Probabilities of a slightly altered dice

August 9, 2010

In art class a few months ago, I made a dice out of clay. This dice isn’t quite your typical dice. Its structure is tampered with ever so slightly, so that just \frac{1}{64} of the material is removed.

Let us model this crude, imperfect dice as a hypothetical, perfect dice. Suppose this dice is represented as a cube, composed of 64 smaller ‘cubelets’ (imagine the cubelets as 1x1x1 and the dice as 4x4x4). Furthermore, we imagine four ‘layers’ of cubelets, such that each layer is a 4×4 square.

In this dice, we remove one of the 16 cubelets from the second layer, like this:

(keep in mind that this is still a solid cube)

Assuming that the cube is made out of uniform density material (and that the missing cubelet has no mass), what is the probability of landing on each of the six sides when the cube is thrown?

A mathematical interpretation

I will solve a problem which may or may not be equivalent to throwing a dice. Instead of throwing the dice, we place the dice on a table with a random orientation, and let go. We then observe which side the dice lands on.

This appears to be a legit and identical interpretation, but there is a flaw. With an uneven dice, it is not guaranteed that each orientation would appear with uniform likelihood. Nevertheless, for our purposes we ignore this and we assume that this is an identical problem.

Which side will it land on? The center of gravity of the dice must be directly on one of the six sides; this side is the side that the dice will fall onto.

Consider a sphere of some radius, centered at the cube’s center of gravity and entirely contained in the cube:

Lines are drawn from the center of gravity to the eight vertices of the cube, essentially dividing the sphere into six pyramid-like parts.

The chance of landing on a side of the cube is proportional to the side’s surface area when ‘projected’ onto the surface of the sphere. This is because the side can be divided into infinitely many small ‘pyramids’, each one of which corresponds to the dice landing on that side when left to drop at that specific angle.

Such a portion of the surface area is called a solid angle, which is measured in steradians (sr). Whereas the entire circle is denoted as 2 \pi radians, the entire sphere is denoted by 4 \pi sr. Likewise, one hemisphere would be 2 \pi sr, and so on.

There is an efficient formula for calculating the solid angle of a tetrahedral projection onto a sphere, given by Oosterom and Strackee. If \vec{a}, \vec{b}, \vec{c} are vectors of three vertices of a tetrahedron (in relationship to the fourth), then the subtended solid angle is given by:

\tan(\frac{1}{2} \Omega) = \frac{|\vec{a} \vec{b} \vec{c}|}{abc + c(\vec{a} \cdot \vec{b}) + b(\vec{a} \cdot \vec{c}) + a(\vec{b} \cdot \vec{c})}

Here a, b, c are magnitudes of the vectors, the dot represents a dot product, and |\vec{a} \vec{b} \vec{c}| is the determinant of the scalar product of the vectors.

The magnitude of the 3-D euclidean distance:

a = \sqrt{a_x^2 + a_y^2 + a_z^2}

The dot product of two vectors is:

\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_x


|\vec{a} \vec{b} \vec{c}| = |a_x b_y c_z + b_x c_y a_z + c_x a_y b_z - a_x b_z c_y - b_x c_z a_y - c_x a_z b_y|

Calculating the probability

Let us position the cube on a 3D coordinate, such that the corners are the origin and (4,4,4). The empty cube is the cube with center (2.5,2.5,2.5).

To find the center of gravity, we just take the averages of the centroids of all the smaller cubes (the missing cube contributing 0 to the sum). This turns out to be (k,k,k) with k= \frac{251}{126} or about 1.9921.

We have k = \frac{251}{126}; now we let k' = 4-k. To facilitate vector manipulations, we will now transpose the cube such that the center of gravity is at (0,0,0):

Because of symmetry, it is obvious that three of the six sides should have one equal probability, and the other three sides to have a different, but again equal, probability. Thus it is not strictly necessary to go through the calculations for all six sides; we only need to do them for two of the sides.

A face forms a solid angle for a square pyramid, not a tetrahedron (for which we have a formula for). To solve this, we note that a square pyramid can be divided into two tetrahedrons, and the solid angle for it is simply the sum of solid angles for the two tetrahedrons.

