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I was playing a game of cards with some friends and wondered :

What's the probability of drawing 4 cards from a normal 52 card deck with all different ranks?

I figured out 3 ways of achieving the answer, but only 2 of them are equivalent.

  1. Treating 4 draws as independent events. We can multiply the probabilities that each card rank selected is one that hasn't been selected yet. Each probability is $\frac{num-cards-of-unselected-rank}{num-cards-to-select-from}$. Each step we subtract 4 from the numerator and 1 from the denominator.

$$\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49} = 0.676$$

  1. I calculated all of the possible sets of 4 without a duplicate rank. First, select 4 of the 13 ranks: $ 13 \choose 4$. Then, for each of those 4 sets, choose one of the 4 suits: $4^4$. Divide the product by the total sets of 4 in a deck of 52: $52 \choose 4$ for the probability of selecting a set of 4 without a duplicate rank.

$$\frac{{13\choose4}*{4^4}}{52\choose4} = 0.676$$

  1. I calculated all the possible sets of 4 with at least one duplicate rank. There are 13 unique ranks. For each rank there are ${4\choose2}=6 $ possible unique pairs. So, this gives me $13*6=78$ possible pairs in a deck. For every pair of cards there are $50\choose2$ unique sets of 4. Divide the product by $52\choose4$ for the probability of selecting 4 cards containing at least 1 pair of duplicate ranks. Subtracting this from 1 should get me the probability of selecting 4 cards without a pair. But it doesn't equal the above 2 values. help!

$$1-\frac{78*{50\choose2}}{52\choose4}=0.647$$

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  • $\begingroup$ I like to use a computer simulation with pseudorandom numbers to double-check my math. You could use that as another, though approximate, approach. $\endgroup$
    – EngrStudent
    Commented Mar 6, 2017 at 16:22
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    $\begingroup$ Good point. I'm convinced that it would probably come out to roughly 0.676 (i.e. the answer that the first 2 solutions led me to). I'll certainly give that a whirl later today, but I'm the most interested in what's wrong with my math in solution 3. $\endgroup$
    – colorlace
    Commented Mar 6, 2017 at 16:34

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The "probability" that you are subtracting is wrong - you are not counting every outcome exactly once - for example 2 kings and 2 aces will be counted twice - once for the kings and once for the aces. On the other hand, you are missing some outcomes - for example 3 kings+1 ace, or 4 kings. The correct way to calculate it is to use the "inclusion and exclusion" formula - see https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

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  • $\begingroup$ Oh good point, and I'll be sure to read through the "inclusion and exclusion" formula. You're right that I'm double counting something like Kheart, Kclub, Aspade, Aclub, BUT from what I can tell I'm not missing 3K+1A or 4K. In fact, by my count- I'm counting 4K 6 times: For all 78 unique pairs, multiply that by 50 choose 2. So, for all 6 unique King pairs, I am adding every possible pair it could be selected with INCLUDING the other 2 Kings. $\endgroup$
    – colorlace
    Commented Mar 6, 2017 at 19:38

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