Equally many cards between aces What is the probability that in a set of $52$ cards, there will be equally many cards between each ace?
I tried this problem, but I am stuck. I know that there are $52!$ possible cases.
If there are $k$ cards between each ace, than this would be likely event $\{A,k,A,k,A,k,A\}$. There is $4!$ ways to permute these aces, and cards between them $k!\cdot k!\cdot k!\cdot 3$, so that is $3\cdot 4!\cdot (k!)^3$ likely events. Another likely case would be $\{m,A,k,A,k,A,k,A\}$. Here it would be $4!\cdot (48-m)!\cdot 3$. 
Total number of likely events would be  $$\frac{3\cdot 4!\cdot (k!)^3+4!\cdot (48-m)!\cdot 3}{52!}$$
Of course, this cannot be right, because I have m and k in expression so it is not good. Where am I making mistake? And if there is some easier approach could you explain it in detail so I could really understand what is happening?
Thanks
 A: It is much easier to segregate the deck into aces and non-aces and to think of choosing the positions for the aces.  There are $52 \choose 4$ ways to place the aces.  The successful placings are defined by the position of the first ace and the number of cards between aces.  If the first ace comes at position $1$ there are $17$ ways to choose the positions of the aces because you can have $0$ to $16$ cards between each pair.  If the first ace comes at positions $2$ through $4$ you have $16$ choices, if the first ace comes at positions $5$ through $7$ you have $15$ choices and so on until if the first comes at positions $47$ to $49$ you have only one choice.  The number of successes is then $17 + 3\sum_{i=1}^{16}i=17+3\cdot \frac 12 \cdot 16 \cdot 17=425$ and the probability is $\frac {425}{52 \choose 4}=\frac 1{637}$
A: Think of the deck as a binary string in which $1$ denotes an ace, e.g.: $$000100...00100...00100...001000000$$
Every binary string of this form represents exactly $4!48!$ different possible decks, so we can confidently use this representation to calculate probabilities.
There are $52\choose 4$ such strings.
If the aces are all together as $1111$, then there can be between $0$ and $48$ cards before them in the deck. That makes $49$ arrangements.
If the aces are one card apart as $1010101$, there can be between $0$ and $45$ cards before them in the deck. That makes $46$ arrangements.
If the aces are two cards apart, there are $43$ arrangements, etc., decreasing by $3$ each time.
Therefore the answer is $$\frac{49+46+43+...+7+4+1}{52\choose 4}$$
Now there are $17$ terms on top (since $49=1+3\cdot 16$), so (adding $2$ to each term),$$\begin{align*}&\ \ 1+4+7+...+49\\=\quad&(3+6+9+...+51)-2\cdot 17\\=\quad&3(1+2+3+...+17)-34\\=\quad&3\tbinom{18}2-34\\=\quad &425\end{align*}$$
The final answer is $\frac{425}{270725}=\frac1{637}$.
