Suppose that $a\_1,\dots,a\_n,b\_1,\dots,b\_n$ are iid random variables each with a symmetric non-atomic distribution.
Let $p$ denote the probability that there is some real $t$ such that $t a\_i \ge b\_i$ for all $i$.
It [was shown](https://mathoverflow.net/a/415074/36721) that
$$p=\frac{n+1}{2^n}.$$


Can this be proved by a combinatorial/symmetry argument?



Can this probability be obtained by a combinatorial/symmetry argument?
If I understand correctly, $c\_i := b\_i/a\_i$ should also be symmetric and non-atomic.
Then the result holds if there exists $t$ so that for all $i$


* $t \geq c\_i$ if $a\_i > 0$;
* $c\_i \geq t$ if $a\_i < 0$.


Reorder the indices so that $|c\_1| < |c\_2| < \dots < |c\_n|$.
There are $2^n$ ways to assign signs to $a\_i$ for each $i \in [n]$.
The only assignments for which an appropriate $t$ exists are those where the signs of $a\_1,\dots,a\_k$ are positive and the signs of $a\_{k+1},\dots,a\_n$ are negative with $k \in \{0,1,2,\dots,n\}$.
There are $n+1$ such assignments, so the desired probability is $(n+1)/2^n$.


