Expected value of number of cards drawn from a deck to get 5 spades. The question: What is the expected number of cards required to be drawn in order to draw 5 spades.
What I have:
Let $X=X_1+\cdots+X_{43}$ (43 because we're examining the case when 4 spades have been drawn and we're waiting for the 5th) with $X_i=1$ iff card $i$ is drawn before the fifth spade.  Then $X=$ the number of cards drawn before the 5th spade.  Let $Y=X+1$.  Then $Y=$ the number of cards drawn before and including the 5th spade.
$P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before the 5th spade $=\frac{9!}{10!}=\frac{1}{10}$ (because there are $9!$ ways of ordering the $i$th card at the front of 9 spades and $10$ ways of ordering 10 cards).
Then $E\left[Y\right]=\frac{43}{10}+1=5.3$.  But this is far too small.  What went wrong?
Thanks.
 A: There is a distribution to cover these waiting time problems without replacement. It is called the Negative Hypergeometric Distribution. It has an expected value of
$ \frac{N+1}{M+1} \cdot k = \frac{265}{14}$
where N ( 52 cards ) is the total number, M ( 13 spades ) is the total number of the target type  and k ( 5 spades needed ) is the number wanted from M.
A: Label the non-spades $1$ to $39$. Let $X_i=1$ if non-spade with label $i$ is drawn before the fifth spade, and $0$ otherwise.
Then the total number $Y$ of cards drawn up to an including the fifth spade is given by
$$Y=5+X_1+X_2+\cdots+X_{39}.$$
I think you know how to find $E(X_i)$. The number of relevant cards is $14$.
A: Fifty two cards are labeled with the numbers 1 through 52 .
Note that there are 13 spades amongst the cards. For the first spade, We may view these as separating the remaining 39 cards into 14 groups of non-spades - those appearing before the ﬁrst spade, between the ﬁrst and second, etc. Each of these groups is equally likely to appear ﬁrst, so 39/14 non-spades are drawn on average. Similarly for the second spade, (39-Avg Drawn for the 1st Spade)/13 non-spades are drawn on average and (39 - Average drawn for 1st and 2nd Spade)/12, (39-Average drawn for 1st, 2nd and 3rd Spade)/11, and (39-average drawn for 1st, 2nd, 3rd and 4th Spades)/10 non spades are drawn on average for 3rd, 4th and 5th spades
Thus the Expected number of cards drawn to have the first five spades 
= $ 5+\frac{39}{14}$(2.785714)
+$\frac{39-2.785714}{13}$ (36.21429/13)
+$\frac{33.4287}{12}$
+$\frac{30.64286}{11}$
+$\frac{27.85714}{10}$
= 5+13.92857 
= 18.9257  This equates to what Andre has suggested.  I am sorry, I thought about it but was not sure.
Thanks
Satish
A: Just so that the answer is here and clear.
Label the non-spades $1\cdots39$.  Let $X_i=1$ iff card $i$ is drawn before the fifth spade (and $X_i=0$ iff card $i$ is drawn after the fifth spade).  Then $X=\sum\limits_{i=1}^{39}X_i=$ the number of non-spade cards drawn before the fifth spade.  Let $Y=5+X$.  Then $Y=$ the number of cards drawn in order to obtain the fifth spade.
$P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before the fifth spade = $\frac{5}{14}$.
Then $E\left[Y\right]=5+E\left[X\right]=5+39*\frac{5}{14}=18.9$
Finally, given any shuffled standard deck of 52 cards, we expect the fifth spade to appear in about 19 draws.
