Probability Discrepancy in drawing 2 cards from a Deck of 52

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.

Solution 1:

Here's a third way: Suppose you draw the two cards one after the other. Then, the probability that the first card is not a diamond is $3/4$. (Or, if you wish, $39/52 = 3/4$.) After that, when you draw the second card from the remaining $13$ diamonds and $38$ non-diamonds ($51$ cards in all), the probability that the second card is also not a diamond is $38/51$. The probability that you've drawn either one or two diamonds is the probability that not both of these happened, and is therefore $$1 - \frac{3}{4} \frac{38}{51} = 1 - \frac{19}{34} = \frac{15}{34}$$ as you got.

Your method, calculating $$\frac{{13\choose1}{39\choose1}}{52\choose2} + \frac{13\choose2}{52\choose2} = \frac{15}{34},$$ is also correct.

As the others have said, your friend's method gives incorrect results. When working with probability (rather than combinatorics), it is not enough to enumerate the outcomes; you must also assign weights (the probabilities) to them (see this question) – and if you are going to assign equal weights, you must be able to justify it.

If you actually do this correctly with your friend's method, such that the outcomes with both suits identical (like "DD") have probability $\frac14\frac{12}{51} = \frac{1}{17}$, and those with the two suits different (like "DH") have probability $\frac14\frac{13}{51} = \frac{13}{204}$, then you'll correctly get, as the probability of one of the seven outcomes "DD", "DH", "DC", "DS", "HD", "CD", "SD", the probability $$\frac{1}{17} + 6 \times \frac{13}{202} = \frac{15}{34}$$ again.

Solution 2:

The discrepancy is due to the fact that in your diagram of all outcomes, not all outcomes are weighted equally. Assigning each outcome a weight of 1/16 is incorrect.

Solution 3:

Your sixteen possibilities do not all have the same probability. The ones with two matching cards are less probable than the ones with cards of differing suits.