railer
railer
Joined: Mar 2, 2012
  • Threads: 11
  • Posts: 44
October 3rd, 2022 at 8:22:19 PM permalink
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.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 31
  • Posts: 2318
October 3rd, 2022 at 9:51:08 PM permalink
1 - (416 * 408) * (400 * 406 + 14 * 407) * 412! / 416! =~ 3.345%

I suspect you double counted the probability of both hands being pairs
Itís all about making that GTA
railer
railer
Joined: Mar 2, 2012
  • Threads: 11
  • Posts: 44
October 3rd, 2022 at 10:21:37 PM permalink
Yes, double counting the instances of two Perfect Pairs was my error.

Thank You.
railer
railer
Joined: Mar 2, 2012
  • Threads: 11
  • Posts: 44
October 11th, 2022 at 8:21:50 AM permalink
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.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 38
  • Posts: 2793
October 11th, 2022 at 9:09:58 AM permalink
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.
teliot
teliot
Joined: Oct 19, 2009
  • Threads: 42
  • Posts: 2498
October 11th, 2022 at 12:20:14 PM permalink
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
Climate Casino: https://climatecasino.net/climate-casino/
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 31
  • Posts: 2318
October 11th, 2022 at 1:09:11 PM permalink
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
Itís all about making that GTA
railer
railer
Joined: Mar 2, 2012
  • Threads: 11
  • Posts: 44
October 12th, 2022 at 8:39:32 AM permalink
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!
Last edited by: railer on Oct 12, 2022

  • Jump to: