Winning Banker with three card total of 7.

Hello Wizards,

My question is related to EZ Baccarat. I wanted to know that how many times a winning Banker with three card total of 7 can appear in an 8 deck shoe?

Over the large sample of say 600 shoes, how many times a winning Banker with three card total of 7 can appear?

What's the formula to calculate this particular appearance of winning Banker over n number of hands?

Thanks

The usual method of working out chances of things happenning in Baccarat is to work through all the permutations of hands.

(i) Rather than looking at all 13 cards, you can simplify it to 10 cards where there are 16 "Tens" in each deck. Also hands such as 97 and 79 can be combined (and multiplied by 2).
(ii) First list all the two-card starting hands for the Player; repeat for the Dealer. This will give a list (and permuations of each) of four cards and their totals for the Player and Banker.
(iii) Some of these will not take an additional card, and some will. Work through them all.
(iv) Using the number of decks in use, work out the permutations for each card drawn; then add up the ones you want to know about.

You can either use a program or a very large spreadsheet!

Quote: sinister1

Over the large sample of say 600 shoes, how many times a winning Banker with three card total of 7 can appear?

This is something better solved through simulation than calculation, as you would have to go through pretty much every possible shuffling of the 416 cards in an 8-deck shoe.

After about 200 million shoes (burn 2-11 cards from the top, and deal all the way through the shoe), I got about 6.3 hands per shoe where the banker won with a three-card 7.

Let me rephrase the question. How many times a Dragon (winning Banker with three card total of 7) comes up in a shoe which pays 40:1 in EZ Baccarat?

It is definitely not 6.3 hands as you worked out as it is not in line with the probability.

Since it pays 40:1, its occurrence will be very limited. I just want to know the no. of such occurrences in a shoe and over a sample of 600 shoes.

Thanks.

To further explain what Charlie has said:

1. If the probability of "a winning banker hand of 7" in the first hand dealt from a fresh 8 deck shoe is P then the average probability of "a winning banker hand of 7" in n hands from an 8 deck shoe is (n x P).

BTW, there are 65 distinct two card hands in Baccarat and so the number of distinct "starting Player vs Banker hands" is (65 x 65 =) 4225.

However, if you want to list all the possible Baccarat hands that result in a "a winning banker hand of 7" it is possible to take these short cuts over the more general method that Charlie describes:

1. whenever the player's two card hand is a 9 or 8 or 7 then it doesn't matter what the banker's hand is because the banker cannot have a winning hand of 7. Of the 65 distinct starting hands for player, 15 of them will be a 9, 8 or 7:

A8, A7, A6, T9, T8, T7, 98, 72, 63, 62, 54, 53, 52, 44, 43

So that's 15 x 65 = 975 different "starting Player vs Banker hands" that you don't need to list. So that initial list of "starting Player vs Banker hands" is reduced from 4225 to 3350.

2. Whenever the banker's' two card hand is 9 or 8, it does not matter what the player's cards are - because you know that the banker will not have a winning hand of 7. So that further reduces the list of "starting Player vs Banker hands" by 50 x 9 =450. So, that initial list of "starting Player vs Banker hands" is further reduced to 2990.

You can keep applying this methodology as you list hands that require a 5th or 6th card. Literally, the only combinations of cards that you want on your list are those in which the dealer makes a two card 7 or a three card 7 (and also for which the player's 3 card hand is 6 or lower.) This kind of methodology will greatly reduce the size of your spreadsheet to something manageable!

Quote: sinister1

Let me rephrase the question. How many times a Dragon (winning Banker with three card total of 7) comes up in a shoe which pays 40:1 in EZ Baccarat?

It is definitely not 6.3 hands as you worked out as it is not in line with the probability.

Since it pays 40:1, its occurrence will be very limited. I just want to know the no. of such occurrences in a shoe and over a sample of 600 shoes.

Thanks.

The Wizard's page on EZ Baccarat can answer your question. For an 8-deck shoe, his probability of a 3-card-7-banker win is about 0.022534.

For 80 hands, the expected number of Dragon 7s is about 1.803.

Quote: sinister1

Let me rephrase the question. How many times a Dragon (winning Banker with three card total of 7) comes up in a shoe which pays 40:1 in EZ Baccarat?

It is definitely not 6.3 hands as you worked out as it is not in line with the probability.

Oh, a 3-card hand of 7... (actually, I knew that all along, but I forgot to take that into account when doing the calculations.)

Now I get 1.85 wins and 82 hands per shoe.

That's the one I was looking for! Thank you so much guys. You truely are math wizards.

Quote: gordonm888

...BTW, there are 65 distinct two card hands in Baccarat and so the number of distinct "starting Player vs Banker hands" is (65 x 65 =) 4225....

Welcome back after your break. Actually I got there are 55 starting hands (10 pairs and 10*9/2 non pairs) giving 3025 lines for the first spreadsheet. This leads to about 103411 lines on the second spreadsheet (or you can split them across two as I did.)

Once you've got the DO loop going you might as well do them all, it also provides a check that the total number of hands considered equals 416*415.... as appropriate. It also means you have something that can be used for other sidebets without recoding the loops. (Mine was last used for players winning with Red 8's!)