# I'm trying to understand what kind of probability distribution I need to use in order to calculate a very simple example using a deck of cards. Assume that there is a standard deck of cards (52 cards): Let X be the number of non-hearts until I get 13 hearts (without replacement). What would be the distribution I need to use?

I'm trying to understand what kind of probability distribution I need to use in order to calculate a very simple example using a deck of cards.
Assume that there is a standard deck of cards (52 cards):
Let X be the number of non-hearts until I get 13 hearts (without replacement). What would be the distribution I need to use?
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procjenomuj
Step 1
Place the cards in a row. If we only discern hearts and non-hearts then there are $\left(\genfrac{}{}{0}{}{52}{13}\right)$ arrangements. Now fix $n\in \left\{13,\dots ,52\right\}$ and notice that there are $\left(\genfrac{}{}{0}{}{n-1}{12}\right)$ arrangements such that the last heart is placed on spot n. So denoting N as the spot of the last heart we come to:
$P\left(N=n\right)=\left(\genfrac{}{}{0}{}{n-1}{12}\right){\left(\genfrac{}{}{0}{}{52}{13}\right)}^{-1}$
Now realize that $X=N-13$ so that $P\left(X=k\right)=P\left(N=k+13\right)=\left(\genfrac{}{}{0}{}{k+12}{12}\right){\left(\genfrac{}{}{0}{}{52}{13}\right)}^{-1}$
This for $k=0,1,\dots ,39$
Step 2
Equality $\sum _{n=13}^{52}P\left(N=n\right)=1$ leads to the observation that $\sum _{n=12}^{51}\left(\genfrac{}{}{0}{}{n}{12}\right)=\left(\genfrac{}{}{0}{}{52}{13}\right)$. More generally it can be shown (e.g. by induction) that $\sum _{k=r}^{n}\left(\genfrac{}{}{0}{}{k}{r}\right)=\left(\genfrac{}{}{0}{}{n+1}{r+1}\right)$.
So a more general setting here would lead to a combinatorial proof of this equality.