TITLE: Probability Discrepancy in drawing 2 cards from a Deck of 52
QUESTION [15 upvotes]: A colleague and I are having a hard time figuring out this probability question and was wondering if anyone could provide an explanation / insight into it.
The question is simple: what is the probability if you draw 2 cards at the same time (so without replacement) from a standard deck of 52, that one of those cards, or both, is a diamond?
He mapped out all the possible outcomes and came to $\frac{7}{16}$, by writing out the following:
DD DH DC DS
HD HH HC HS
SD SH SC SS
CD CH CC CS
Since 7 out of those 16 outcomes have a diamond in them, that is the probability. Trying mathematically though, the answer is not equivalent. We both believe that the formula is not right, but we can't figure out where the problem is.
This is the formula we've tried: $\displaystyle\frac{{13\choose{1}}{39\choose1}}{52\choose{2}}+\frac{13\choose2}{52\choose2}$; the left term is for 1 diamond, another card different, and the right term is for 2 diamonds. This sum comes out to $\frac{15}{34}$ or approximately .44, which is close, but not exact to the answer we'd expect above. Which is correct - or where is the mistake? An explanation of the discrepancy would be much appreciated.
Thanks a lot.

REPLY [2 votes]: Here's another method, which is more straightforward but also more tedious than ShreevatsaR's.
The event that either one or both cards is a diamond can be broken down into the following three mutually exclusive events -- (a) both are diamonds, (b) first card is a diamond, second isn't, (c) first card isn't a diamond, second card is.
The overall probability is then just the sum of the individual probabilities
$$P = \frac{1}{4}\times \frac{12}{51} + \frac{1}{4}\times\frac{39}{51} + \frac{3}{4} \times\frac{13}{51} = \frac{15}{34}$$
You can actually simplify slightly by combining the first and second events to get (a) the first card is a diamond, the second can be anything and (b) the first card is not a diamond, the second one is, to get
$$P = \frac{1}{4} + \frac{3}{4} \times \frac{13}{51} = \frac{15}{34}$$