I, like many other AP Statistics students, just spent an entire class flipping Hershey's Kisses and trying to determine the empirical probability of landing it on its base. This led me to wonder if there was a way to actually calculate the probability. To simplify, I decided that instead I would flip a right circular cone with $d = h,$ ignoring the wrapper and any imperfections in the shape. My first idea was to simply use surface area, getting $$\dfrac{\pi r^2}{\pi r(r + \sqrt{r^2 + h^2})} = \dfrac{r}{r + \sqrt{r^2 + (2r)^2}} = \dfrac{1}{1 + \sqrt{5}} \approx \boxed{0.309}$$
Obviously, there are some flaws with this approximation. Is there a better way to solve this? I have a sneaking suspicion that moment of inertia comes into play. If this is the case, should I post this in a different forum? Note that the experimental probability is 0.35.
Modelling the Hershey's Kiss as a cone with uniform density, base radius $r$ and height $h$, we find its centre of mass $G$ is at a point $\frac{h}{4}$ above the centre of the base. A cross-section through the apex and the centre of the base is shown below:
If we assume the cone's orientation when it lands is random, the probability it lands on its base is the angle $\angle AGB$ divided by $2\pi$, ie $$p=\frac{1}{\pi} \tan^{-1}\frac{4r}{h}$$
In the case $h=2r$, this works out to be $p=0.3524$ (which is ridiculously close to the numerical value found!).
I would suggest the key assumption here is about how the cone lands. It is, of course, spinning, and will have angular momentum; so the orientation at the moment it lands alone isn't enough to work out if it will fall onto the base. You can imagine cases where its momentum causes it to topple and land on the curved surface. However, this happens in the other direction too so, although things won't cancel out exactly, the error will be mitigated.