Drawing from a deck with replacement, expected number of cards drawn to see all 4 suites Similar questions have been asked before but I am looking for a specific way to solve this question (namely using a recurrence relation).
Assume we have a fair deck of $52$ cards. In each iteration, we will draw one card, record the suite, and then replace it in the deck. The cards are drawn uniformly and randomly from the deck. How many times would we expect to draw cards before we see at least $1$ of each suite?
I want to solve this using a recurrence relation so something of the form:
$X = 1 + \frac{3}{4}(x) + \frac{1}{4}(x+1)...$
#This is not correct just the format I want to solve the question in.
The solution is 8$\frac{1}{3}$.
As a reference, I am trying to use the same logic as presented here:
https://www.geeksforgeeks.org/expected-number-of-coin-flips-to-get-two-heads-in-a-row/
 A: As @GregMartin referenced in his comment, this is an example of the "coupon collector problem", whose solution can be found by using linearity of expectation.
Suppose we perform a binomial experiment with probability of success $p$. Let $E(S)$ be the number of expected trials until we succeed. At any point in the experiment, if we have not yet succeeded, then if the next trial fails (with probability $1-p$), the expected number of further trials until success will still be $E(S)$. This leads to the recurrence
$$E(S) = 1 + (1-p)E(S)$$
which yields $E(S) = \frac{1}{p}$.
Now applying this to the question at hand. Let $p_i$ be the probability of drawing the $i$th different suit, and $X_i$ be the expected number of draws to get the $i$th different suit. Then clearly $p_i = \frac{4-(i-1)}{4}$, so by the above result, we have that $E(X_i) = \frac{4}{4-(i-1)}$. Then by linearity of expectation, we get that
$$E \left[ \sum_{i=1}^4 X_i \right] = \sum_{i=1}^4 E(X_i) = \frac{4}{4} + \frac{4}{3} +\frac{4}{2} +\frac{4}{1} = 8\frac{1}{3}$$
A: Ignore the ranks, then we have $13$ cards of each suit. Since this is with replacement, we might as well have a deck of only $4$ cards, one of eah suit. Then, supposing $n$ suits have not been seen, the number of cards that must be drawn to see a new suit is geometrically distributed with success probability $\frac n4$, with expectation $\frac4n$.
The expected number of draws to see all four suits is then
$$\frac41+\frac42+\frac43+\frac44=\frac{25}{3}$$
