Distribution of infinity-norm over the unit sphere I need to compute probabilities of the form
$P( \Vert X \Vert_\infty < r ),$
where $X$ is a random variable of dimension $n$, drawn with a uniform distribution on the unit sphere $\mathcal{S}_{n-1}$. Clearly, the distribution of $Z = \Vert X \Vert_\infty$ is supported by the interval $[\frac{1}{\sqrt{n}},1]$.
Any hint to get a simple formula (ideally a closed-form expression or a one-dimensional integral) will be greatly appreciated !
Note:
Using the fact that $X \sim \frac{Y}{\Vert Y\Vert_2}$ for $Y \sim \mathcal{N}(0,1)$, at first I thought I could first compute the probabilities
$P_i:=P( |X_i| < r \Vert X \Vert_2)$, which can be expressed as a quantile of the F-distribution, but then I realized that those events are not independent, so 
$P( \Vert X \Vert_\infty < r ) \neq \prod_{i=1}^n P_i$.
 A: Edit: As I write at the end. This is wrong for low dims, but approximately correct for more than a few hundreds, when I guess the coordinates become less dependent.
You can start by uniformly selecting points on the (n − 1)-sphere and then ask what is the infinity norm of such points.
Given a vector $Z$ with $n$ coordinates taken i.i.d. from the normal distribution $Z_i \sim \mathcal{N}(0, 1)$. The CDF of the absolute maximum of those is given by considering the CDF of $n$ points drawn from the half normal distribution  $F_{||Z||_{\infty}}(x)={\operatorname{erf}\left( \frac{ x }{\sqrt 2} \right) }^{n}$. Now we need to normalize the points, so they'll be on the unit sphere. Given that the positive squared root of the sum of squared normal variables has a Chi distribution with $n$ degress of freedom, $||Z||_2\sim \chi_n$, we have:
$$P(||X||_{\infty} < r\;|\;||X||_2=1) = \\P(||Z||_{\infty}/||Z||_2 < r) = $$ [wrong transition! (Though it still works in high-dims)]
$$\int_{0}^{\infty} P(||Z||_{\infty}/ ||Z||_2 < r\; |\; ||Z||_2 = x)  f_{\chi_n}(x)\;dx = \\ 
\\\int_{0}^{\infty} F_{||Z||_{\infty}}(x \cdot r) f_{\chi_n}(x)\;dx = \\\int_{0}^{\infty} {\operatorname{erf}\left( \frac{ x \cdot r}{\sqrt 2} \right) }^{n}  \cdot \dfrac{x^{n-1}e^{-x^2/2}}{2^{n/2-1}\Gamma\left(\frac{n}{2}\right)}\; dx$$
Update
Alternatively, I'm not entirely sure we can do this, but I think we can just consider the expected scale of $r$, instead of calculating it as above. It is the mean of $\chi_n$, so the above should be the same as:
$$
F_{||Z||_{\infty}}(\mu(\chi_n) \cdot r)
$$
Update2
These formulations seem to be correct for large enough $n$, but not in general:


A: There is a trick to reduce these kinds of questions to questions about independent normals, as in my ancient preprint. For large $n,$ concentration of measure will presumably give you easy estimates.
