Question

Is there a more efficent, possibly more mathematical and less algorithmic way of achieving a similar random number distribution to this?

unsigned int weighted_random_UINT()
{
    float r2 = 1;
    while(rand() % 4 != 0) // 3/4 chance
    {
        r2 *= fmod(
            ((float)rand()/RAND_MAX)+1, // random float between 1 and 2
            (float)UINT_MAX
        );
    }
    return (unsigned int)r2 - 1;
}

Below is a less safe but more easily readable version of the inside of the while.

r2 *= ((float)rand()/RAND_MAX)+1;

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The distribution visualized:

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Comparison between the smoother solution in the question (1st graph) and the faster solution in the best answer (2nd graph):
comparison http://with-logic.co.uk/a/graph.png

Answer 1

I think you don’t have to loop through it, but once is enough, like so:

unsigned int weighted_random_UINT()
{
    float r2 = ((float)rand()/RAND_MAX)+1; // random float between 1 and 2
    unsigned int k = 0;
    while(rand() % 4 != 0) // 3/4 chance
    {k = k < UINT_MAX ? k + 1: UINT_MAX;}
    return (unsigned int)fpow(r2,(float)k) - 1;
}

The first part is a geometric distribution, and the last one is an uniform distribution.
And you want (1+U(0,1))^G(3/4).

It should be possible to find some faster way to find G(3/4) though.

Edit:
I found it on wikipedia:
http://en.wikipedia.org/wiki/Geometric_distribution#Related_distributions

G(p)=floor(ln(U)/ln(1-p))

Thus you want:

U^floor(ln(U)/ln(1-3/4))

Which should be just two calls to rand.