFfaces), as it's the one to use in your chart.

I highly suspect the 3 identical faces will be a lot rarer than the 3 to a straight flush, and 3 identical should be your top pay, not the SF as you have it listed. That's because, in your 416 card deck, there are only total 8 cards in the pool of availability (for each face) and you must match 3 of them to win. With the SF, with any one card (of that same subset), there are 8 cards available for one part of it and another 8 cards available for the other. I could be wrong.

Now, to determine a HE, you make a spreadsheet.

Column 1 is your hands. Yours in row order are, Suited 3OAK Faces, Suited SF Faces, Any 3 Faces (not including those above), All Other.

Column 2 is your rate of occurrence. Yours might be (swagging numbers), 324, 1826, 426,000 , 98,355,466,400. Those MUST add up to the total number of baccarat hands possible in a game, both 2 and 3 card hands (which is shown at the bottom line, under totals). Use the Summation function on the totals line to verify your hands add up correctly.

Column 3 is your probability. This is where you will divide each of those numbers by the total bac hands possible, for a decimal that must add up to 1 in that column. The guys here carry this number out to 6 decimal places. (FWIW but not on your table: Each number multiplied by 100 represents the percentage of times you will see that hand). Again, use the summation function on the totals row to ensure your numbers add up to 1. Note: you'll want to tell the program to display this info (under format column, for this and Column 5) to display 6 decimal places in numbers format (default is unusable; you get exponentials and stuff).

Column 4 is how many units you will pay for each 1 unit bet. This is where you'll list your paytable amounts, which might be 500, 100, 5, and -1 for your 4 elements.

Column 5 is the reason you're doing this on an excel spreadsheet, so you can play around with the numbers. For each row, you will multiply the number in Column 3, probability, by the number in Column 4, units paid. When you make those, use the box number over the box number(e.g. =c10*d10), don't just type the numbers themselves, so that you can play with the pays and it will update each time.

At the bottom of that column, you will sum total the numbers in that column (summation of the boxes above it again). That will give you your house edge. If it favors the casino, the number will be negative (because of the -1 for any other hand loses). If it favors the player, the number will be positive (and your pays are too high). Keep adjusting your pays column until you get a total in that last column that's something like: -.032977, which is a house edge FOR the casino (negative) of 3.297%. Your target for a sidebet is somewhere from 2 to 10% (me, I think anything over 5% is too high to play, but that's me, and another discussion), and in no case higher than -.25xxx (because some places are not allowed to offer a bet with a higher edge, and this is too high to be viable in most cases anyway).

Personally, if I were doing this bet, I would add in any 3oak and maybe also any 3 to a straight (faces in both cases), which is one step more complicated for you to figure but gives 5 pay elements. But since those two both fall within the any 3 faces subset, it doesn't change the hit frequency, it would just change the pays and maybe make more sense at the table (that you would pay 3oak/any a bit more than just any 3 faces, for example). It would be especially useful if you need to fine-tune your pays to get a good HE, because say 5 to 1 is too high a HE for any 3 faces, but 6 to 1 is a player advantage. Both the other top hands are going to be rare enough that you'd have to crank them to a scary amount (to the casino) to get the HE in line, but the other 2 hands I'm suggesting will be more common than your rare hands, so they'll have more effect on your HE if you insert them in the middle.

Hope this makes sense and helps. Math guys are welcome to correct any bad info I'm providing, but I feel pretty good about these instructions.

EDIT: I misread one of your pays, thinking you were looking for 3OAK suited where you said any 3 faces same suit. Sorry about that. Any 3 faces same suit would happen more often than 3SF. The rest above is still good, just adjusting what you calculate as subsets. 3SF will be a subset of 3Suited Faces.

Quote: tommyngo215

ahhh thank you Don!!! Now I understand the math behind it now.

i'm so sorry if I repeat myself again but let's say the pay out is following:

any 3 face cards=30 to 1
any 3 same suit face cards= 50 to 1
any 3 face card straight flush= 100 to 1

How do I calculate the house edge for this particular scenario?

You had the calculation right - it's 100% x ((the probability of winning) x (the amount won) - (the probability of losing) x (the amount lost)).

