Any pair probability

Say a 8 decks baccarat, keep dealing, probability of any pair will change whenever the combinations of cards in shoe changed.
But if it is a full 8 decks shoe, take out one card randomly, the probability of any pair keep unchanged.
I tried some numbers of decks, from 1/4 deck, 1/2 deck, 1 deck, 4 decks, 8 decks, 18 decks, ..., same thing happens, take out one card from the full shoe, probability of any pair keep unchanged.
And this only happens from full shoe to full shoe -1.
Why ?

Quote: 7up

Say a 8 decks baccarat, keep dealing, probability of any pair will change whenever the combinations of cards in shoe changed.
But if it is a full 8 decks shoe, take out one card randomly, the probability of any pair keep unchanged.
I tried some numbers of decks, from 1/4 deck, 1/2 deck, 1 deck, 4 decks, 8 decks, 18 decks, ..., same thing happens, take out one card from the full shoe, probability of any pair keep unchanged.
And this only happens from full shoe to full shoe -1.
Why ?

Assuming that's correct, the probability of each of the other possible pairs goes up by 1/12th the amount that the probability of the removed card pairs goes down.

Effect of card removal. When you're always starting with a fresh deck/shoe/etc you don't have any knowledge. Even if it's 8D and you didn't see the first 4D played out, the math works out the same. What changes is when you KNOW information, then you can change the math.

Example:

Single deck of cards, shuffled, and no cards missing (so no information). If you pull an Ace first card, what's the probability you'll get another Ace for a pair? 3/51... 5.88%

Now say you take those 2 aces and put them in the discard tray. Now you're left with 50 cards, and 2 aces missing. So now say you draw a 10. What are the odds you'll draw another 10 and get a pair? 4/49 = 8.16%. Why did this number go up? Since we knew more information (the missing aces) we can adjust the math accordingly.

Thus, whenever we don't know more information, then we cannot make mathematical assumptions. Unseen cards are just like cards that have yet to come out of the shoe. So if you have 8D with 4D dealt, you use the 8D math since you have no idea what's in the discard. All possibilities are equally likely.

I have a different way of describing this.

When you remove two cards from the deck the chances of a pair do indeed change, like this:

- If the two cards you remove are of the same rank, i.e. a pair, then the chance of pairing in the remainder of the deck will increase.
- If the two cards you remove are of different ranks, i.e. not a pair, then the chance of pairing in the remainder of the deck will decrease.

So you can look at the two cards you remove and notice whether they are paired or unpaired - you now have more information than you did before they were removed and thus the pairing probability for the remaining deck can be different.

HOWEVER, when you remove only one card you will always take one card away from one rank - there is no other "state" that the remaining deck can be in. Since you knew that was going to be the situation before you actually removed a card, we say that the removal of the one card does not give us any new information about the remaining deck (relative to pairs probability.) And without new information, there is no reason to expect that the probability of getting a pair from the remaining deck will be different.

Give yourself a backslap if you understood that.

Sorry for not having make a clear description.
8 decks baccarat,
New shoe 416 cards, Any-pair%=0.0746988
Dealt an Ace, 415 cards remained, Any-pair%=0.0746988(same as new shoe)
So what is the special of this? ...When a card is dealt, combinational analysis, banker win%, player win%, Tie%, all changed, but only any-pair% unchanged. (Same thing as in Blackjack, one card is dealt, EV changed, anypair% unchanged.)
And this anypair% unchanged thing, only happens at "from new shoe to one card is dealt". We have all information of which cards remain in the shoe.
If two cards are dealt from a new shoe, pair% changed.
If one cards is dealt from the middle of the shoe, pair% changed(compare with before a card is dealt, unless all the cards, A,2,3...K have the same number of cards before dealing a card).

