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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.

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    $\begingroup$ $15/34$ is right. $\endgroup$ – André Nicolas Mar 4 '14 at 19:03
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    $\begingroup$ $7/16$ would be correct if you had an infinite shoe. $\endgroup$ – Nick T Mar 4 '14 at 20:03
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    $\begingroup$ For anyone (like me) who didn't understand the previous comment: en.wikipedia.org/wiki/Shoe_(cards) A shoe is apparently a device for holding multiple decks of cards. It is true that if you are drawing from multiple decks of cards rather than just one, then the difference in probability between outcomes like "DD" and "DH" will get smaller: in the limit (infinitely many decks of cards), both have probability $1/4 \times 1/4$, rather than (as here) $1/4 \times 12/51$ and $1/4 \times 13/51$ respectively. $\endgroup$ – ShreevatsaR Mar 5 '14 at 5:43
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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.

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  • $\begingroup$ Of course! What a silly mistake, I didn't even consider when mapping them out that the probabilities would be different. Everyone's answers were great - but I'm accepting this one because of the detailed explanation + the other methods of calculating the result. $\endgroup$ – Marco Mar 4 '14 at 20:17
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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.

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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.

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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

$$P = \frac{1}{4} + \frac{3}{4} \times \frac{13}{51} = \frac{15}{34}$$

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  • $\begingroup$ +1, another way: consider the three events (A) "first card is a diamond", (B) "second card is a diamond", (C) "both cards are diamonds". Then calculate $\Pr(A) + \Pr(B) - Pr(C) = \frac14 + \frac14 - \frac14\frac{12}{51} = \frac12 - \frac1{17} = \frac{15}{34}$. $\endgroup$ – ShreevatsaR Mar 7 '14 at 8:58
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Your counting argument for 7/16 does not work since the events listed there have not the same probability. For example, the probability of getting DD is $(13/52)(12/51)=(1/4)(12/51)$ since after extracting the first D only 12 diamonds remain, while the probability of getting, say, DH is $(13/52)(13/51)=(1/4)(13/51)$ since after extracting the first D still 13 H remain.

I do not know if this is an homework, so I do not want to post a complete solution, but I would suggest you in this case (and other similar ones, where the second probability depends on the first outcome) to solve the problem by conditioning on the first event; that is, if $s_1$ and $s_2$ the suite of the first and second card and $E$ is the event you are interested in write

$$ P[E] = P[E | s_1=D] P[s_1=D] + P[E | s_1 \ne D] P[s_1\ne D] $$

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