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Given $k\in\mathbb{N}^+$ and $P = \{p : p \in [0, 1]^k \wedge \sum_{i=1}^k p_i = 1 \}$, we choose a random element $q$ of $P$. What is the probability distribution of $S = \sum_{i=1}^k q_i^2$?

I have made some plots with synthetic examples and all I have figured out is that the mean seems to be something similar to $1/k$ (although I could be wrong) and the kurtosis appears to decrease with $k$.

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    You need to define a distribution on $p$ first. I assume you're taking it to be uniform on the simplex $P$? – 2017-02-22
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    @PepeSilvia Thanks, I meant uniform yes. I have rephrased the question, I hope it's less ambiguous now (or at least not more). – 2017-02-22

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A common distribution on the simplex is the Dirichlet Distribution. See that your example we set all $\alpha_i=1$ to get $\mathbb{E}[p_i^2]=\frac{1}{k^2}+\frac{k-1}{k^2(k+1)}=\frac{2}{k(k+1)}$, and so $\mathbb{E}[S]=\frac{2}{k+1}$. As far as distributions go the $p_i$ are beta which means their squares don't have an elementary distribution. However if you choose your distribution on the simplex so that $\alpha_i=\frac{1}{k-1}$ then $p_i\sim\text{Beta}(\frac{1}{k-1},1)$ so that $p_i^2\sim\text{Beta}(\frac{1}{2(k-1)},1)$. However sums of betas are not well-understood.