For 3 face cards, it's (1/83.37178 x 30) - ((1 - 1/83.37178) x 1) = -0.628171, or 62.8171%.
For 3 face cards of the same suit, it's (1/1471.3636 x 50) - ((1 - 1/1471.3636) x 1) = 96.6%
For JQK of the same suit, it's (1/5816.4844 x 100) - ((1 - 1/5816.4844) x 1) = 98.2635%.

Those payouts seem really low considering what the probabilities are.

^^^^^^ You're absolutely right about the probabilities and payouts... Thus a 57% House Edge. I wouldn't think this would even be legal in the states.

Quote: tommyngo215

Romes,

Thank you for your reply. I really appreciate that. At the moment, I'm just trying to get the House Edge % down for these 3 scenarios, and from this number perhaps I will have a different Tiers for the appropriate payouts.

For example:
Any 3 face cards= 30 to 1 (69.1% H/A)
Any 3 same suit face cards= 50 to 1 (? H/A)
And 3 face cards Straight Flush any suit= 100 to 1 (?H/A)

Would you please show me how to calculate the H/A for all three payouts above? and From there I can use your math try to figure out different payout as necessary for other scenarios..

You'd do them like the Wizard does on any of his Expected Return tables... Sum up all the possibilities by their frequency, win amount, etc, etc. Here are some other house edges calculated by the Wizard.

So for your side bet:

Event	Pays	Probability	Return
Any 3 Face Cards	30	.012	0.36
Any 3 Suited Face Cards	50	0.0006786	0.0339
3 Face Card Straight Flush	100	0.0001719	0.01719
Any Other (Loss)	-1	0.9872	-0.9872
Total		1	-0.5761

Thus, this side bet has a 57.61% House Edge. What is this side bet called? Is it even in a real casino? US or not? Perhaps the Wizard could update his baccarat side bets page to say NEVER to play this side bet lol.

Quote: ThatDonGuy

You had the calculation right - it's 100% x ((the probability of winning) x (the amount won) - (the probability of losing) x (the amount lost)).

For 3 face cards, it's (1/83.37178 x 30) - ((1 - 1/83.37178) x 1) = -0.628171, or 62.8171%.
For 3 face cards of the same suit, it's (1/1471.3636 x 50) - ((1 - 1/1471.3636) x 1) = 96.6%
For JQK of the same suit, it's (1/5816.4844 x 100) - ((1 - 1/5816.4844) x 1) = 98.2635%.

Those payouts seem really low considering what the probabilities are.

Don,

so if that's the calculations for them. Then please correct me if I'm wrong for these scenarios below:

For 3 face cards, non-suit 3 of a kind
(96/416) x (31/415) x (30/414) = 0.00124914373
therefore if the payout is 40 to 1 then the HE is
(0.00124914373 x 40) - ((1-0.00124914373) x 1)=94.8%

For 3 face cards, suited 3 of a kind
(96/416) x (7/415) x (6/414) = 0.00005641294
therefore if the pay out is 100 to 1 then the HE is
(0.00005641294 x 100) - ((1-0.00005641294) x1)=99.4%

For 3 face cards (straight non suited)
(96/416) x (64/415) x (63/414)=0.0054156425031229
therefore if the pay out is 90 to 1 then the HE is
(0.0054156425031229 x 90) - ((1-0.0054156425031229) x1)=50.7%


Don, Romes yea the HE is staggering. Thank to you guys now I know what I have to work with in order to set up a more appropriate payout table for these side bets. Of course, it's gonna have to be a higher amount in order to justify the ridiculous HE they provide. Heck, perhaps this will go further as a WAP (Wide Area Progressive) jackpot idea per se.

I think I have a cool idea for a new exciting game for U.S. But I have no idea whether it will work or not. By crunching these numbers, it certainly help me a lot to adjust the game and develop it more appropriately along with the side bets. To answer your questions Romes, this side bet is not in any U.S casinos lol. They have it in Macau as listed in the 3 card baccarat section on this website, in a few joints but just 3 face cards, no multi-tiers, no WAP.

beachbumbabs: thank you for your suggestions. I agreed with you that it's great to have a 5 different payout elements like that. Oh boy, it's way lights years too advance for me lol my IQ is not high at all haha i'm afraid it's out of reach for me to make such table. I wouldn't even know where to begin.

