Probability of finding a random mass

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: $$ \mathbb{E}[x_i] = \mathbb{E}[x_{i-1}] + \frac{2 \sum_{j=1}^{n+i-2}\mathbb{E}(x_j)}{(n+i-1)(n+i-2)} $$