Why is $pdf(\Psi) = \frac{1}{2\pi}$ when picking from hemisphere uniformly? I'm trying to figure out why the probability density function for picking a direction uniformly on the hemisphere is $\frac{1}{2\pi}$.

Something tells me this is related to the number of steradians in a hemisphere being $2\pi$, but I can't see how this is connected. We could come up with another measure for solid angle ("blibs" for example) where the number of blibs in the hemisphere would be $8000\pi$, but I'd expect the probability for picking a direction uniformly to be unchanged.

EDIT: I understand this now. My confusion arose because I wasn't thinking clearly about the units of the probability density function, i.e. $\frac{1}{sr}$. I got thrown off by thinking of area, and confusingly thought the pdf units were inverse area (or something)!
It's like this because the surface area of a unit sphere is $4\pi$. As ratchet freak points out, the integral of a probability distribution over its domain has to be 1. Put another way, the probability of choosing some direction is 1. The surface area of the hemisphere is $2\pi$. The constant that you integrate over a $2\pi$ domain to get 1 is $\frac{1}{2\pi}$.

The connection to steradians is that 1 steradian is defined as the solid angle that subtends a surface of area 1 on a unit sphere. That is, a unit sphere has $4\pi$ steradians and a surface area of $4\pi$.

It's just the same in two dimensions. A radian is chosen so that a 1 radian arc of a unit circle has length 1, so there are $2\pi$ radians in a circle, and $\pi$ radians in a semicircle. If you choose 2D directions uniformly on a semicircle, the probability distribution is $\frac{1}{\pi}$.