Quote: tommyngo215

Don,

so if that's the calculations for them. Then please correct me if I'm wrong for these scenarios below:

For 3 face cards, non-suit 3 of a kind
(96/416) x (31/415) x (30/414) = 0.00124914373
therefore if the payout is 40 to 1 then the HE is
(0.00124914373 x 40) - ((1-0.00124914373) x 1)=94.8%

For 3 face cards, suited 3 of a kind
(96/416) x (7/415) x (6/414) = 0.00005641294
therefore if the pay out is 100 to 1 then the HE is
(0.00005641294 x 100) - ((1-0.00005641294) x1)=99.4%

For 3 face cards (straight non suited)
(96/416) x (64/415) x (63/414)=0.0054156425031229
therefore if the pay out is 90 to 1 then the HE is
(0.0054156425031229 x 90) - ((1-0.0054156425031229) x1)=50.7%

If the first and third ones specifically have to be "not all in the same suit," then those have to be calculated differently; count the total number of sets of 3 cards that are 3 of a kind or a straight, as appropriate, and then subtract how many suited 3s of a kind/straights there are.
Also, for the 3s of a kind, are they limited to Jacks / Queens / Kings?

Also, something Romes posted makes me think we're going about this slightly wrong. I was under the assumption that the three different hands were three different side bets, but Romes seems to be under the assumption that it's a single bet, which makes much more sense.
If you know all of the different paying hands and what the payouts are for each one, then the overall house edge can be calculated; for each possible result, including losing, multiply the probability of that result by how much profit (or loss) you make on that result, then add the numbers up and divide by the bet (which is usually 1 to make calculating easier).

Yea Don, it's a single side bet per hand. Each hand gets 3 cards like 3 card baccarat in Macau. It's limited to Jacks/Queens/Kings only. 10 has zero value. 3 face card would be the best hand instead of 9...the rest of Bac rules apply to the game.

Thank you very much for your reply.

Quote: tommyngo215

Yea Don, it's a single side bet per hand. Each hand gets 3 cards like 3 card baccarat in Macau. It's limited to Jacks/Queens/Kings only. 10 has zero value. 3 face card would be the best hand instead of 9...the rest of Bac rules apply to the game.

In that case, the problem is completely different. If you have a list of all of the different paying hands and what they pay, then we can show you how to calculate the house edge.

Quote: ThatDonGuy

In that case, the problem is completely different. If you have a list of all of the different paying hands and what they pay, then we can show you how to calculate the house edge.

Please see your PM =P

Are you sure about those payouts? I am getting a player advantage of about 26%.

No haha

this is where I'm trying to figure out how to find appropriate payout table for them with a reasonable house edge. Were they too high? What are you suggesting?

Thanks,
Tommy

Would this be more feasible?

3 any face card (any J, any Q, any K not in any order) pays 30 to 1
3 any face card that makes a straight non-suited pays 50 to 1
3 any face cards Flush (suited) pays 75 to 1
3 any face cards 3 of a kind (non-suited) pays 100 to 1
3 any face cards Straight Flush pays 125 to 1
3 any face cards 3 of a kind (suited) pays 150 to 1

Quote: tommyngo215

Would this be more feasible?

3 any face card (any J, any Q, any K not in any order) pays 30 to 1
3 any face card that makes a straight non-suited pays 50 to 1
3 any face cards Flush (suited) pays 75 to 1
3 any face cards 3 of a kind (non-suited) pays 100 to 1
3 any face cards Straight Flush pays 125 to 1
3 any face cards 3 of a kind (suited) pays 150 to 1

I showed you exactly how to do the math, so you tell me =).

Quote: tommyngo215

Would this be more feasible?

3 any face card (any J, any Q, any K not in any order) pays 30 to 1
3 any face card that makes a straight non-suited pays 50 to 1
3 any face cards Flush (suited) pays 75 to 1
3 any face cards 3 of a kind (non-suited) pays 100 to 1
3 any face cards Straight Flush pays 125 to 1
3 any face cards 3 of a kind (suited) pays 150 to 1

That's a house edge of 44.95868%

Here are the hand counts:

In each case, any of the 96 face cards can be the first card

Suited three of a kind: 7 cards for the second card x 6 cards for the third x 96 = 4032

Straight flush: 16 cards for the second (8 cards of each of the other two ranks of the same suit) x 8 cards for the third (the 8 cards of the remaining rank of the same suit) x 96 = 12,288

Unsuited three of a kind = all threes of a kind - the suited ones:
31 cards for the second card x 30 for the third card x 96 - the 4032 suited threes of a kind = 85,248