Now by doing these calculations, we get a probability of 16.740% of landing on each of the three heavier sides, and 16.594% on each of the three lighter sides (sides closer to the missing cubelet).

By comparison, the probability of landing on any side on a fair dice is 16.667%. As only \frac{1}{64} of the dice’s mass is removed, the resulting difference is slightly less than 1%.

Throwing a rock off a cliff: Calculating the optimal angle

August 7, 2010

You want to throw a rock far, as far as you can possibly throw. Using all your energy, you throw the rock into the sky, and it lands some distance away a few seconds later. But at what angle should you throw it?

It is easy to see that the optimal angle to throw a rock when you are on flat ground is 45^\circ. That is, to make the rock travel as far as possible, one should throw the rock halfway between vertical and horizontal. Any other angle and you do worse than throwing it at a 45^\circ angle.

But what about throwing a rock off a cliff into, say, the ocean? Is 45^\circ still the optimal angle, or is it different? Perhaps it varies?

Suppose you are standing on the edge of a cliff overlooking the sea. It is h meters above the sea, and you can throw a rock with a velocity v, at an angle \theta between 0^\circ and 90^\circ. You are trying to maximize d, the distance thrown.

We assume that there is no friction or air resistance, and we let g be the gravitational constant, 9.81 m/s^2.

When throwing a rock at angle \theta, the velocity of the rock is a vector with a horizontal component of v \cos \theta and a vertical component of v \sin \theta.

Then the distance traveled can be expressed partially in terms of \theta:

d = vt \cos \theta .

The next equation we can get by substituting our values into one of the motion equations:

-h = (v \sin \theta)t - \frac{1}{2}gt^2

The unknown in the two equations is t, or the total time traveled. From the first equation we have

t = \frac{d}{v \cos \theta} .

Substituting into the second equation, we get

-h = (v \sin \theta) (\frac{d}{v \cos \theta}) - \frac{1}{2}g(\frac{d}{v \cos \theta})^2

-h = d \tan \theta - \frac{gd^2 \sec^2 \theta}{2v^2}

From here we want to find the maximum for d, or when \frac{dd}{d \theta}=0. To solve for \frac{dd}{d \theta}, we first differentiate both sides of the equation with respect to \theta:

0 = \frac{d}{d \theta} (d \tan \theta) - \frac{g}{2v^2} \frac{d}{d \theta} (d^2 \sec^2 \theta)

0 = d \sec^2 \theta + \tan \theta \frac{dd}{d \theta} - \frac{g}{2v^2}(\frac{dd^2}{d \theta} \sec^2 \theta + 2d^2 \tan \theta \sec^2 \theta)

As \frac{dd^2}{d \theta} = 2d \frac{dd}{d \theta}, we can isolate \frac{dd}{d \theta}:

\frac{dd}{d \theta} = \frac{\frac{gd^2 \tan \theta \sec^2 \theta}{v^2}-d \sec^2 \theta}{\tan \theta - \frac{gd \sec^2 \theta}{v^2}}

Since \frac{dd}{d \theta} = 0, we eliminate the denominator and are left with

0 = \frac{gd^2 \tan \theta \sec^2 \theta}{v^2} - d \sec^2 \theta

0 = d(\frac{gd \tan \theta \sec^2 \theta}{v^2} - \sec^2 \theta)

Ignoring the trivial case where d=0, we get

gd \tan \theta \sec^2 \theta = v^2 \sec^2 \theta

d = \frac{v^2}{g \tan \theta}

We have the optimal distance; we can substitute this back into a previous equation:

-h = (\frac{v^2}{g \tan \theta}) \tan \theta - \frac{g (\frac{v^2}{g \tan \theta})^2 \sec^2 \theta}{2v^2}

-h = \frac{v^2}{g} - \frac{v^2 \sec^2 \theta}{2g \tan^2 \theta}

-h = \frac{v^2}{g}(1 - \frac{1}{2\sin^2 \theta})

Finally we solve for \theta:

\frac{1}{2 \sin^2 \theta} = \frac{hg}{v^2} + 1

\sin^2 \theta = \frac{v^2}{2hg+2v^2}

\theta = \arcsin(\frac{v}{\sqrt{2hg+2v^2}})

This is the final formula for \theta in terms of v, h, and g.


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