Probability density for velocity in mechanical energy To be sure my basic physics isn't rusty...
Consider a 2-D bowled shaped classical potential well within which a classical particle of mass m is rolling.
In this system the conservation of energy holds so the particle of mass m would roll from one end to another.
Because conservation of energy holds, we expect the mechanical energy to be

$E=T + U = \frac{1}{2}mv^{2}-mgh$
where v is the velocity of the particle
g is the gravitational acceleration 
and 
h is the heigh relative to the ground. 

In classical physics, the maximum velocity of the particle occurs when the particle is at $r=\left ( x,h=0 \right )$  and the minimum velocity occurs when the particle is at some position $r=\left ( x,h \right )$ \exists h on both end of the well such that its kinetic energy T is 0 and potential energy U is at maximum. 
Again, this follows from the conservation of energy: 

$\Delta T= - \Delta U$ 

Now, I would like to construct a mathematical equation describing the probability of finding this particle of mass m as a function of its velocity.
Intuitively, the greater the velocity of the particle at some point the lower the probability to find the particle and the smaller its velocity is the higher the probability to find the particle.
Solving $ E=T + U = \frac{1}{2}mv^{2}-mgh$ for v:

$v=\sqrt{\frac{2 \left ( E+mgh \right )
}{m}}$

If we want to explictly determine the probability of finding the particle as a function of its velocity, we should expect the probability density as a function of velocity to be of the form 

$P=P\left ( v \right ) \propto \frac{1}{v}$

which comports to our common sense intuition. 
How can I go about constructing a more explicit and informative equation that would enable to me determine the probability of finding the particle as a function of its velocity?
Any help is appreciated.
 A: I think with the information given, you cannot construct such a probability density function, because it very much depend on the shape of the 2-D bowl in which the ball is moving. 
Compare the 'bowls' A and B with shape $y=x^2$ and $y=\min((x+1)^2, 0, (x-1)^2)$, respectively. Bowl B is similar to bowl A, except that its 'bottom' is on the interval $[-1, 1]$ rather than a single point. In both cases, the maximum speed of the balls will be the same (given that the balls started from the same height). But the speed profile will be different, i.e. the ball in bowl B will maintain its maximum speed for a while. As a result the probability density function $P(v)$ will be different.
A: It's probably late to post the answer, but this answer may help others.
Your bowled-shaped-potential can approximately deduce the result $P(v)\propto v^{-1}$, not exactly.
This result is acturally based on an old friend model, harmonic oscillator. 
As you may know, the harmonic oscillator can be visualized as a 2-D rotation on the circle (with constant angular velocity $\omega$) projected on to the horizontal $x$ axis.
We can define the time-averaged probability density on the circle orbit, which is  $1/2\pi R$ and independent of the angular $\theta$.
After the projection to $x$ axis, we get 
$$\rho(x)\propto 1/\sqrt{R^2-x^2}\propto 1/\sqrt{E-V}\propto 1/v.$$
