Flipping a Hemisphere
Take a solid ball and cut in in half. Label the flat side "Heads" and the
curved side "Tails." What are the probabilities of Heads and Tails when
flipping the hemisphere?
Source: Original.
Solutions were received from Nick McGrath. His solution is below.
(I didn't leave this puzzle up
very long, because I was excited to get the next puzzle up.
If others have enhancements to add, please feel free to
send them.)
Nick's Solution:
I suspect that, in practice, this is a very complex problem dependent on
the coefficient of restitution of the hemisphere and surface. i.e what
happens after the first bounce. Does the hemisphere rotate so that any
part of it, with equal probability, may hit the ground the next time,
or is its rotation a function of where on its surface it impacts the
first, second or subsequent times. Who knows?
For the purpose of this question, we must assume that the hemisphere
impacts the ground for the last time (i.e. does not bounce again) at
any orientation with equal probability.
It is an easy excercise in Calculus to show that the centroid of a
hemisphere is 3/8 of its radius from the center of the flat face.
The hemisphere will land Heads if the impact point is anywhere on the
curved side. It will also land heads if the impact point is on the edge
(i.e. on the circumference of the flat face) and the centroid is on the
same side of the impact point as the curved side of the hemisphere.
i.e the hemisphere will land heads if the angle of
impact = 90+arctan(3/8) = 110.55605 out of a possible orientation of
180 degrees.
=> probabilty Heads = 110.55505/180 = 0.6142
=> prob tails = 1-0.6162 =.3858
Mail to Ken