After qmech's answer, I'm afraid the best answer to this...

what I'm hoping to find is a formula that can approximate the probability of, for example, Moneylender pairing up with a Copper (but it could be any card that cares about another specific card) when you also have something like a Laboratory in your deck.

... is as so often

`1/N * sum_{i=1}^N f(X),`

where N is large, X is a random hand of your deck after you played all of your Labs, and f is 1 if you have a Moneylender and a Copper in hand and zero if not...

Edit: But let's try the analytic way. Say you have L Labs, 1 Moneylender, C Coppers and a total of N cards.

You want to either collide the ML with the Copper directly while having no Labs in hand,

XOR collide with having exactly 1 Lab in the first 5 cards,

XOR collide in 6 cards with having exactly 1 Lab in the first 5 and exactly 1 Lab in the first 7 cards,

XOR collide in 7 cards with having exactly 1 Lab in the first 5 and exactly 1 Lab in the first 7 and 1 Lab in the first 9 cards

...

XOR collide in 5+L with having exactly 1 Lab in the first 5, and ..., and 1 Lab in the first 5+(L-1)*2 cards.

Hope I didn't forgot something or counted something twice.

The probability to collide in i cards with having exactly 1 Lab in 5, ...., 1Lab in 5+(i-1)*2 cards is.... ... ....

The Labs is I think 5 positions for the first, 6 for the second etc until 4+i for the ith, times permutations should be

`L!*5*6*...*(4+i)/(N*(N-1)*...(N-i+1)) * ADDITIONAL FACTORS BECAUSE OF "EXACT i LABS"`

Colliding in 5+i positions is 5+i positions for the M.

`(5+i)/(N-i)`

Copper is probably now easier to calculate that it will not collide, therefor all C Coppers must stay out of the 4+i remaining cards of my hand, that means they have to be in the remaining N-2i-5-1 cards out of N-i(Labs)-1(ML) remaining

`(N-5-2*i)/(N-i) * (N-5-2i-1)/(N-i-1) * ... * (N-5-2i-C+1)/(N-i-C+1)`

so that althogethe we come to

`L!*5*6*...*(4+i)/(N*(N-1)*...(N-i+1)) * (5+i)/(N-i) * (1 - (N-5-2*i)/(N-i-1) * (N-5-2i-1)/(N-i-2) * ... * (N-5-2i-C+1)/(N-i-C) )`

... and sum this for i from 1 to L.

Probably with some error somehow, someone please iterate.

** If some of the brackets (N-2i) or whatever is negative, it is defined to be zero, and 0/0:=1**Edit: Forgot about the "EXACT" above for the Labs, but no time to correct