Anyone got something on this?
Overall, the odds are worse to make a pair once you've removed cards of that rank.
This. And Peter, why would knowing the first card help you? The only way this could be helpful is if you are counting ranks remaining in the deck, but as Ibeatyouraces points out, when you split the pair, you just lost one of the cards you can pair with.
Just noticed that in Macau it's possible to place a perfect pairs bet after splitting pairs. It can't be the same math as it is on the original two cards but I can't find the math behind it anywhere. There must be some discount for throwing away your original winning pair but better odds by knowing your first card before making the second pair bet.
Anyone got something on this?
I don't recall ever seeing a Perfect Pairs Paytable, even though I may have. Do you happen to know it?
For the time being, I will say that it puts you at a huge disadvantage, even with knowing the dealer's card and the hit card on the first split hand. I'm going to put something that is probably simpler than perfect pairs to give you an idea.
Two Cards: Probability of Pair in single deck: (52/52 * 3/51) = 0.05882352941
Probability of Pair, Six Decks: (312/312 * 23/311) = 0.07395498392
As you can see, the EoR (Effect of Removal) is not as strong with multiple decks. The reason why is because, even removing one of the Pair cards, the overall percentage of Pair Cards remaining is greater with multiple decks.
Thus, we already know that the EoR is greater with fewer decks.
Now, that applies both ways: If you remove a Non-Pairing card from a single deck, then the difference in percentage of pairing cards will be greater than removing a single non-pairing card from a multiple deck, but removing the Pairing card will still result in a lesser probability of Pairing and a greater disadvantage due to the lower probability of success.
Okay, so if you have been dealt 8-8 and Split, the dealer's card is not an Eight and the card the first Eight gets is not an Eight...what is the probability the next card will Pair it up?
52 Cards - 4 Known Cards = 48 Cards
2/48 = 0.04166666666
Of course the casino would let you make that bet with the same PayTable!!!!
Now, let's do it for Six Decks:
312 Cards - 4 Known Cards = 308 Cards
22/308 = 0.07142857142
Again, the EoR is not as strong with multiple decks, which is why a card-counter at BJ may prefer single-decks---all else remaining equal---but you see you would still be at a greater disadvantage making the bet after you know a non-pairing card has been removed.
Again, of course they'd let you make that bet!!!!
Give me a Paytable, and I will give you more specific numbers, such as Expected Value...but the principle shall remain the same.
I'll get around to it in the late p.m. or early a.m. hours, EST, have to get ready for work now...
Okay, so the adjustment is actually pretty easy. We have eight decks, and we're going to assume (after splitting a pair) that neither the first card on the player's first split hand is a Pairing Card, nor shall the Dealer's Card be a Pairing Card.
We start with 416 Cards and we know what Four of them are.
There are three different scenarios, because there are three different pays. It was either a Perfect Pair, a Colored Pair, or a Red/Black Pair, so we will determine this based on Splitting each one of them:
The Base HE is 4.1% on Paytable A for Eight Decks.
Splitting Perfect Pairs
This means there are still six PP cards left, there are eight CP cards left, and there are 16 RBP cards left.
(6/412 * 25) + (8/412 * 12) + (16/412 * 6) - (1 * 382/412) = -0.09708737864077676 or 9.71% HE
Splitting Colored Pairs, Imperfect
This means there are still seven PP cards left, there are seven CP cards left, and there are 16 RBP cards left.
(7/412 * 25) + (7/412 * 12) + (16/412 * 6) - (1 * 382/412) = -0.06553398058252435 or 6.55% HE
Splitting Colored Pairs, Red/Black
This means there are still seven PP cards, eight CP cards, and fifteen RBP cards left.
(7/412 * 25) + (8/412 * 12) + (15/412 * 6) - (1 * 382/412) = -0.050970873786407855 or 5.1% HE
Why not? Can't resist. To emphasize my earlier point, we're going to look at the EoR for two decks against a Base HE of 22.3% on PayTable A, so you see that removing one card which you absolutely must start with already drives up the HE.
You start with 104, we know four of them, so 100 cards.
Zero PP Cards, two CP Cards, Four RBP Cards:
(2/100 * 12) + (4/100 * 6) - (94/100) = -0.45999999999999996 or a HE of 46%, YIKES!
One PP Card, One CP Card, Four RBP Cards:
(1/100 * 25) + (1/100 * 12) + (4/100 * 6) - (94/100) = -0.32999999999999996 or a HE of 33%, Oh, the pain!
One PP Card, two CP Cards, Three RBP Cards:
(1/100 * 25) + (2/100 * 12) + (3/100 * 6) - (94/100) = -0.27 or a House Edge of 27%!!!
Same as my first post in this thread. The fewer decks, the worse it gets. The added information doesn't help you, the only added information you need is that you were dealt a pair, so don't make the bet!!!
(24/1299 * 25) + (25/1299 * 12) + (50/1299 * 6) - (1200/1299) = 0
It's exactly a break-even proposition.
And at 26 decks:
(25/1351 * 25) + (26/1351 * 12) + (52/1351 * 6) - (1248/1351) = 0.0007401924500370915
The player is at a 0.07% advantage, with the advantage increasing the more decks you add.
Such as 100 Decks:
(99/5199 * 25) + (100/5199 * 12) + (200/5199 * 6) - (4800/5199) = 0.014425851125216437
With the advantage still increasing the more decks you add.
An infinite deck can be construed as a single-deck with no EoR, so:
(1/52 * 25) + (1/52 * 12) + (2/52 * 6) - (48/52) = 0.019230769230769273 or a 1.923% Player Advantage.
It can get no better than that, for proof, consider 1,000,000 decks:
(999999/51999999 * 25) + (1000000/51999999 * 12) + (2000000/51999999 * 6) - (48000000/51999999) = 0.019230288831351694