Math
Just because you've asked for it
Okay, so the first step is declaring the possibility of n card pairs when two readers draw 6 cards. As I see what the first person drew is irrelevant - it's just a random set of 6 different cards so we shouldn't calculate with that. We can declare it fixed.
The next step is to declare the possibilities of the next person drawing 'n' of the cards the first person has already drawn. We'll use combination for that (
http://en.wikipedia.org/wiki/Combination) I'll abbreviate "n choose k" as (n,k) for simplicity's sake from now on.
So all the possible six card combination is (78,6)=256851595
Now the chance that the second person will have n cards the same as the first one was is: (6,n)*(72,6-n) as we need to choose n cards from those six of the first person and 6-n from the remaining cards. The result is the amount of cases when the second person has exactly n pairs with the first person. It's
pairs cases
0 156238908
1 83949264
2 15431850
3 1192800
4 38340
5 432
6 1
By dividing the number of possible cases with the number of all cases we get the probabilities, for having exactly that amount of pairs:
0 60,83%
1 32,68%
2 6,01%
3 0,46%
4 0,01%
5 0,00%
6 0,00%
Now it becomes tricky. Cause at this point comes the third person, but the amount of 'free' cards (those which don't pair with the first two's cards) depends on how many pairs did the first two have. So we have two variables here:
k: the number of pairs in the first two hands
l: the number of repeating cards in the third hand
The way to calculate the amount of cases is: (12-k,l)*(78-(12-k),6-l)
Explanation: 12-k is the amount of cards with which the third person's card can make pairs, 6-l is the amount of non-paired cards in his hand.
Calculating it for all the l-s and k-s gives us 36 numbers - the number of possible cases for each ... case
(the number of possible amount of pairing cards in the third person's hand considering the amount of pairs in the first two person's hand)
These numbers should be divided by the amount of all cases to get probabilities, but beware! These are only conditional probabilities for each setup of the first two persons (so the numbers show what is the probability of the third person has 2 pairing cards with the first two SUPPOSING they had only 1 pair).
To get the whole probability field, you have to multiply the conditional probabilities with the actual probabilities of each case of the first two persons(listed above), and then sum up all those cases which contain the same amount of pairs. The result looks like this:
0 21,52% 100,00%
1 38,10% 78,48%
2 27,34% 40,38%
3 10,40% 13,04%
4 2,30% 2,64%
5 0,31% 0,33%
6 0,02% 0,02%
Where the first column shows the amount of pairs/matches in the three hands, the second column contains the probability of having EXACTLY that amount of matches and the third shows the probability of having at least that amount of matches.
Of course it could be continued to 12 - as the maximum amount of matches is 12, 6 cards in the second person's hands and 6 in the thrid one's hands, but the probability of anything above six matches is so close to zero that it doesn't even worth mentioning.
I hope I could make it clear - writing math text in english is not easy at all