Find thickness of a coin

Question

This is one of the question asked in a written test conducted by a company. The question sounded stupid to me. May be its not.

"Given the area of the coin to be 'A'. If the probability of getting a tail, head and the edge are same, what is the thickness of the coin?

Answer 1

If you assume that the probability of getting tail, head or edge only depends on the surface area that those regions of the coin have, then you see that the area of the edge must also equal $A$. If you denote by $T$ the thickness of the coin, then the area of the edge is given by $T \cdot C$, where $C$ is the circumference of the coin. Now, as you are given $A$, you can determine $C = 2 \sqrt{A \pi}$ and thus you get $T = \frac{A}{C} = \frac{A}{2 \sqrt{A \pi}} = \frac{\sqrt{A\pi}}{2\pi}$

Answer 2

I think that the question is stupid, because it does not take into account the fact that even with a really thick coin it's easier to tilt it when it lands on the edge than when it lands on a face.
This said, as Sebastian wrote down in formulas you should think at the edge of the coin as a surface, and not at the single line where the coin ends up.

Answer 3

This problem is also considered in the book Fifty challenging problems in probability with solutions by Frederick Mosteller.

For a more in depth study on bias in coin tossing you could read: http://comptop.stanford.edu/preprints/heads.pdf

Answer 4

I assumed that the probability of getting a head, tail or edge depended on the angle from the centre of the coin that the side lies in. So the head, tail and edge must each occupy 120 degrees when viewed along the axis of rotation.

In the diagram above the angles at the centre are all (meant to be) 60 degrees and the radius of each face is $\sqrt{A/\pi}$. A small amount of trigonometry later and I found the edge length to be $\sqrt{\frac{4A}{3\pi}}$.