Odds of Baccarat Perfect Pairs Sidebet

Hello all. This is a splitting-hairs question, but it's driving me nuts.

So, I'm trying to figure out the odds on the perfect pairs sidebet. The bet is paid when either the Player or Banker's first two cards match exactly.

In an 8-deck shoe...

There's = 52 (different ranks and suits) * 8 * 7 = 2,912 ways to make a perfect pair. Out of a possible 416*415 = 172,640 two card hands. The probability is .01686 = 1.686%

Double this number because there's two chances at it (Player or Banker). 1.686% *2 = 3.373%

The Wiz has it at 3.345%....I'm just curious as to the discrepancy.

Thanks.

1 - (416 * 408) * (400 * 406 + 14 * 407) * 412! / 416! =~ 3.345%

I suspect you double counted the probability of both hands being pairs

Yes, double counting the instances of two Perfect Pairs was my error.

Thank You.

1 - (416 * 408) * (400 * 406 + 14 * 407) * 412! / 416! =~ 3.345%

Still trying to get my head around this formula.... 416 is the number of cards in 8 decks... the rest I' m lost on.

This formula translates well into computer code which is what I'm looking for.

Care to elaborate? Thank You.

It uses the 1-Pr(of no Perfect Pair) logic.
First Player has to mismatch
(a) 416 = any card in 8 decks (say As)
(b) 408 = any other card, not being As, = 8*51 (say Ks)
Second Banker has to mismatch. Two possibilities for the first card.
EITHER
(c) 400 = any card other than one of those dealt to player (i.e. not As nor Ks, so say Qs)
(d) 406 = any other card, not being Qs, given two (As Ks) have already gone = 406
OR
(e) 14 = any card which was given to the player (i.e. As or Ks, say As)
(f) 407 = any card not matching (e) given that one non-matching card (i.e. Ks) has already gone.
Divided by the number of ways of getting four cards = 416*415*414*413.

Quote: railer

1 - (416 * 408) * (400 * 406 + 14 * 407) * 412! / 416! =~ 3.345%

Still trying to get my head around this formula.... 416 is the number of cards in 8 decks... the rest I' m lost on.

This formula translates well into computer code which is what I'm looking for.

Care to elaborate? Thank You.
link to original post

You can find my code (in VBA) for many different Baccarat side bets, including the Pairs, in my Baccarat Edge Calculator here:

https://researchers.one/doc/61a90587ec1251c54f18ea35

Approximately 1 - (51/52)^2 =~ 3.81%

The effect of removal is significant in this case since there are only seven ways to match a card in an eight deck shoe. Still not bad for a calculation you can do in your head

Quote: charliepatrick

It uses the 1-Pr(of no Perfect Pair) logic.
First Player has to mismatch
(a) 416 = any card in 8 decks (say As)
(b) 408 = any other card, not being As, = 8*51 (say Ks)
Second Banker has to mismatch. Two possibilities for the first card.
EITHER
(c) 400 = any card other than one of those dealt to player (i.e. not As nor Ks, so say Qs)
(d) 406 = any other card, not being Qs, given two (As Ks) have already gone = 406
OR
(e) 14 = any card which was given to the player (i.e. As or Ks, say As)
(f) 407 = any card not matching (e) given that one non-matching card (i.e. Ks) has already gone.
Divided by the number of ways of getting four cards = 416*415*414*413.
link to original post

Thanks for the explanation. This makes a lot of sense to me.

So, if I was to take this process and apply it to a penetrated shoe, the formula would be a little different, because I would have to consider the effect of removal of each individual card. So I would calculate the probability of mis-matching each individual rank and suit, then sum all the probabilities.

For instance.... to take a simple example...let's say there's 12 cards left in an 8-deck shoe and, there's 5 As and 4 Ks and 3 Qs remaining....

1 -
(

5 (As) * 7(any other card) * 3(Remaining Ks) * 7 (Any card minus the Ks and one As) / 12*11*10*9(Total ways to deal 4 cards) +
5 (As) * 7(any other card) * 2(Remaining Qs) * 8 (Any card minus the Qs and one As) / 12*11*10*9(Total ways to deal 4 cards) +

4 (Ks) * 8(any other card) * 4(Remaining As) * 6 (Any card minus the As and one Ks) / 12*11*10*9(Total ways to deal 4 cards) +
4 (Ks) * 8(any other card) * 2(Remaining Qs) * 8 (Any card minus the Qs and one Ks) / 12*11*10*9(Total ways to deal 4 cards) +

3 (Qs) * 9(any other card) * 4(Remaining As) * 6 (Any card minus the As and one Qs) / 12*11*10*9(Total ways to deal 4 cards) +
3 (Qs) * 9(any other card) * 3(Remaining Ks) * 7 (Any card minus the Ks and one Qs) / 12*11*10*9(Total ways to deal 4 cards)

)

A lot of arthimetic for situations where the subset of cards contain 52 different ranks and suits... I guess that's why computers were invented.

Thanks again!