Probability of finding a random mass Suppose we have a list of distinctive elements: 
$$X_0=\{x_1,x_2,x_3,\cdots,x_n\}$$
Each element has mass 1. Suppose we take two elements at random and make a new element with appropriate mass. For example we take $x_1$ and $x_n$ and we make a new elements $(x_{n+1})$ of mass two, we put the new element back in the list so, 
$$X_1=\{x_1,x_2,x_3,\cdots,x_n,x_{n+1}\}$$
Now if we repeat this procedure $t$ times we will have 
$$X_t=\{x_1,x_2,x_3,\cdots,x_n,x_{n+1},x_{n+2},x_{n+3},\cdots,x_{n+t}\}$$
Surely as the process of taking two elements and making a new one is random the masses also will be distributed randomly. Now I was wondering can anyone use probability theory and compute the mass of particle $x_i$ at a given time? 
 A: One could try to solve the problem recursively. There may be better ideas and there may be a simple expression. This is just a thought.
Define $x_i \hat{=} x_{n+i}$. Denote the expectation of the mass of an element $x_i$ by $\mathbb{E}[x_i]$. Then the mass of $x_i$ is generated by two elements $x_j$ with $0 \leq j \leq i-1$. Denote the event that $x_{i-1}$ is chosen to calculate the mass of $x_i$ by $A_i$. Then $$\mathbb{E}[x_i] = \mathbb{E}[x_i | A_i] \mathbb{P}(A_i) + \mathbb{E}[x_i | A_i^c] \mathbb{P}(A_i^c)$$
We already know that if $x_{i-1}$ is not used to calculate the new mass then the expectation should stay the same, thus $\mathbb{E}[x_i | A_i^c] = \mathbb{E}[x_{i-1}]$. The probability of $A_i^c$ is also easily computed by $\mathbb{P}(A_i^c) = \frac{n+i-2}{n+i-1}\frac{n+i-3}{n+i-2}$ and $\mathbb{P}(A_i) = 1 -\mathbb{P}(A_i^c)$.
$\mathbb{E}[x_i | A_i]$ can be calculated by $$ \mathbb{E}[x_i | A_i] = \mathbb{E}(x_{i-1}) + \mathbb{E}(X_{i-2}) = \mathbb{E}(x_{i-1}) + \frac{n + \sum_{j=1}^{i-2}{\mathbb{E}(x_j)}}{n+(i-2)}$$
Putting it together:
$$ \mathbb{E}[x_i] = \mathbb{E}[x_{i-1}] + \frac{2 \sum_{j=1}^{n+i-2}\mathbb{E}(x_j)}{(n+i-1)(n+i-2)} $$     
