Constructing a 5-Sided Pyramid Die for Equal Probability

[Yoni]

I want to construct 5 sided dice in the shape of a square-based pyramid.
When rolled, I want the die to have equal probability to fall on all sides. That is, I want the odds that the die would fall base down to be 0.2.

I insist on a pyramid shaped die (not tetrahedron). Other solutions are not relevant. Can this be done? Which pyramid dimensions would provide this? (Height to base diagonal ratio, or angle of the tip)

My logic: This should be possible. If tip angle is wide and reaches pi, the die becomes flat and chance to flip base down is 0.5. If tip is narrowed reaching 0, the chance for rolling the die base down falls to 0 as well. A point must exists that provides a chance of 0.2 for throwing the die base down.

Also, I wonder if this pyramid proportion is highly sensitive or not. Will small changes to angle tip throw the chance drastically away from 0.2? What is the full dependence of this chance on tip angle (or height to base ratio)?

Also, does the die quasi-randomness maintained? In other words, is sensitivity to initial conditions similar to the regular cubic die? Can a normal person learn to throw the dice in a way as to alter the chance of outcome?

 

[Yoni]

Note, a patent for 5-sided dice exists. However, it is not a pyramid, and only empirical evidence vouch for it's fairness.

 

[rcgldr]

As an alternative, a 20 sided die, icosahedron, with 4 sets of numbers 1 to 5 could be used.

 

[A.T.]

Is there a general method to find the probabilities of such an irregular die? Something like: Project the faces onto a sphere centered at the center of mass. Or do the mass distribution, moments of inertia also play a role?

 

[Chronos]

A pentagonal trapezohedron might be best. It has 10 sides, but, you can duplicate 1-5. It is one of the few die, aside from 6, that is perfectly symmetric

 

[dipole]

I think moment of inertia must be important. One starting point might be to calculate the moment of inertia about every edge as a function of the parameters you list, and see if it's possible to make them all equal.

 

[A.T.]

I think moment of inertia must be important.

I'm afraid so too. Considering just the geometry and center of mass might give you the answer for putting the die on a table at rest with a random orientation, and letting it fall over onto one of the faces. But with arbitrary initial conditions rotational dynamics will play a role, and make the problem much more difficult. It might even be that the shape depends on the frictional parameters.

 

[NascentOxygen]

I think moment of inertia must be important. One starting point might be to calculate the moment of inertia about every edge as a function of the parameters you list, and see if it's possible to make them all equal.

Constructed as a hollow die with 5 cutouts from sheet plastic, it might be possible to make some cutouts of different thickness to even out all probabilities? Or even cut the sides from sheet plastic of tapered thicknesses?

No, I'm not offering ...

 

[CWatters]

My logic: This should be possible. If tip angle is wide and reaches pi, the die becomes flat and chance to flip base down is 0.5. If tip is narrowed reaching 0, the chance for rolling the die base down falls to 0 as well. A point must exists that provides a chance of 0.2 for throwing the die base down.

You can apply the thinking to other shape dice such as a "thick coin" - eg How thick does a coin have to be before there is an equal probability of it landing on it's edge? Papers have been written on that..

http://arxiv.org/pdf/1008.4559.pdf

 

[epenguin]

A pentagonal trapezohedron might be best. It has 10 sides, but, you can duplicate 1-5. It is one of the few die, aside from 6, that is perfectly symmetric

But as two faces would be equally 'up' you'd have to have a rule about which one counts, like the clockwise one looking down the axis.

 

[jbriggs444]

But as two faces would be equally 'up' you'd have to have a rule about which one counts, like the clockwise one looking down the axis.

If you look at the picture on http://en.wikipedia.org/wiki/Pentagonal_trapezohedron it is clear that exactly one face is unequivocally 'up'.

 

[epenguin]

If you look at the picture on http://en.wikipedia.org/wiki/Pentagonal_trapezohedron it is clear that exactly one face is unequivocally 'up'.

Ah yes now I see!

 

[Haborix]

Here is a wikipedia page which may help answer the question (All its faces being congruent and being face-transitive implies that you have a fair dice, but from the discussion so far perhaps the implication does not run in the opposite direction).