Yes, sum of the "ups" = "down"
But why? And why only from a new shoe?

in shoe probability in shoe probability
A 32 0.005746 A 32 0.005774
2 32 0.005746 2 32 0.005774
3 32 0.005746 3 32 0.005774
4 32 0.005746 4 32 0.005774
5 32 0.005746 5 32 0.005774
6 32 0.005746 6 32 0.005774
7 32 0.005746 7 32 0.005774
8 32 0.005746 8 32 0.005774
9 32 0.005746 9 32 0.005774
10 32 0.005746 10 32 0.005774
J 32 0.005746 J 32 0.005774
Q 32 0.005746 Q 32 0.005774
K 32 0.005746 K 31 0.005413
416 0.0746988 415 0.0746988

"K" is dealt, all the individual pair% changed, but the sum of them, AnyPair% unchanged!

Quote: 7up

Yes, sum of the "ups" = "down"
But why? And why only from a new shoe?

in shoe probability in shoe probability
A 32 0.005746 A 32 0.005774
2 32 0.005746 2 32 0.005774
3 32 0.005746 3 32 0.005774
4 32 0.005746 4 32 0.005774
5 32 0.005746 5 32 0.005774
6 32 0.005746 6 32 0.005774
7 32 0.005746 7 32 0.005774
8 32 0.005746 8 32 0.005774
9 32 0.005746 9 32 0.005774
10 32 0.005746 10 32 0.005774
J 32 0.005746 J 32 0.005774
Q 32 0.005746 Q 32 0.005774
K 32 0.005746 K 31 0.005413
416 0.0746988 415 0.0746988

"K" is dealt, all the individual pair% changed, but the sum of them, AnyPair% unchanged!

Technically, it's not, "ONLY,' from a new shoe, it just usually is. The reason why it always happens from a new shoe is because the cards start out equally distributed, which means the same amounts of each rank from a new shoe. That's why it happens regardless of the number of decks in a shoe, because the number of each rank is perfectly distributed before you remove the card.

That's why I say it would usually be from the beginning of a shoe. It's possible for the distribution of the ranks to be equal later on in the shoe, as well, it just doesn't happen very often.

Let's look at 32 cards left in a deck distributed as follows:

A-3 (3/32 * 2/31) = 0.00604838709
2-3 (3/32 * 2/31) = 0.00604838709
3-3 (3/32 * 2/31) = 0.00604838709
4-3 (3/32 * 2/31) = 0.00604838709
5-3 (3/32 * 2/31) = 0.00604838709
6-3 (3/32 * 2/31) = 0.00604838709
7-1---0
8-2 (2/32*1/31) = 0.00201612903
9-1---0
10-3 (3/32 * 2/31) = 0.00604838709
J-2 (2/32*1/31) = 0.00201612903
Q-2 (2/32*1/31) = 0.00201612903
K-3 (3/32 * 2/31) = 0.00604838709

Total Pair Probability: (0*2) + (0.00604838709*8) + (0.00201612903 * 3) = 0.05443548381

Okay, so the difference between this and a, "Fresh shoe," is that not all of the ranks are equally likely to pair to begin with prior to removing one of the cards. For example, the 7's and 9's cannot pair, because there is only one of each. Therefore, if we remove a 7 or 9, the pair probability is going to go up because we're taking away a card that screws up pairing.

Removing one of the cards that there are only two of (8's, J's, Q's) is going to cause the overall pair probability to go up also, as you will see, because the result is three remaining ranks that cannot pair...but the 8's were not all that likely to pair to begin with. I guess we'll just have to see what happens when we remove one of the cards from the ranks there are three of.

Seven Removed

A-3 (3/31 * 2/30) = 0.0064516129
2-3 (3/31 * 2/30) = 0.0064516129
3-3 (3/31 * 2/30) = 0.0064516129
4-3 (3/31 * 2/30) = 0.0064516129
5-3 (3/31 * 2/30) = 0.0064516129
6-3 (3/31 * 2/30) = 0.0064516129
7-0----0
8-2 (2/31*1/30) = 0.00215053763
9-1---0
10-3 (3/31 * 2/30) = 0.0064516129
J-2 (2/31*1/30) = 0.00215053763
Q-2 (2/31*1/30) = 0.00215053763
K-3 (3/31 * 2/30) = 0.0064516129

Total Pair Probability: (0.0064516129*8) + (0*2) + (3*0.00215053763) = 0.05806451609

As predicted, the total pair probability increased, because we removed a card that could not pair anyway. If you were betting on a pair to come, we took away a, "Bad Card," leaving open more possibilities of pairing.

