Distribution of urns for non-uniform distribution Let $X_1,\dots, X_n$ be i.i.d. random variables on a set $X$. Let $j$ be a uniformly random number from $1$ to $n$; define a new random variable
$$ Y(j) = \frac{\left\vert \{i: X_i=X_j\}\right\vert}{n}.$$
That is, $Y(j)$ should follow a distribution of how many samples have the same value. 
If the $X_i$ are uniformly random on some finite set, then this is the classic problem of a monkey throwing balls into urns. I'm curious about the case where $X_i$ have other distributions. My specific application is a discretized normal distribution (i.e., divide the real numbers into regular bins, sample a normal distribution, and output which bin it fell into), but I'm curious about this in general.
I think what I am asking could equivalently be phrased as: what is the distribution of the sizes of the bars in a histogram?
Do these distributions have a name/known properties?
 A: To make the problem easier to read, I am going to use slightly different notation to you.  I will confine attention to the case where the set of interest $\mathscr{X}$ is a finite set.  Without loss of generality, take $\mathscr{X} \equiv \{ 1,...,m \}$ and let the values of interest have a categorical distribution:
$$X_1,...,X_n \sim \text{IID Cat}(\mathbf{p})
\quad \quad \quad \mathbf{p} \equiv (p_1,...,p_m).$$
To facilitate the analysis, let $W_J \equiv | \{ i=1,...,n| X_i=X_J \} |$ denote the number of sample values that are equal to the $J$th sample value.  This can be written in alternative form as:
$$W_J = \sum_{i=1}^n \mathbb{I}(X_i = X_J) = 1 + \sum_{i \neq J} \mathbb{I}(X_i = X_J).$$
If we condition on $X_J=x$ then we have:
$$\begin{aligned}
W_J 
= 1 + \sum_{i \neq J} \mathbb{I}(X_i = x)
\sim 1 + \text{Bin}(n-1, p_x). \\[6pt]
\end{aligned}$$
Since $J \sim \text{U}(1,...,n)$ you have $\mathbb{P}(X_J = x) = p_x$, and so application of the law of total probability gives:
$$\begin{aligned}
\mathbb{P}(W_J = w) 
&= \sum_{x=1}^m \mathbb{P}(W_J = w | X_J=x) \cdot \mathbb{P}(X_J=x) \\[6pt]
&= \sum_{x=1}^m \text{Bin}( w-1 | n-1, p_x ) \cdot \ p_x \\[6pt]
&= {n-1 \choose w-1} \sum_{x=1}^m p_x^{w} (1-p_x)^{n-w}. \\[6pt]
\end{aligned}$$
As you can see, the random variable $W_J-1$ has a binomial-mixture distribution, and so $W_J$ has a closely related distribution.  We can confirm that the mass function we have derived is valid by checking that it sums to one.  Using the binomial theorem we have:
$$\begin{aligned}
\sum_{w=1}^n \mathbb{P}(W_J = w) 
&= \sum_{w=1}^n {n-1 \choose w-1} \sum_{x=1}^m p_x^{w} (1-p_x)^{n-w} \\[6pt]
&= \sum_{x=1}^m p_x (1-p_x)^{n-1} \sum_{w=1}^n {n-1 \choose w-1} \Big( \frac{p_x}{1-p_x} \Big)^{w-1} \\[6pt]
&= \sum_{x=1}^m p_x (1-p_x)^{n-1} \Big( 1 + \frac{p_x}{1-p_x} \Big)^{n-1} \\[6pt]
&= \sum_{x=1}^m p_x (1-p_x)^{n-1} \Big( \frac{1}{1-p_x} \Big)^{n-1} \\[6pt]
&= \sum_{x=1}^m p_x =1. \\[6pt]
\end{aligned}$$
This confirms the validity of the density.  In the uniform case where $\mathbf{p} = (\tfrac{1}{m},...,\tfrac{1}{m})$ you have:
$$\mathbb{P}(W_J = w) = {n-1 \choose w-1} \frac{(m-1)^{n-w}}{m^{n-1}}.$$
This gives you the distribution for $W_J$ and so the corresponding distribution for $Y_J = W_J/n$ is a simple scaled variation of this.  I do not recognise this distribution by name, but it should not be too difficult to derive its properties.  As stated, $W_J-1$ has a binomial-mixture distribution; mixtures of binomial distributions are examined in Blischke (1964) and various other papers.
