Power 8’s Baccarat Side Bet

On Wizard of Odds site there is a table on Power 8’s Baccarat side bet listed a Three Suited Eights event with 666,369,134,592 combination.

https://wizardofodds.com/games/baccarat/side-bets/power-8s/

I want to know and to understand how the 666,369,134,592 combination was derived.

Looping through all the way the cards come out that cause three suites eights to win.

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.

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.