Power 8’s Baccarat Side Bet discussed in Math/Questions and Answers at Wizard of Vegas

777

777

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Oct 7, 2015

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So you apply the traditional Baccarat drawing rules, go through 6 loops each representing 52 cards in a deck (52^6 iterations at a minimum). Am I correct so far? What I would like to know is the mathematical formulation or procedure that would yield the necessary 8-deck combinatorial combinations when the looping iterations are at the “three suites eights” events.

ThatDonGuy

ThatDonGuy

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Jun 22, 2011

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Thanks for this post from:
777

I think I can clarify.
There are 416 cards in the deck, and the table is counting all 416 x 415 x 414 x 413 x 412 x 411 possible ways to deal six cards in a particular order.
The deals have to be divided into four groups:
1 – neither the player nor the banker draw a third card; count the number of deals where the 4 dealt cards include 3 suited 8s and no other 8s (of any suit – otherwise, it would be counted as 4 8s In Initial Deal instead), and multiply by 412 x 411 (the number of possible permutations for the fifth and sixth cards).
2 – the player draws a third card, but the banker does not; count the number of deals where the 5 dealt cards include 3 suited 8s and no other 8s, and multiply by 411 (the number of possible cards that can be the sixth card).
3 – the banker draws a third card, but the player does not; count the number of deals where the 5 dealt cards include 3 suited 8s and no other 8s, and multiply by 411.
4 – both the player and the banker draw a third card; count the number of deals where the 6 dealt cards include 3 suited 8s and no other 8s.
Add up those four numbers.

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