Possible Baccarat Permutations

Happy New Year! I'm a math enthusiast and I stumbled on your site when I started studying baccarat recently (thanks for the great site!).

I have a question: in your tables at /games/baccarat/basics/ you show that there are 4,998,398,275,503,360 possible combinations for a 8 deck game. I noticed that this equals P(416,6) = 416!/410!.

This seems imply that both the Player and Banker each get 3 cards every time, ignoring the 3rd card draw rules. With the 3rd card draw rules, there should be far fewer possibilities, right? Can you please help me understand what I'm missing? Thank you!

Quote: kane

Happy New Year! I'm a math enthusiast and I stumbled on your site when I started studying baccarat recently (thanks for the great site!).

I have a question: in your tables at /games/baccarat/basics/ you show that there are 4,998,398,275,503,360 possible combinations for a 8 deck game. I noticed that this equals P(416,6) = 416!/410!.

This seems imply that both the Player and Banker each get 3 cards every time, ignoring the 3rd card draw rules. With the 3rd card draw rules, there should be far fewer possibilities, right? Can you please help me understand what I'm missing? Thank you!

It’s so the combinations are weighted properly.

Thanks for the reply. So those permutations include weights of zero?

If there are weights of zero, then the probability of an outcome should not equal the number of combinations for that outcome divided by the total combinations, right?

Some games require six cards, where both the Player and Dealer draw a third card.
Some games only require five cards, where only one party draws a third card.
Some games require four cards, where both the Player and Dealer stand.

The chances of the first four cards being P(Qc Qd) D(Jc Jd); P(Qc 9d) D(Jc Jd) are the same but in the first case both parties will draw a third card, therefore one has to work through all the possible 5th and 6th cards. In the second case no further cards are drawn, so one has to multiply its probability (or combinations) by the equivalent of drawing all possible 5th and 6th cards.

As an example consider a three-set tennis match. Assume the players have equal ability what are the chances of A winning in two sets.

A	A
A	B	A
A	B	B
B	A	A
B	A	B
B	B

The total number of permutations are 8, but you need to multiply AA by 2 (AAA AAB) and BB by 2 (BBA BBB).

Quote: kane

Thanks for the reply. So those permutations include weights of zero?

If there are weights of zero, then the probability of an outcome should not equal the number of combinations for that outcome divided by the total combinations, right?

There are no "weights of zero." Instead, the hands that would have weights of 410, or 411 x 410, are spread out into the corresponding number of weights of 1.

A deal of six specific cards would have a weight of 1.
A deal of five cards (because, e.g,, the banker took a third card but the player did not) would have a weight of 411, to account for each of the 411 cards that would have been dealt had both sides taken a third card.
A deal of four cards (e.g. because the player's two-card total was 8) would have a weight of 412 x 411, to account for each of the 412 cards that would have been dealt to the player, and, for each of those, each of the 411 that would have been dealt to the banker.

In other words, assume there are always six cards dealt, even if one or two of them are burned because the player or banker didn't need a third card. Since the calculation assumes a full shoe, the actual number of cards dealt is irrelevant to the calculation. (If you were trying to calculate, say, the average number of hands per shoe, then it would matter, but in this case, it does not.)

Got it. I see where my logic was wrong. This was the key part:

"Since the calculation assumes a full shoe, the actual number of cards dealt is irrelevant to the calculation. "

Thanks guys!