1
$\begingroup$

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!

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

$\endgroup$
  • $\begingroup$ 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 Mar 6 '17 at 16:22
  • 1
    $\begingroup$ 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. $\endgroup$ – timwiz Mar 6 '17 at 16:34
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – timwiz Mar 6 '17 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.