[cschmid]

This is pretty obvious, no? If you choose a (uniformly) random point inside an n-sphere, the components of the vector will go to zero as n gets larger -- after all, the sum of their squares has to be less than 1.

A (uniformly) random point in an n-dimensional cube will have random coordinates from zero to 1, with no other constraint on their size.

[reikonomusha]

No, it is not obvious to people who haven't been trained in college-level mathematical thinking, or people who don't think about higher-dimensional objects using the handlebars of abstraction.

[jakear]

Pick N random numbers between -0.5 and 0.5. Square them. Then add them up. How does the probability that {sum of these N positive values exceeds 1} change as the N increases?

I think anyone with a basic understanding of arithmetic should be able to answer that.

[thomasahle]

Yes, this is a simple way to think about it:

If you choose n uniformly random values in [-1/2, 1/2] (the unit cube) then the sum of squares will be concentrated around n/12 (just take the variance).

This is way more than 1, which is what you would need to stay inside the unit ball.

[jstimpfle]

It's a nice explanation (maybe leave out the "obvious" part?). How do you explain that the volume goes up first, and peaks at N=5?

[j7ake]

This is nice way to think of the problem. Although I think for unit hyper sphere , I think of the sphere centred at the origin, so the distance from point to origin is always less than 0.5