Average number of cards drawn without replacement when first, second, third, and fourth aces appear I have a standard $52$ deck of cards, shuffled randomly.
If I flip cards off the top until a Ace is shown, what is the average number of cards expected to be flipped over, including the Ace?
Further to this, what would be the average numbers of cards expected to be flipped for the second Ace, then the third Ace, and then finally the fourth?
My personal thoughts were that it would be $4/52$ chance, so take the reciprocal of this and it's $13$ cards on average.
Take the number of cards left, and so, for example, let's say it was $13$ cards the first time. $52-13=39$. $1/(3/39)$ is the average number next time, which is $13$ cards, so this seems to fit the pattern.
However when I actually do this, the number on average has been much lower. So I'm assuming there is something I'm missing?
 A: Re: Replacement
The expected number of draws with replacement until getting the first ace would be $13$. However, without replacement, which is what applies here, the expected number of draws is smaller. This makes sense because the non-aces are not being replaced to be drawn a second time. 

Without Replacement
The probability that the $n^\text{th}$ ace is at position $k$ is the number of ways for $n-1$ aces to be in the first $k-1$ positions and $4-n$ aces to be in the last $52-k$ positions. That is,
$$
\frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}}
$$
The mean position of the $n^\text{th}$ ace is
$$
\begin{align}
\sum_{k=1}^{52}\frac{\binom{k-1}{n-1}\binom{52-k}{4-n}}{\binom{52}{4}}k
&=\sum_{k=1}^{52}\frac{\binom{k}{n}\binom{52-k}{4-n}}{\binom{52}{4}}n\\
&=\frac{\binom{53}{5}}{\binom{52}{4}}n\\[3pt]
&=\frac{53n}{5}
\end{align}
$$
A: @robjohn already answered in full.  Here I provide 2 different views:
(1) You asked "So I'm assuming there is something I'm missing?" - The answer is that 13 cannot possibly be correct.  If 13 cards on average show up before each ace, then on average the 4th ace would have to be the last card.  Clearly wrong.
(2) Your "interval" idea is not without merit, but there are in fact five intervals.  Imagine you insert a 5th ace (which is actually a joker).  Now there are 53 cards, and you deal them out into a circle.  The 5 aces chop them up into 5 intervals.  By symmetry each interval is equi-distributed.  Since they add up to 53, by linearity of expectation each interval is of expected length 53/5.  Now cut the circle open by removing the joker.  You now have a linear sequence of 52 cards, and the cards before the 1st ace = the cards in the interval to the right of the original joker.
