What is the expected value of this sum (drawing slips from a bag)? Suppose there are thirteen slips in a bag, labelled 1-13. If I draw a 10, 11, 12, or 13 then I stop adding to the sum and return the sum. If not, I add the number I draw to the current sum and place the slip back in the bag. 
What is the expected value of the sum?
My Attempt:
S: total sum
Xi: the ith drawn slip
Probability of drawing a slip from 1-9: 9/13
Have a 9/13 chance of getting # from 1-9 and expect to add average of 5 and trying again...
Ending with a 10, 11, 12, or 13 each has 1/4 chance
$$
E[S] = \frac{1}{4}(\frac{9}{13}(5+E[S]) +10) + \frac{1}{4}(\frac{9}{13}(5+E[S]) + 11) +... +\frac{1}{4}(\frac{9}{13}(5+E[S])+13)
$$
which gives E[S] = 48.625
I wasn't sure how to factor in the four different ways that we could end the drawing (i.e. ending with a 10, 11, 12, 13). The answer doesn't seem right to me. 
Additionally, let Y be the number of times a number of a particular number is drawn. What is E[Y]? I have no idea how to start with this with this part.  

We have
$$E(S)=\sum_{k=1}^{13}E(S\mid\hbox{first draw is $k$})P(\hbox{first draw is $k$})\ .$$
But if the first draw is $k=1,\ldots,9$ then the expected sum is $k$ plus the overall expected sum; if the first draw is $k=10,\ldots,13$ then the expected sum is just $k$.  So
$$E(S)=\frac{1}{13}\sum_{k=1}^9(k+E(S))+\frac{1}{13}\sum_{k=10}^{13}k\ ,\tag{$*$}$$
and you can solve this to find $E(S)$.  Specifically,
$$E(S)=\frac{45+9E(S)+46}{13}$$
and so
$$E(S)=\frac{91}{4}\ .$$
Comment.  I'm not quite clear from your question whether a $10,11,12$ or $13$ gets added to the sum and then stops the count, or if it just stops the count.  I have assumed it is added.  If not, remove the second sum in $(*)$: the answer will be $E(S)=\frac{45}{4}$.
