Suppose that I have a deck of $T$ cards, with some set of size $X$ of them having some property. If it helps you to think of a specific property, in a poker deck they could be all the hearts, in a MTG deck they could be mana cards, etc.
If I shuffle the deck, and pull cards one at a time into my "hand," what is the expected number of cards I need to draw to get one card of type $X$? What sort of distribution do the first occurrences fall into.
Here is where I have gotten so far. I may be on the completely wrong path, and I would appreciate being told so, even if you don't have the solution.
I tried to compute some sort of weighted average, multiplying each draw number by the chance that an $X$ card is drawn first on that turn.
The first draw has an $X\over T$ chance of being an $X$ card. The first term in my sum is $$1 * \frac{X}{T}$$
The second draw has an $\frac{X}{T-1}$ chance of being an $X$ card, and there is a $1-{X\over T}$ chance that the first draw was not an $X$ card. Now my sum reads: $$1 * \frac{X}{T} + 2 * (\frac{X}{T-1}) * ({1-{X\over T}}) $$
The chance that the the third draw is an $X$ card is $\frac{X}{T-2}$, the chance that the second draw was not the first occurrence is $(1- (\frac{X}{T-1}) * ({1-{X\over T}}))$. This must be multiplied by the chance that draw #$1$ was not the first, which is $(1-\frac{X}{T})$. All in all, the sequence with the third term reads:$$1 * \frac{X}{T} + 2 * (\frac{X}{T-1}) * ({1-{X\over T}} + 3 * (\frac{X}{T-2})(1- (\frac{X}{T-1})({1-{X\over T}}))(1-\frac{X}{T})$$ This is starting to look rather unwieldy. I have no clue how to continue beyond this method, which I am pretty sure is wrong. Any help would be much appreciated. Is there a general formula for the expected number of cards one must draw before the $n$th
card of type $X$ is drawn?
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EDIT:
The answer which Lulu gave does appear indeed to be correct. I simulated via computer 100000 shuffles of a random deck and checked the first occurrence of a heart. The average was $3.79157$. $53 \over 14$ is $3.78571428571$. Those numbers are close enough for me to be satisfied.
Think of it this way: Add an extra card $C$ to the deck and (for now) include it in the subset $X$. Now arrange all the $T+1$ cards randomly in a circle. By uniformity, we expect there to be $\frac {\#T+1}{\#X+1}$ cards in between each of the cards in the extended $X$ (counting the first element of $X$ that you reach). Now remove $C$ from the circle and thereby define the start of the shuffled deck. We then see that we expect $\frac {\#T+1}{\#X+1}$ until we reach the first element of $X$.
Sanity check: $\#T=2,\#X=1$. Then the two possible shuffles are $XY, YX$ so the answer is $\frac 32$ which matches the result given by the formula. Similarly, for $\#T=n, \#X=1$, there are $n$ equally probable places for the unique $X$ to occur so the answer is $\frac {n(n+1)}2\Big / n=\frac {n+1}2$ which again matches the formula.
As another example, if $\#T=3,\#X=2$ then the possible shuffles are $XXY,XYX,YXX$ so the answer is $\frac 43$, again confirming the formula.