WHAT ARE THE ODD'S

What are the odds of getting matching cards (both value and suit) in a hand of Baccarat for both the Bank and Player hands in a 8 deck shoe? For example:

Player dealt - Ace of Diamonds and Jack of Diamonds on first 2 cards.

Banker Dealt same first 2 cards - Ace of Diamonds and Jack of Diamonds

Then each hand draws and both Banker and Player draw the 5 of Diamonds (resulting in a tie hand of course).

All cards are an exact match for each hand! WHAT ARE THE ODDS mathematically and verbally (such as 1 in 28 billion)?? How many hands would have to be dealt to result in this expectation? HAS TO BE ASTRONOMICAL!!!! This actual hand was dealt at the Beau Rivage Hotel & Casino Biloxi, MS

Quote: rck58

What are the odds of getting matching cards (both value and suit) in a hand of Baccarat for both the Bank and Player hands in a 8 deck shoe? For example:

Player dealt - Ace of Diamonds and Jack of Diamonds on first 2 cards.

Banker Dealt same first 2 cards - Ace of Diamonds and Jack of Diamonds

Then each hand draws and both Banker and Player draw the 5 of Diamonds (resulting in a tie hand of course).

All cards are an exact match for each hand! WHAT ARE THE ODDS mathematically and verbally (such as 1 in 28 billion)?? How many hands would have to be dealt to result in this expectation? HAS TO BE ASTRONOMICAL!!!! This actual hand was dealt at the Beau Rivage Hotel & Casino Biloxi, MS

Not as easy as asking 'what are the odds of one person being dealt two cards and another person being dealt the same two cards, then each getting a third card which is the same.' Because some two card matches would not require a third card draw using the rules of baccarat. Also, certain hands which themselves have repeats (5 of hearts, 5 of hearts) would be rarer of course for the dealer to match. But a smart WoVer will figure it out for you. I'll gues around 1 in a million.

I’m getting roughly 102,976 / 52^6 or about 1 in 192,000.

I used infinite deck, which has always been very accurate (within a couple absolute basis points) compared to 8 decks.

I took a few other minor shortcuts as they simplified the calculations though I believe they’d have a very small effect on the result. So this is an estimate

Quote: Ace2

I’m getting roughly 102,976 / 52^6 or about 1 in 192,000.

I used infinite deck, which has always been very accurate (within a couple absolute basis points) compared to 8 decks.

I took a few other minor shortcuts as they simplified the calculations though I believe they’d have a very small effect on the result. So this is an estimate

Not knowing how you figured it out, infinite deck has to be off quite a bit from 8 deck. Only 7, not 8 of whatever card we are taking about available.

Quote: SOOPOO

Not knowing how you figured it out, infinite deck has to be off quite a bit from 8 deck. Only 7, not 8 of whatever card we are taking about available.

Yes I suppose you’re right in this case since we are talking about duplicating cards at least 3 times.

So the chance of banker duplicating a certain card on the next draw is closer to 1 in 59 for 8 deck instead of 1 in 52 for infinite. I should revise my estimate by a factor of (59/52)^3 but I’m sticking with my original estimate since I just fixed an error in it that more or less cancels the infinite/8 deck discrepancy. As you can probably tell this is a fairly rough estimate...I’d be satisfied to be within 5% of the exact number since I didnt spend much time on this. It probably wouldn’t be rocket science to calculate the exact number but quite tedious due to effect of removal.

For the calculation I just ran through the possible combinations using infinite deck. The player must have 0-5 for his first two cards, can draw anything to a 0-2, anything but 8 to a 3, 2-7 to a 4 and 4-7 to a 5. For instance there are (16 x 4 + 4 x 4 x 4) x 16 x (2 x 2) ways for player to get 5 on first two cards, draw 4-7 then have the banker duplicate. 16 x 4 ways for 0,5 plus 4 x 4 x 4 ways for non-zero combinations times 16 ways to draw 4-7 times 2 x 2 to account for the fact that the initial two cards can come in two orders for both player and banker. Add them all up for starting hands of 0-5, divide by 52^6 and by (59/52)^3

I just made an exact calculation which was much easier than I expected. I get:

(32 * 32 * 2 * 7 * 7 *2 *7) * (120 * 7 + 128 + 368 * 7 + 376 + 400 * 36.25 + 184 * 7 + 176 * 1.5) +
(32 * 32 * 2 * 7 * 7 * 6 * 2 * 5) * 67 +
(32 * 7 * 6 * 5 * 7) * (184 * 2 + 408 * 7) +
(32 * 7 * 6 * 5 *4 * 3) * 54

= 28,618,648,576 permutations where player hand matches banker.

divide by (416 * 415 * 414 * 413 * 412 * 411) total permutations for a probability of ~ 1 in 174,655.

The first line counts perms with no duplicates in player hand. Second part accounts for the different 3rd cards allowable to fit the problem
The 2nd line counts perms when player 3rd card matches one of first 2 cards
The 3rd line counts perms when player's first 2 cards match
The 4th line counts perms when all 3 player's cards match

I should a propose side bet for player hand matches banker paying 100,000 to 1. Nice 42% house edge .

Quote: Ace2

I should a propose side bet for player hand matches banker paying 100,000 to 1. Nice 42% house edge .

Sure at the start of the shoe. Wonder is it would be countable enough.

Ace2,
Thank you very much for the answer!! It was truly amazing to see this hand actually dealt. Best Regards, rck58

Hello Ace,
Just wondering if the calculation took into account that all the cards were the same suit (not just matching cards)?