Convex-chain median concentration ($\mathbb{P}(L_n\ge m+t)\le 2e^{-t^2/(4(m+t))},\mathbb{P}(L_n\le m-t)\le 2e^{-t^2/(4m)}$)

Let $m$ be an integer median for the longest convex chain length of independent uniform points in a triangle. If $L_n(x)\ge a$, fix a certificate of $a$ coordinates. Every configuration with $L_n\le b$ must disagree with at least $a-b$ of those coordinates; equal weights $1/\sqrt{a}$ prove Talagrand convex distance separation at least $(a-b)/\sqrt{a}$. Talagrand's convex distance inequality then gives $\mathbb{P}(L_n\ge a)\mathbb{P}(L_n\le b)\le e^{-(a-b{)}^2/(4a)}$. Set $b=m$, then $a=m$, in the two respective applications to obtain the displayed tails for positive integer $t$. The exact chain-counting bound gives $m\le C{n}^{1/3}$. Hence any interval centred at $m$ with length $\omega (n)n^{1/6}$, where $\omega (n)\to \infty$, contains $L_n$ with high probability. No upper restriction on the growth of $\omega$ is required.