I'm trying to figure out probabilities of certain hands for a game I've been considering. An example is the probability of having a card that is an Ace or King and a card that is a Heart or a Diamond, but it can't be just a single Ace of Diamonds alone. There must be a card from each set. It can be an Ace of Diamonds and an Ace of Hearts, though. I.e. one fulfilling the rank and one fulfilling the suit.
I can figure the combinations that fulfill either set. Assuming a 7 card hand:
Ace, King: ${52 \choose 7} - {52-4*2 \choose 7}$
Heart or Diamond: ${52 \choose 7} - {52-13*2 \choose 7}$
My issue is figuring at least one from each set, excluding ones where there is only one match, total, common between both sets.
I tried finding the combinations of the exclusive sets, the union with two, the union and each exclusive, but the non-exclusiveness and overlapping I've not been able to figure out. I've gone through my probability and statistics books and tried searching the web, but the examples I could find were all for exclusive poker hands, etc. I'm frustrated because I'm convinced it is so simple and yet I can't see it.
I'm also trying to figure out the probabilities of this success in various different sized hands to see what a reasonable hand size is. I also have other success sets in mind, like the one above but also requiring say at least one Queen, etc., but I figure I can glean how to figure that if I knew how to think this one through the logic.
Thank you for considering this.
EDIT: re true blue anil, AH with AD would be a success. As long as there is at least one unique card fulfilling each set, it is a success.
EDIT: My current line of attack is dividing the space into unique groups of hands: no good cards, 1+ cards in Set A but none in B, 1+ cards in set B but none in A, 1+ cards in A-B and 1+ cards in B-A, 1 card in A intersecting B but none outside of it, etc. Then just add up the success hands. I don't know if that will work, but I'm at a loss.
EDIT: See the accepted answer below, but wanted to post what I had figured as well for posterity.
I broke the space into four sets.
A $[2, .., Q] \oplus [\heartsuit, \diamondsuit]^T$
B $[A, K] \oplus [\spadesuit, \clubsuit]^T$
C $[A, K] \oplus [\heartsuit, \diamondsuit]^T$
D $[2, ..., Q] \oplus [\spadesuit, \clubsuit]^T$
Good hands are:
1) two or more C
$${A + B + C + D \choose 7} - \left( {A + B + D \choose 7 } + {C \choose 1} \times {A + B + D \choose 6} \right) $$
I count the number of ways to get only hands with no card in C at all and hands with one of the C cards and 6 cards from anything but C. I subtract this from the total hands, and I have the hands with two or more C cards.
2) one C and any from A or B
$${C \choose 1} \times \left({A + B + D \choose 6} - {D \choose 6} \right)$$
Here, I count the number of hands having one of the C cards with the rest of the hand selected from and remaining cards not in C, but excluding hands that are only made of cards from D (leaving subsets with at least one A or B)
3) none from C and at least one from A and B each
$${A+B+D \choose 7} - \left( {A + D \choose 7} + {B + D \choose 7} \right) + {D \choose 7}$$
Here I create my sub set of hands from all hands that don't contain C. I then remove any hands that do not contain a B and any hands that do not contain an A. I have to add back the same number of hands that contain only cards from D because I subtracted those hands in removing the A and B exclusions twice.
The accepted answer is much more concise, but I wanted to also show my work in case it helped someone somehow on a search of this.
