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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?

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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:

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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.

alt text

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 $\sqrt{\frac{4A}{3\pi}}$.

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I don't think that's sufficient; given an angle the coin falls from, you have to consider whether the barycenter is above the point of contact – Mechanical snail Aug 23 '11 at 0:26
@Mechanical snail: The initial point of contact will almost certainly be one of the four 'corners' shown above. Consequently the centre of mass will almost certainly be to the left/right of the point of contact, and will cause the coin to fall on a head/tail/edge (at which point it will be above the point of contact). – tttppp Aug 23 '11 at 7:39

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.

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"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." I have to disagree, consider the case of thickness >> face area, i.e. a dowel rod shape. Here, P(heads) ~= 0, P(edge) ~= 1 – cobaltduck Sep 27 '10 at 12:05
ok, I meant "really" in the sense of "a coin which satisfies the hypotheses of the question", not in "a coin thick enough". – mau Sep 28 '10 at 8:41

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 $T = \frac{A}{C} = \frac{A}{2 \sqrt{A \pi}} = \frac{\sqrt{A\pi}}{2\pi}$

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Why should the probability depend on the surface area? Does embossing the face with a pattern, increasing its surface area, increase the chances of it landing on that face? – Rahul Sep 27 '10 at 13:15
@Rahul: I'm indeed assuming that all surfaces are plane. In a very simple model, one could argue that dropping the coin corresponds to randomly picking a point on the coin, namely the point where it hits the ground. Of course, this doesn't take into account any physics: the coin might for example tumble on the ground after it has hit it for the first time. – Sebastian Sep 27 '10 at 15:27

It is designed to be a not well posed question. The "correct answer" to this simple question should be to reference:

Bertrand's Paradox and Maximum Ignorance Principle.

They're looking for you to think outside of the box and apply theory.

How hard is the coin? How elastic? How hard/elastic is the table? How sharp are the edges of the coin. Is the weight of the coin evenly distributed? Is the coin coated? Will the coin be mechanically flipped? How high? How much spin? Spin on which axis? Will be be randomly flipped?

In practice you'll find the solution lies somewhere between a thickness of .35 and .6 of the diameter.

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