TITLE: Can this probability be obtained by a combinatorial/symmetry argument?
QUESTION [16 upvotes]: 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 that
$$p=\frac{n+1}{2^n}.$$
Can this be proved by a combinatorial/symmetry argument?

REPLY [24 votes]: 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 $(n+1)/2^n$.