Flush = all flushes minus suited threes of a kind minus straight flushes
23 cards for the second card x 22 for the third card x 96 - the 4032 suited threes of a kind - the 12,288 straight flushes = 32,356

Straight = all straights minus straight flushes
64 cards for the second card (32 cards in each of the remaining two ranks) x 32 cards for the third card x 96 - 12,288 = 184,320

Other win - any of the 96 x 95 x 94 sets of three face cards not already counted = 539,136

Losses - any of the 416 x 415 x 414 sets of three cards besides the 539,136 sets of three face cards = 70,615,680

To calculate the house edge, multiply each payout by the number of hands of that type, then subtract 70,615,680 (the number of losses), and divide by 714,729.6 (there are a total of 71,472,960 three-card hands, but then you would multiply the result by 100 to get a percentage).

Here's one to consider, and the house edge is still a comfortable 7%:
Suited Three Of A Kind 5000-1
Straight Flush 500-1
Flush 250-1
Unsuited Three Of A Kind 100-1
Straight 50-1
Other 3 Face Cards 25-1
Notice that I have a flush paying more than an unsuited three. This is because there are over 2 1/2 times as many unsuited threes as there are flushes. Remember that there are 32 face cards of each rank, but only 24 of each suit.

Thanks so much Romes & Don. You guys have been a great help.

Quote: ThatDonGuy

You're close, but you're misapplying the single-deck math to the 8-deck game.

There are 52 x 8 = 416 cards in an 8-deck shoe. Of these, 4 x 8 = 32 are Jacks, 32 are Queens, and 32 are Kings (there are 8 of each particular rank and suit - e.g. 8 Kings of Spades), so there are 96 face cards.

The probability that the first three cards are face cards = 96/416 x 95/415 x 94/414 = about 1/83.37178.

For three face cards of the same suit, 96 cards of the 416 available for the first card are face cards; for the second card, there are 23 face cards remaining of that suit (8 Jacks, 8 Queens, and 8 Kings, minus the one that was the first card) out of the 415 remaining; for the third card, there are 22 face cards remaining of that suit out of 414, so it's 96/416 x 23/415 x 22/414 = about 1/1471.3636.

For J,Q,K of the same suit, again any of the 96 face cards out of the original 416 can be the first card; however, there are only 16 cards of the first card's suit that don't match the first card's rank (e.g. if the first card is the King of spades, the second can be the Jack or Queen of spades, and there are 8 of each) out of the 415 remaining, and there are only 8 cards (whichever of J, Q, K is not in the first two cards) out of 414 for the third card, so it's 96/416 x 16/415 x 8/414 = about 1/5816.4844.

and Romes

Quote: Romes

P(3 face cards) = P(1st face card) * P(2nd face card) * P(3rd face card) = (96/416) * (95/415) * (94/414) = 0.012, about 1 in 84.


Now on to problem 2: P(3 face cards same suit) = P(1st face card) * P(2nd face card SAME SUIT) * P(3rd face card SAME SUIT) = (96/416) * (23/415) * (22/414) = 0.0006786, about 1 in 1500.

*For problem 2 I took it as 3 suited face cards... Thus Kh-Kh-Kh is valid.


Lastly, problem 3: P(KQJ Straight Flush) = P(1st face card) * P(2nd face same suit diff rank) * P(3rd face same suit last rank) = (96/416) * (16/415) * (8/414) = 0.0001719, about 1 in 5825.

Hello,

Hope you guys have a wonderful weekend. I got everything for the most parts now. However, I do have another question. How do you guys figure out the frequent hit rate like this? 1 in 84, 1 in 1500 and 1 in 5825

Thank you in advance for your replies =D

Quote: tommyngo215

How do you guys figure out the frequent hit rate like this? 1 in 84, 1 in 1500 and 1 in 5825

What's "frequent hit rate"?

If you want to know how we got 1/84 (well, I got 1/83,37178). it is explained in both of our replies that you quoted above. I don't know how else to say it.

In Baccarat, what are the impacts on win-loss-tie ratio when the number of cut cards varies? Some casinos cut more, some less...

Quote: UCivan

In Baccarat, what are the impacts on win-loss-tie ratio when the number of cut cards varies? Some casinos cut more, some less...

None that are remotely noticeable. You are not the first person to ask this.