# How to physically model/construct a biased coin?

A perfectly unbiased coin is one that has the same probability for heads and tails (i.e., 50%/50%).

A perfectly biased coin is one that has (as the name suggests) different probabilities for head than for tails.

The design of a perfectly unbiased coin is pretty straightforward: a cylinder with height << radius (h << r).

However, I’m wondering how the design for a perfectly biased coin would be. Although I’m not a mathematician, I can intuitively think that a perfectly biased coin (with 60% for heads and 40% for heads) would have the heads circle surface area 60% higher than that of the tails’ circle surface area.

I don’t know how true this is and, if true, I’d like to reach the same conclusion using a mathematical approach.

Would this problem be much more difficult if we had, for example, a perfectly biased dice (i.e. cube)? How can I start constructing a mathematical model for biased, throwable, generic objects?

#### Solutions Collecting From Web of "How to physically model/construct a biased coin?"

One way to construct a biased coin is to use a solid and uniformly dense spherical cap. Obviously there are only two ways it can land: on the base or on the spherical surface. To model the probabilities of each requires some integral calculus.

The cap is formed from the unit sphere centred on (1, 0, 0) and is bounded by x = h, the cap’s height. The cap’s radius can be found to be $\sqrt{h(2-h)}$ and the volume V is given by
$$\int_0^h\pi x(2-x)dx=\pi h^2(1-\tfrac 13h)$$
The centre of mass (CM) must now be calculated. Since the cap is symmetrical about the x axis it may be reduced first to an infinitely thin rod along that axis, where the density at x = a is the cap’s cross-sectional area along that plane ($\pi a(2-a)$). The reduction to a single point is another integral:
$$\frac 1V\int_0^h\pi x^2(2-x)dx=h\frac{2-\frac 34h}{3-h}$$
This is the CM’s distance from the origin, so it is also $h-h\frac{2-\frac 34h}{3-h}=h\frac{1-\frac 14h}{3-h}$ from the cap’s base (this last value is z in the diagram).

To determine the probability of this “coin” landing on its base now requires finding the solid angle subtended by the base at the CM. This analysis works because the result of letting this coin fall in a particular orientation can be taken equivalent to the face intersected by a ray from the CM in the direction of gravity. In this case the base forms a cone with the CM of height z, so the solid angle is $2\pi(1-\cos\theta)$ where $\theta$ is half the cone’s apex angle. The fraction of the sphere this is, which is also the probability the coin lands on the base, is
$$\frac{1-\cos\theta}2=\sin^2\frac{\tan^{-1}x}2=\frac12-\frac1{2\sqrt{x^2+1}} \text{ where }x=\frac{\sqrt{h(2-h)}}{h\frac{1-\frac 14h}{3-h}}=\frac{(3-h)\sqrt{h(2-h)}}{h(1-\frac 14h)}$$
and this simplifies to
$$\frac12-\frac1{2\sqrt{1-\frac{16(h-2)(h-3)^2}{h(h-4)^2}}}$$
The plot of this expression as a function of h is shown below. It exhibits the expected behaviour at both ends: fair in the limit of a flat, infinitesimally thin coin (h = 0) and completely one-sided as a sphere (h = 2).

In particular, if the desired distribution is 60/40 the ratio of cap height to sphere radius is very close to 1:2.

To determine the dice distribution of an arbitrary solid with an arbitrary weight distribution:

• Find its centre of mass.
• For each face of the convex hull of the solid, determine the fraction of the sphere it subtends at the CM.
• If the solid is unstable (or has an unstable equilibrium) on one face and naturally rolls to another, absorb the former’s solid angle into the latter.

Hence your idea of a coin split 60/40 between its two faces’ surface areas will give a distribution closer to fairness.