You draw four cards from a standard deck, with replacement. How many possible hands are there where at least one card appears multiple times?

Here's what I have so far: to count the ways using the Fundamental Counting Principle it looks like 52*1*52*52 because you have 52 options for the first draw, only one option for the second draw so it matches the first, and then the third and fourth don't matter. But, now I think I need to divide because the order of the draws don't matter. Should I divide by 4 since the 1 above could be in 4 different positions? Should I divide by 4! since any 4 card hand can be arranged in 4! ways? But, what about the hands where the same card is repeated 3 times? Are those hands included in how I'm counting?

Is there a simple and organized way to keep track and count the possibilities?

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    $\begingroup$ It's easier to count the ways to draw $4$ distinct cards $\endgroup$ – lulu Jan 25 at 13:59

I assume you are disregarding the order of the cards in the hand.

The difficulty here is that the total number of possible hands isn't clear (trusting that, as usual, the order of the cards in a hand is not important).

Stars and Bars gives us a handy way to do that count. Here we are looking for the number of $52-$ tuples of non-negative integers that sum to $4$. (a coordinate in such a $52-$ tuple tells us how many of that card appears in the hand). The standard result shows that this number is $\binom {52+4-1}{4}=\binom {55}4$

As there are clearly $\binom {52}4$ ways to choose a hand of $4$ distinct cards, the answer is the difference: $$\binom {55}{4}-\binom {52}4=\boxed {70,330}$$

  • $\begingroup$ This is an elegant solution. Thank you. $\endgroup$ – jontail Jan 28 at 15:15

I would approach this from the other persepctive - look at how many possible hands there are and remove those which have no repeated cards.

We have $52\times 52\times 52\times 52 = 7,311,616$ possible combinations of all the cards (provided we care about the ordering).

We now need to see how many of those hands had no repeated cards in, which we know is equal to $52\times 51\times 50\times 49 = 6,497,400$ (again, provided we care about the ordering)

Therefore, there are $$7,311,616-6,497,400=814,206$$ hands where at least one card appears multiple times.

  • $\begingroup$ Unfortunately, it isn't true that "If ordering does not matter then we have counted each hand $4$ times." You only count $\{A\spadesuit, A\spadesuit,A\spadesuit,A\spadesuit\}$ once, for example. $\endgroup$ – lulu Jan 25 at 14:10
  • $\begingroup$ @lulu That's is true I have removed that half of my answer $\endgroup$ – lioness99a Jan 25 at 14:15

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