If we draw n cards from a 52 cards deck (without jokers), I was wondering what was the probability of having 2 cards of the same rank (for example a double 5, a double ace etc.) amongst these n cards. I have tried the following hypergeometric distribution, using $\begin{pmatrix}4\\2\end{pmatrix}$ as the number of possible places for the double and multiplicated by 13 as there is 13 different card ranks. Since this appears to be superior to one according to the www.desmos.com website, I guess it's false. $$\frac{13\cdot\begin{pmatrix}4\\2\end{pmatrix}\cdot\begin{pmatrix}48\\n-2\end{pmatrix}}{\begin{pmatrix}52\\n\end{pmatrix}}$$ I was also wondering the probability of getting a triple or a quad, but also of having 2 different doubles, 2 triples, 1 double and one triple, etc.

Added: As expected the error comes from multiple counting yes ... (how could it be fixed ?) And as I count three of a kind as a failure I couldn't use your answer Eric (which is much easier I agree, I remember trying something like $1-1\cdot\frac{13-1}{13}\cdot\frac{13-2}{13} ... \frac{13-(n-1)}{13}$ where n is the number of cards drawn which is I think what you have suggested)

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    $\begingroup$ Your formula counts pairs multiple times. For example $\{2,2,J,J\}$ comes up as two $2's$ and again as two $J's$. $\endgroup$
    – lulu
    Mar 27 '16 at 18:59
  • $\begingroup$ It's much easier to find the probability of drawing $n$ cards, no two of which are the same rank. $\endgroup$ Mar 27 '16 at 19:00
  • $\begingroup$ Do you count three of a kind as a success or failure? $\endgroup$
    – lulu
    Mar 27 '16 at 19:00

If we draw $n$ cards from a deck of $52$, the probability that the first card is a new rank is $1$; the probability that the second card is a new rank is $\frac{48}{51}$ (there are $51$ cards remaining, but $3$ of them are the same rank as the card we've already chosen). Following this logic for $n$ cards, $$P(\mbox{no repeated ranks in } n \mbox{ cards}) = \prod_{i=0}^{n-1} \frac{52 - 4i}{52-i},$$ for any $n \leq 52$. Note that when $n\geq14$, there is a term with $i=13$, giving $\frac{52-4(13)}{52-13} = 0$. This is correct, since you can draw at most $13$ cards from a standard deck without repeating ranks. Also note that for $i > 13$, the terms in the product above are meaningless, though the product is still correctly $0$ for $14 \leq n \leq 52$.

The probability that some rank is repeated is just $1$ minus the probability given above.


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