Eight Removed

Now, we are going to remove an eight from the deck, which will leave only one eight. Because of that, the probability of pairing should drop a little bit because there are now three cards that could come out and make a pair impossible....at least, one would think...

A-3 (3/31* 2/30) = 0.0064516129
2-3 (3/31* 2/30) = 0.0064516129
3-3 (3/31* 2/30) = 0.0064516129
4-3 (3/31* 2/30) = 0.0064516129
5-3 (3/31* 2/30) = 0.0064516129
6-3 (3/31* 2/30) = 0.0064516129
7-1---0
8-1---0
9-1---0
10-3 (3/31* 2/30) = 0.0064516129
J-2 (2/31*1/30) = 0.00215053763
Q-2 (2/31*1/30) = 0.00215053763
K-3 (3/31* 2/30) = 0.0064516129

(0.0064516129*8) + (0*3) + (2*0.00215053763) = 0.05591397846

That's interesting, isn't it? The probabilities of any individual pair remained the same as a seven being removed (except 8's, which became impossible) but our overall probability of pairing went UP compared to before we removed a card. Again, that's because the 8's weren't terribly likely to pair to begin with, as there were only two of them...so the likelihood of the ranks of three each (and remaining ranks of two each) goes up more than enough to compensate.

Ace Removed

A-2 (2/31*1/30) = 0.00215053763
2-3 (3/31* 2/30) = 0.0064516129
3-3 (3/31* 2/30) = 0.0064516129
4-3 (3/31* 2/30) = 0.0064516129
5-3 (3/31* 2/30) = 0.0064516129
6-3 (3/31* 2/30) = 0.0064516129
7-1---0
8-2 (2/31*1/30) = 0.00215053763
9-1---0
10-3 (3/31* 2/30) = 0.0064516129
J-2 (2/31*1/30) = 0.00215053763
Q-2 (2/31*1/30) = 0.00215053763
K-3 (3/31* 2/30) = 0.0064516129

Total Pair Probability: (0.0064516129*7) + (0.00215053763*4) + (2*0) = 0.05376344082

In this case, the overall pair probability went DOWN compared to the starting state. The reason for this is because we drastically reduced the probability (tied for highest) of the aces pairing up and the remaining card pair probabilities simply did not increase enough to compensate.

Conclusion

Anyway, I did this with a more extreme example this time in order to highlight the difference and have the change in overall probability being recognizably meaningful without having to go more than a few decimal places to see the difference.

As you remove the cards, you no longer have a uniform starting state in which all pairs are equally likely. Removing cards that are more likely (composition) based on the number of those cards left to pair will generally cause the overall pair probability to go down. If not always, then generally. Removing cards that cannot pair (if only one card of that rank) or cards that are less likely to pair will generally cause the overall pair probability to go up.***

***It might always cause it, I just didn't feel like doing an eight-deck example to see if there are exceptions.

In Your Example

In your example, you can either remove another king as the second card to be removed, or you can remove a card that is not a king. Let's see what happens in each:

A 32 0.005746 A 32 0.005774
2 32 0.005746 2 32 0.005774
3 32 0.005746 3 32 0.005774
4 32 0.005746 4 32 0.005774
5 32 0.005746 5 32 0.005774
6 32 0.005746 6 32 0.005774
7 32 0.005746 7 32 0.005774
8 32 0.005746 8 32 0.005774
9 32 0.005746 9 32 0.005774
10 32 0.005746 10 32 0.005774
J 32 0.005746 J 32 0.005774
Q 32 0.005746 Q 32 0.005774
K 32 0.005746 K 31 0.005413
416 0.0746988 415 0.0746988

