I've got the same answer to my question via probability and combinations, but I don't know why! I am trying to find the answer to the following question:
If I am playing poker and am dealt two cards neither of which are spades, what are the chances of two or more of the flop cards (the flop is the first three community cards) being spades.
I answered this question in the following way:
As I have no spades, there are still 13 spades unseen and a total of 50 cards remaining unseen (52 cards in the pack less my two cards which I've been dealt.
If the probability of the first flop card being a spade is p(A) and the second p(B) and the third p(C) then I need to find:
p((A∩B)∪(A∩C)∪(B∩C))
so given the formula:
p(X∪Y∪Z) = p(X) + p(Y) + p(Z) - p(X∩Y) - p(X∩Z) - p(Y∩Z) + p(X∩Y∩Z)
I calculated:
p(2 or more spades on flop) = 3(13/50 x 12/49) - 3((13/50 x 12/49) x 11/48) + (13/50 x 12/49 x 11/48)
which gives 16.18%
What I wanted to do next was solve the question using combinations. Through a process of trial and error I got the same answer, but am a bit confused as to why it works this way. My method was:
13C2 * 48 = 3744
3744 - (3 * 13C3) + 13C3 = 3172
total number possible 3 card combinations on flop = 50C3 = 19600
3172 / 19600 = 16.18%
So I get the total number of 2 card combination of spades by 13C2 = 78. Then there are 48 other cards which can form combinations with each of these 2 card combination so 13C2 * 48.
Now I'm losing my grasp of whats happening. Why does 13C2 = p(A) + p(B) + p(C)?
All I have done to follow is find the number of combos for p(A∩B), p(A∩C), p(B∩C) and p(A∩B∩C) and then crunched the numbers as in the calculation shown above. But even though I (think I've) got the right answer, I've come to it in a try-it-and-see way and can't understand for instance why 13C2 * 48 doesn't give the right answer straight off. Surely that gives the number of 3 card combinations on the flop with 2 or more spades which is what I'm looking for, doesn't it? But I don't get the same answer as with the original calculation if I use that figure.
Sorry for the waffle. I hope someone can see where my thinking is going wrong and can help explain the combinations part a bit better for me.
Many thanks in advance.
 A: OK, I think I've got the answer. I tripped myself up initially because I didn't really understand what I was doing when I was multplying ${13\choose2}$ by 48. I'd reasoned that if I had Ace, King of spades then I could times by 48 because there were 48 other cards to make 48 combinations with AK spades in.
But what I now realise is that multiplying the whole of ${13\choose2}$ by 48 causes an overlap in counting. For instance if I have A7 of spades then one of the other 48 cards is the King of spades. And in fact I need to apply the Inclusion/Exclusion rules just the same as if I'd been using probability to come to the right answer.
So ${13\choose2}\times48$ does equal p(A) + p(B) + p(C) and I need to use the inclusion exclusion rules to take away any combinations that have been counted twice. SO:
p(A∩B) = p(A∩C) = p(B∩C) = ${13\choose3}$ = 286
p(A∩B∩C) = ${13\choose3}$ = 286
and as:
p(A∪B∪C) = p(A) + p(B) + p(C) - p(A∩B) - p(A∩C)- p(B∩C) + p(A∩B∩C)
= (${13\choose2}\times48$) - (3 x ${13\choose3}$) + ${13\choose3}$
= 3744 - 858 + 286
= 3172
3172 / 19,600 (total number of possible flops) = 0.1618 = 16.18%
