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Given a point ${\bf x} = (x_1,x_2,\dots,x_n)$ in the orthosimplex $K = \{(x_1,x_2,\dots,x_n)\ : \ 0 \leq x_1 \leq x_2 \leq \dots \leq x_n \leq 1\}$, what proportion of a ball of radius $\epsilon$ around ${\bf x}$ lies in $K$, for small positive $\epsilon$? (This will be independent of $\epsilon$ once it's small enough.)

For instance, with $n=2$, $K$ is an isosceles right triangle. The proportion in question is 1/8 at two of the vertices of $K$, 1/4 at the remaining vertex of $K$, 1/2 for ${\bf x}$ lying on an edge of $K$, and 1 for ${\bf x}$ lying in the interior of $K$.

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    $\begingroup$ Recall that $n!$ copies of the orrhosimplex tile the unit cube. This should provide at least some relations on the solid angles. Perhaps, they are even enough to determine everything? $\endgroup$ Oct 3, 2020 at 16:41

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Note that $x+\varepsilon y\in K$ for small $\varepsilon$ if and only if $y$ satisfies the inequalities $y_i\leqslant y_{i+1}$ if $x_i=x_{i+1}$; $y_1\geqslant 0$ if $x_1=0$; $y_n\leqslant 0$ if $x_n=0$. We may think that $y_i$ are i.i.d. Gaussian, then the probability of a chain of an inequalities like $0\leqslant y_1\leqslant y_2\leqslant \ldots \leqslant y_k$ equals $2^{-k}/k!$ ($2^{-k}$ is the probability that all $y_1,\ldots,y_k$ are positive, $1/k!$ the probability that they are ordered this way, these two events are independent). For the chain $y_a\leqslant \ldots \leqslant y_b$ the probability equals $1/(b-a+1)!$. The probabilities for disjoint chains must be multiplied.

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