[antonylee]

I noticed the following when doing some simulations...

Say you reach a 5 card ending with the opponents still holding 6 clubs and 2 diamonds, as well as 2 spades are known to be held by your RHO. Let's say you want to compute the probability (with no extra information, as usual) that diamonds split 1-1.

There are 8 unknown cards, 3 of which go to your RHO, which makes a total of C(8,3)=56 possibilities. If diamonds split 1-1, you have 2 choices for your RHO's diamond and C(6,2)=15 choices for his (two) clubs, i.e. 30 possibilities. So the probability of a 1-1 diamond split seems to be 30/56=53.4%. (As a check: if diamonds are 2-0, there are C(6,3)=20 possibilities for RHO's three clubs, and if diamonds are 0-2 there are C(6,1)=6 possibilities for RHO's single club, adding up to 56.)

Now you cash a club (you already know a priori both will follow). At that point, there are 6 unknown cards and 2 go to RHO, so C(6,2)=15 possibilities. If diamonds split 1-1, you have again two choices for RHO's diamond and C(4,1)=4 possibilities for his single club, i.e. 8 possibilities. (If diamonds are 2-0, there are C(4,2)=6 possibilities for RHO's clubs, and if diamonds are 0-2 then there is only one possibility, again adding up to 15). So now it seems that the probability of diamonds 1-1 went down to 8/15=53.3%.

Either I got the math wrong, or there is some weird restricted choice argument that I don't understand there...

[nigel_k]

The short answer is that your arithmetic is fine but it's not valid to use vacant spaces when some small cards in a suit have been played and others have not. Instead, work out the probabilities taking into account only known high cards and suits whose distribution has been completely determined, but without any cards have been played in other suits. Then assume those probabilities still hold after further small cards appear.

[antonylee]

nigel_k, on 2012-October-04, 00:08, said:

Then assume those probabilities still hold after further small cards appear.

That is the gist of my question: Say instead of dealing out a full deck I just dealt the remaining cards. Is it really the case that the probabilities are going to be different than from the case where I dealt all the cards and then came down to a five-card ending?

[EricK]

This page covers this sort of issue:
http://www.rpbridge.net/7z75.htm

[antonylee]

That's a very interesting read.

I think I start to get it. Say I start with 11 spades between dummy and me. At trick 1 then I know the 1-1 split is barely favored (52%). Now let's say instead that the opponents start by cashing 10 tricks (sigh), never touching spades, and I win trick 11. Now I'm down to two spades in both hands. Assume that neither opponent pitched in that suit. I have extracted no information, so the the 1-1 split should still be at 52%. If instead I just reasoned based on the two-card ending then there are C(4,2)=6 possibilities and the 1-1 split occurs in 4 of them, or 67%.

Now, why was I confused? I basically made the following thought experiment: assume that you just have a deck of 26 cards, 2 of which are black and the rest are blank. You deal them to your opponents and look at the probability that the black cards are split. It is 52%. Now assume that someone peeks at the cards while you deal them, and throws away all the non-black cards except two of them as you are dealing. This shouldn't change the probability of the 1-1 split and I thought this would be the same as computing the probability in the end position. But actually this procedure is wrong, because it doesn't ensure that (after all but four cards get thrown away) each opponent gets two cards.

Instead, a proper protocol would be to have 2 black and 2 red cards in the deck (and 22 blank cards). Again, somebody peeks at the cards while you deal them and throws away all the blank cards, but not the colored ones. Now clearly sometimes you will end with the opponents having different numbers of cards; so what you can do is reshuffle everything and redeal if that happens, keeping only the cases where both opponents have two cards. But by doing so, you are artificially increasing the frequency of 1-1 splits (precisely from 52 to 67%), because a deal where the black cards are 1-1 is more likely to result in both opponents having two cards than a deal where the black cards are 2-0, and thus more likely to be kept.

[phil_20686]

Quote

Instead, a proper protocol would be to have 2 black and 2 red cards in the deck (and 22 blank cards). Again, somebody peeks at the cards while you deal them and throws away all the blank cards, but not the colored ones. Now clearly sometimes you will end with the opponents having different numbers of cards; so what you can do is reshuffle everything and redeal if that happens, keeping only the cases where both opponents have two cards. But by doing so, you are artificially increasing the frequency of 1-1 splits (precisely from 52 to 67%), because a deal where the black cards are 1-1 is more likely to result in both opponents having two cards than a deal where the black cards are 2-0, and thus more likely to be kept.

Take your example of two black cards and 24 blank cards, and you remove blank cards until there are only two cards left in each hand. Clearly, the numbers that you should use are the apriori numbers.

Suppose instead that there are three colours, 2 black 10 red and 14 blank. You again come down to four cards, and in the process discover that the red cards were divided 2-8 (all red cards are discarded). Now you want to know the apriori numbers, given that the other cards were divided 2-8. The appearance of blank cards gives you know information about the layout of the blank or the black cards, since regardless, blank cards must appear. This is the justification for vacant spaces.

[bluecalm]

Quote

If diamonds split 1-1, you have again two choices for RHO's diamond and C(4,1)=4 possibilities for his single club, i.e. 8 possibilities. (If diamonds are 2-0, there are C(4,2)=6 possibilities for RHO's clubs, and if diamonds are 0-2 then there is only one possibility, again adding up to 15).

The problem is that you want to calculate probability by dividing:
(interesting layouts) / (all possible layouts)

If you don't know the whole suit distribution there is a problem with this though because you don't know how often you get given layout. If they have Q65432 in clubs and followed with 3 and 2 you are tempted to assume that now say:
Q65 - 4 and
Q64 - 5

have the same probability (times number of combinations in side suits) but it's not the case.
It's not the case because maybe your opponents never discard a 2 from Q652 and then Q65 is not possible at all. Or maybe they do that only rarely. You can't know for sure and without that knowledge you can't determine what probability of Q65 is.
What you could do is to say how probable Q652 was before they played any cards to that suit or how probable Q65 is now if you know for sure their strategy of following suit.

For example if they always follow with the lowest card available then your reasoning in OP is valid (assuming you take into consideration only layouts consistent with that).

Finally you can assume they always follow with random low (as in not significant for trick taking) card. If you go through calculations you will find out that probabilities don't change at all. Once you write that down and understand you will see that's where restricted choice principle comes from