Index where first half exceeds second half Let $n$ be a positive integer, and independently randomize numbers $x_1,\dots,x_n,y_1,\dots,y_n$ from $(0,1)$ uniformly. Let $i(x)$ be the least index such that $$x_1+\dots+x_{i(x)}>x_{i(x)+1}+\dots+x_n.$$ Define $i(y)$ similarly. As $n$ grows, is it true that the probability that $i(x)=i(y)$ approaches $0$?
Since the number of indices grows with $n$, the probability that $i(x)=i(y)$ should go down because it is unlikely that they will coincide. But how can this be shown formally?
 A: Since $x$ and $y$ are independent, 
$$
\mathsf{P}_n\big(i(x) = i(y)\big) = \sum_{k=1}^{n-1}\mathsf{P}_n\big(i(x) = i(y)=k\big) = \sum_{k=1}^{n-1}\mathsf{P}_n\big(i(x) = k\big)^2\\\le \max_{j}\mathsf{P}_n\big(i(x) = j\big)\sum_{k=1}^{n-1}\mathsf{P}_n\big(i(x) = k\big) = \max_{j}\mathsf{P}_n\big(i(x) = j\big).
$$
Thus, it is enough to prove that the latter quantity vanishes. There are many ways to do this, I'll sketch one of them. It seems clear that $\max_{j}\mathsf{P}_n\big(i(x) = j\big)$ is attained$^*$ for $j_n = \lceil n/2\rceil$. Now
$$
\mathsf{P}_n\big(i(x) = j_n\big)\le \mathsf{P}_n\left(0\le \sum_{k=1}^{j_n} x_k - \sum_{k=j_n+1}^n x_k\le 2\right)\le \mathsf{P}_n\left(\left| \sum_{k=1}^{n-j_n} (x_k-x_{n+1-k})\right|\le 2\right).
$$
The variables $z_k = x_k-x_{n+1-k}$ are iid and centered, so by the central limit theorem,
$$
\mathsf{P}_n\left(\left| \sum_{k=1}^{n-j_n} (x_k-x_{n+1-k})\right|\le 2\right) = \mathsf{P}_n\left(\frac{1}{\sqrt{n-j_n}}\left| \sum_{k=1}^{n-j_n} (x_k-x_{n+1-k})\right|\le \frac{2}{\sqrt{n-j_n}}\right)\\
= O\left(\frac1{\sqrt{n-j_n}}\right)  = O\left(\frac1{\sqrt{n}}\right), \quad n\to\infty.
$$
(I think that the true rate of convergence is $O(1/n)$.)

$^*$ The argument still works if $j_n = n/2 + o(\sqrt{n})$, $n\to\infty$, which does follow from CLT.
