Number of 'walks' which stay above 0. Consider a set of distinct $n$ numbers where $a_i \in \mathbb{R} $ and $$\sum_{i=1}^{n} a_i = 0$$ A walk is defined to be the sum of the numbers, so that the $k$th position is the partial sum to $k$. How many of all possible walks are always above or at $0$?
I've conjectured this to be $\frac1n$ of the $n!$ total walks and proven it for cases up to $n=4$ by hand and MATLAB has verified this for larger cases. I cannot seem to find a way to solve this generally though. At first I thought this was a probability problem as it is similar to the Ballot problem but am now quite sure it is a combinatorial one.
If we restrict this problem to integers, it will also be true for rationals as we can multiply a set by the largest denominator. It should then extend into the reals as the rationals are dense in the reals    
 A: The conjecture is true if no non-empty proper subset of $\{a_1,\ldots,a_n\}$ sums to $0$. Hereafter I will make that assumption.
Let $\langle b_0,\ldots,b_{n-1}\rangle$ be any permutation of the numbers $a_1,\ldots,a_n$. Extend it to an infinite periodic sequence $\langle b_k:k\in\Bbb N\rangle$ with period $n$ by setting $b_{\ell n+k}=b_k$ for $k=0,\ldots,n-1$ and $\ell\in\Bbb N$. For $k\in\Bbb N$ let
$$s_k=\sum_{i=0}^kb_i\;.$$
Now plot $s_k$ against $k$: the graph is periodic, with $s_{\ell n}=0$ for each $\ell\in\Bbb N$. The assumption implies that there is a unique $m\in\{0,\ldots,n-1\}$ such that $s_m=\min\{s_k:k\in\Bbb N\}$. Shift the permutation $m$ circularly places to the left and recompute to get partial sums $s_k'$; the resulting graph of partial sums has the same rises and falls as the original one, but its vertical origin is now at the minimum $y$-coordinate of the original graph. Thus, the graph remains above the $x$-axis except at the ends of each cycle, where $s_{\ell n}'=0$ for $\ell\in\Bbb N$. The uniqueness of $m$ means that the graph of any other circular permutation has a higher origin and therefore drops below the $x$-axis in each cycle.
Thus, for each circular permutation of the numbers $a_1,\ldots,a_n$ there is a unique starting point that yields positive partial sums, so the fraction of walks staying above $0$ is $\frac1n$.
If some non-empty proper subset of $\{a_1,\ldots,a_n\}$ sums to $0$, you’ll get more than $\frac1n$ of the walks that remain non-negative, and fewer than $\frac1n$ that remain strictly positive.