A-31 (31/414 * 30/413) = 0.00543916903
2-32 (32/414 * 31/413) = 0.0058017803
3-32 (32/414 * 31/413) = 0.0058017803
4-32 (32/414 * 31/413) = 0.0058017803
5-32 (32/414 * 31/413) = 0.0058017803
6-32 (32/414 * 31/413) = 0.0058017803
7-32 (32/414 * 31/413) = 0.0058017803
8-32 (32/414 * 31/413) = 0.0058017803
9-32 (32/414 * 31/413) = 0.0058017803
10-32 (32/414 * 31/413) = 0.0058017803
J-32 (32/414 * 31/413) = 0.0058017803
Q-32 (32/414 * 31/413) = 0.0058017803
K-31 (31/414 * 30/413) = 0.00543916903

Total Pair Probability: (0.00543916903*2) + (11* 0.0058017803) = 0.07469792136

Because we took away a card more likely to pair, as opposed to less likely to pair, the any pair probability went down very slightly (because there are so many cards) compared to your 415 card state with only one king removed. Let's see what happens when we return our ace and just remove a second king:

A-32 (32/414 * 31/413) = 0.0058017803
2-32 (32/414 * 31/413) = 0.0058017803
3-32 (32/414 * 31/413) = 0.0058017803
4-32 (32/414 * 31/413) = 0.0058017803
5-32 (32/414 * 31/413) = 0.0058017803
6-32 (32/414 * 31/413) = 0.0058017803
7-32 (32/414 * 31/413) = 0.0058017803
8-32 (32/414 * 31/413) = 0.0058017803
9-32 (32/414 * 31/413) = 0.0058017803
10-32 (32/414 * 31/413) = 0.0058017803
J-32 (32/414 * 31/413) = 0.0058017803
Q-32 (32/414 * 31/413) = 0.0058017803
K-30 (30/414 * 29/413) = 0.0050882549

Total Pair Probability: (0.0050882549) + (0.0058017803*12) = 0.0747096185

It went up! And, again, the reason it went up is because, though the probability of getting a pair of kings was reduced, the probability of any other pair coming went up more than enough to compensate.

On and on forever and ever this will continue...with the only exception being a, "Fresh shoe," or a naturally occurring (though rare) state in the current shoe where all ranks are evenly distributed prior to the removal of a card again.

Final Conclusion

The fresh shoe is NOT the cause. The fact that all ranks are equally distributed prior to one card being removed is the cause. It just happens to be a characteristic of a fresh shoe that all ranks are equally distributed before a card is removed.

Final Conclusion

The fresh shoe is NOT the cause. The fact that all ranks are equally distributed prior to one card being removed is the cause. It just happens to be a characteristic of a fresh shoe that all ranks are equally distributed before a card is removed.
+++

Agree.
So I said at the beginning, 1/4 deck, 1/2 deck, 1 deck,...

[the probability of each of the other possible pairs goes up by 1/12th the amount that the probability of the removed card pairs goes down.]
Why the goes up is the same as goes down? Is it a coincidance?
And this is quite surprising to me. Because in Blackjack or Baccarat, whenever a card is drawn, all the EV changed. And I never expected the any-pair EV will keep unchanged.

The EV is changed in those other games because the EoR is not strictly conditional to the probability of there being a pair.

Quote: Mission146

The EV is changed in those other games because the EoR is not strictly conditional to the probability of there being a pair.

If the bet is not "any-pair" but "pair of A's", "pair of 2's",..., the EoR applied, because if a card is drawn, any card, any time, the EV changed.
For blackjack, if one card is drawn, EV of each of the player's hand vs dealer's upcard, changed. But if the overall EV for all the combinations added up keep unchanged, it will surprise me too.

