Probability: Finding the expected time between events A deck with 52 cards shuffled has each card turned up in sequence until the deck is empty.  Let $X_1$ be the location of the first ace, and so on until $X_4$ represents the placement of the last ace.  Define $\Delta_1$ to be the number of cards before the first ace, $\Delta_2$ the number of cards between the first and second, and so on until $\Delta_5$ represents the cards after the last ace.  I want to find $\mathbb{E}X_2$ using the fact that $\mathbb{E}\Delta_i = \mathbb{E}\Delta_j$ for $1\leq i,j\leq 5$.  
My reasoning is that $4+\sum_{i=1}^5\Delta_i = 52$ so $4+\sum_{i=1}^5\mathbb{E}\Delta_i = 52 = 4+5\Delta = 52$ so 
$$\Delta = 48/5$$
From this I reason that $\mathbb{E}X_2$ should be $\mathbb{E}\Delta_1 + 1 + \mathbb{E}\Delta_2+1= 2 + 2\cdot 48/5$.
First, since I've never made an argument that worked quite like this, I'm just a little unsure of whether even this is right.  But assuming it is, I wanted to check it against a more direct calculation of the quantity.  However, the only way I could find to approach the problem directly was a nightmare (which is perhaps the point of the exercise).  In effect, it seems to me to bifurcate into a mess of cases:
$P(\Delta_2 = 0)$ already bifurcates into $P(\Delta_2 = 0 | \Delta_1 = 0)P(\Delta_1=0) + \ldots$
Is there any better direct approach or is the only good way to do this through the indirect approach above?
 A: To see that the five intervals have the same distribution, add a fifth "ace of crowns" or similar to make a 53-card deck. Shuffle the deck and consider the cyclic order of the cards. In the cyclic order, there are five gaps between "aces". Since these gaps are preserved by permuting the aces, each gap is equally likely to be the gap immediately after the ace of crowns, each is equally likely to be the second gap after the ace of crowns, and so on. Thus for each $i$, the $i$th gap after the ace of crowns has the same distribution (so the same expected value; since their sum is $48$ each expected value is $9.6$).
Now create a linear order on the remaining 52 cards starting with the card immediately after the ace of crowns and continuing in the cyclic ordering. If the original ordering was uniformly random, so is this ordering, and it has $\Delta_i$ equal to the $i$th gap after the ace of crowns in the cyclic ordering.
We have $X_2=2+\Delta_1+\Delta_2$, so its expectation is $2+2\times 9.6$.
