# 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.

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Think of a string of 50 bits, where 1 means 'spade' and 0 means 'non-spade'. There are $50 \choose 13$ such strings, and ${3 \choose 2}\times{47 \choose 11}$ have 2 spades in the first 3 positions and 11 in the other 47 positions. – rgrig Aug 12 '10 at 15:10
That can't be right, surely. You say there are ${50\choose13}$ strings, but that is not how many different 50 bit strings there are, that would be 50! ie 50 factorial. ${50\choose13}$ is the number of combinations of 13 items you can have from a superset of 50, isn't it? And ${3\choose2}\times{47\choose11}$ gives 5.22 x 10^10 which is a huge number and not right given there are only a maximum 19,600 possible combos for the first three cards (given by ${50\choose3}$. – Joe Aug 12 '10 at 16:18
50! is the number of rearrangements if all 50 items look different. In the case of bits, where 0 and 1 mean spade and nonspade, you cannot distinguish between different rearrangements within each class. So we divide by 13! and 37!, the ways to rearrange the 1's and 0's. – Larry Wang Aug 12 '10 at 17:12
Joe, assuming you understood Kaestur's explanation, I want to add that my point was that, when dealing with probabilities, it is safest to work with elementary events. Otherwise you have to be careful about independence/incompatibility of events. Sometimes looking at elementary events becomes cumbersome, but in this case it is straightforward. – rgrig Aug 13 '10 at 8:08

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%

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