Maths behind Punto Banco House edge

How do you calculate the house Edge on Punto Banco 6 deck 5% commission when bank wins??
The maths behind the house Edge on Bank, Player and Tie.

"The long way." Seriously.

It's easier to describe with "infinite decks."
There are 100 pairs of possible (player first card, player second card) values:
(0,0), (0,1), (0,2), ..., (0,8), (0,9)
(1,0), (1,1), ..., (1,9)
(2,0), (2,1), ..., (2,9), and so on through (9,9)
The probability of a card being a zero-value is 4/13, so the probability of (0,0) = 4/13 x 4/13 = 16/169
The probability of a card being any particular nonzero value is 1/13; the probability of, say, (0,3) = 4/13 x 1/13 = 4/169
The probability of, say, (9,6) = 1/13 x 1/13 = 1/169
The same applies to the bank's first two cards.
The probability of the player being dealt (6,2) and the bank being dealt (3,0) = 1/169 x 4/169 = 4/287,131
In this case, the player has a natural 8 and the bank has 3, so the hand ends with a Player win.
If a pair of two hands is not a natural for either Player or Bank, then, if the Player draws, check each possible Player card draw, and if the Bank then draws, check each possible Bank card draw.
Add up the probabilities of Player wins for the overall Player win probability.
Add up the probabilities of Ties for the overall Tie probability.
Add up the probabilities of Banker wins to get the overall Bank probability.
To determine the house edge for each bet, add up the probabilities of the player losing the bet, then subtract (the probability of the player winning the bet multiplied by the odds that the bet pays).

If you want 6-deck specifics, it gets a little harder, as the probabilities change for each card. For example, there are 312 cards, 24 of which are 6s, so the probability of the Player being dealt (6,6) is 24/312 x 23/311; if the Player is dealt (6,5), then the probability that the Bank is dealt (2,6) is 24/310 (there are only 310 cards left in the deck) x 23/309. A full deck is always assumed, as otherwise there are far too many possible numbers to calculate.

Thank you very much for your Reply. How would that look when you would have to present it as a chart. Does These calculations exist as a chart?

The chart would be rather large. Even if you treated a Player's (2,3) the same as (3,2), which, technically, it is (so a (2,3) has a 2/169 chance of being dealt), and neither the Player nor the Bank ever drew a card, there would be 55 x 55 = 3225 rows.
Now, look at the real Player (3,2) case: the Player draws a card, so, for every one of the 55 Bank pairs of cards, there are 10 rows (1 for each value from 0 to 9 that the Player can be dealt), and for each one of those, you have to check to see if the Bank would draw, and if it does, that's 10 rows for those 10 values.

For example, if the Player is dealt (3,7) and the Bank is dealt (5,6), both have to draw a third card, so that is 100 rows (Player third card 0, Bank third card 0; 0 and 1; 0 and 2; and so on through 9 and 9).

If the Player is dealt (3,7) and the Bank is dealt (8,7), the Player draws a third card; the possible results are (0,X), (1,X), (2,X), (3,X), (4,0) through (4,9), (5,0) through (5,9), (6,0) through (6,9), (7,0) through (7,9), (8,X), and (9,X), where the first number is the Player's third card and the second is the Bank's, where X means the Bank does not draw a third card; that is 46 rows.

I count slightly more than 100,000 rows would be needed to cover every possibility.

I previously did infinite deck calculations here:

https://wizardofvegas.com/forum/gambling/tables/26936-baccarat-house-edge-calculation/#post553381

I once did it in excel, but you need some smart programming.
One idea is to look at all 4-card starting positions. Some of these won't need any additional cards. So their perms and results will be C(1)*C(2)*C(3)*C(4)*(say) 308*307.
Otherwise copy the ones needing additional cards and work through all the cards and whether the player/dealer takes them. (You can optimise this it you like.)

So you might get
0 0 0 0 - more cards needed
to
9 9 9 9 - no more cards
Also note Dealer getting 5 then 4 can be considered same as 4 then 5 but multiply by 2.