Exact formula for $P_n(r)$, the probability of drawing a card of rank $r$ from a standard shuffled deck at the $n$-th draw without replacement. My question is rather easy to state, and perhaps easy to answer, but I'd need a pointer in the right direction. Suppose you have a standard 52-cards, well-shuffled deck, which contains $4$ cards for each rank (the rank of a card is the 'number' it displays, from $1$ to $13$). You draw one card at a time, for $n$ times. Cards already drawn are not put back into the deck, but just set aside. I would like to have a formula for the probability $P_n(r)$ that a card of rank exactly $r$ is drawn at the exact $n$-th extraction (i.e. after the first $n-1$ cards have been set aside).
Is anyone able to help me? 
 A: The probability of a $4$ on the $17$-th card is $1/13$, since each of the $13$ ranks is equally likely to be the rank chosen at that draw (since you are talking about absolute, not conditional, probability). And the answer is the same with or without replacement.
And similarly, of course, the probability of rank $r$ on draw $n$ is $1/13$, for the same reason.
A: We can split this problem into two steps: The first one is figuring out the probability that $x$ cards of rank $r$ are drawn in $n-1$ draws. The second step is to compute the probability that the $n$-th card is of rank $r$ given that $x$ cards of rank $r$ have already been drawn.
For the first step: The probability for drawing $x$ cards of rank $r$ using $n-1$ draws follows a hypergeometric distribution: https://en.wikipedia.org/wiki/Hypergeometric_distribution
For the second step: This is straight forward: Consider for example $x=2$. This means that $2$ cards of rank $r$ have already been drawn. Thus $2$ of the remaining $52-(n-1)$ cards are of rank $r$. Then the probability to draw a card of rank $r$ given that $2$ cards of rank $r$ have already been drawn is $\frac{2}{52-(n-1)}$.
Having figured out the solution to both steps you can use the theorem of total probability (in German "Totale Wahrscheinlichkeit", hope I did get the translation right) to combine the results and obtain your desired distribution.
Assume $n>4$: Then we obtain $P_n(k) = \Sigma_{i=0}^{4}p_X(i)\cdot\frac{4-i}{52-(n-1)}$ where $X \text{~ Hypergeometric}(4, 52, n-1) $. You can find the formula for the $p_x$ on Wikipedia using the link above. Use the parameters as I have given them: $K = 4, N = 52, n_{hypergeometric} = n-1$.