It is interesting. I just worked it out given any deck that is equally distributed, you have the identical probability of a pair if you remove 1 card. It doesn't just work with pairs, but with any number of cards, so for instance, the probability of quads is the same with a full deck or with 1 card removed. It doesn't matter how many ranks there are either, as long as they are equal.

I discovered this years ago and wrote about it in a book that I have never published:

Symmetry of Poker Hands dealt from the same deck
For poker hands, if you hold one pair in your hand ( say, a pair as a two card hand in Texas Hold'em, or a pair + singleton in 3 card poker or a pair + 3 singletons in 5 card poker) then the chance that your opponent has a similar one pair hand is higher than it would have been if his hand had been dealt from a fresh 52 card deck..

If you hold a flush (3 suited cards or 5 suited cards or whatever) in your hand then the probability that your opponent has been dealt a similar flush is higher than it would have been if his hand had been dealt from a fresh deck/shoe.

If you have a straight ((3 adjacent cards or 5 adjacent cards or whatever) in your hand then the probability that your opponent also has a straight will be increased by the fact that your hand has been removed from the deck.

In Pai Gow Poker, if you hold a 7 card Pai Gow hand (no five card flushes, no five card straights and no pairs) then the probability that the dealer also will have a 7 card Pai Gow hand will be increased due to the removal of your hand from the deck.

However, if your remove one or more Aces from a fresh deck then the probability of another player holding an Ace is, of course, lower -the opposite of the above .

The reasons underlying the above rules are surprisingly profound. Some properties, like abundance of a rank, are simple properties of a deck that derive from the fundamental definitions of rank and suit of a card. But the following properties of poker hands are only meaningful in the context of a group of cards because they depend upon the "degree of coordination" of some property of individual cards within the group of cards. The most common examples from poker are:

- pairing of ranks to form one pair, two pair, trips, boats, quads etc. (I think of this as pairing coordination)
- coordination of suits to form flushes
- coordination of ranks to form a 5 adjacent ranks or a straight

Again, these are properties of a group of cards and such group properties have a degree of correlation in groups that are dealt from a common deck.

So, starting with a fresh deck of m ranks and n suits (or a fresh shoe with p decks) and given that m is greater than the number of cards needed to form a straight or flush :

Removal of two groups of x cards (each) from the fresh deck will cause the two groups to have a higher than random chance of being coordinated in the same way.

Example: Remove two seven card hands from a fresh 52 card deck. If the first hand has 3 pairs+ 1 singleton, then the second hand will have a higher probability of being 3 pairs + 1 singleton then if it had been dealt from a fresh deck.

Straights are kind of funky because of the idiosyncratic rules for adjacency of ranks and straight formation in poker; that is ,because an Ace is defined to be both high and low and adjacent to both a King and a Two and because a hand such as QKA23 is defined to not be a straight. These quirky rules cause all sorts of mathematical problems, because it means that different ranks will have different propensities for forming straights. Despite this issue, straights and other patterns based on adjacency will usually follow the rules that I am stated.

Ex. Suppose you remove two four card hands from the deck and find that the first hand has a 3 card straight + 1 nonadjacent singleton such as : 4-5-6-K. You will find that the probability of the second 4 card hand having a similar 3 card straight +1 nonadjacent singleton will be higher than if the 4 cards had been dealt from a fresh deck.

Ex: Imagine a 7 card table game like High Card Flush. You are dealt 7 cards and find that you have a 6 card flush. However, the probability that the dealer has a winning 7 card flush is now higher than if his hand had been dealt from a 52 card deck.

So again, what I am claiming is that when two groups of cards (two hands) are randomly dealt from a well-ordered deck; e.g., from a fresh 52-card deck, that the two groups of cards will have a higher probability of being coordinated in the same way (pairing, suit coordination, and adjacency coordination) then if the two groups/hands had been dealt from two separate 52-card decks.

This is a complicated subject with some ideas that I believe may be new and for which I don't know the correct mathematical terminology (if it exists). I'll try to write it up as best I can and probably post it as a separate thread.