I am generating a flat Dirichlet distribution such that $\alpha_{1},....,\alpha_{k} = 1$. The distribution looks even for values of $ k < 5$ and 1.000.000 draws. However when I increase the value of $k$ for the same number of draws I struggle to reach extreme values of $x$, like $x > 0.95$. Thus I would like to know how to calculate the minimum number of draws such that $$p(x_i > 0.95) > 90 \%$$
Alternatively could I decrease the value of $\alpha$ to achieve the same goal?
Thank you
Fortunately all of the extremes are disjoint (and will remain so if you replace 0.95 with anything above 0.5), so there are no complications in calculating this.
Starting with the $\alpha = 1$ case:
For each $i$, \begin{align} P &= P(x_i \gt 0.95) \\ &= \int_{x_i \gt 0.95} \mathrm{d}x \end{align}
Instead of the standard simplex, we've clipped off a tiny corner that is a smaller simplex. It's $0.05^{K-1}$ smaller, so we immediately get that $P = (0.05)^{K-1}$, without even having to think about any normalization factors. You'll note that this becomes tiny as $K$ grows, even though the samples continue to be distributed over the $K$-simplex.
The chances of any of them being in their own extreme region is just $K$ times this, or $KP = K(0.05)^{K-1}$. Finally, for none to be selected in $n$ tries is $(1 - KP)^n$, and for at least one, $1 - (1 - KP)^n$. Set that to .9, solve for $n$ as a function of K, and you're done.
You can indeed influence things by using a different $\alpha$; a smaller $\alpha$ is more concentrated in the corners. The math is hairier for general $\alpha$, as we no longer have a constant density, but I'm busy so I'll come back and finish this answer later.