What's the smallest number of regular polyhedral dice necessary to "throw 1 in 7"?

Question

A regular cubic dice can be marked to give 2, 3, or 6 equally likely events. A tetrahedron can give 2 or 4 equally likely events. A Isocahedron can give 2, 4, 5, 10, 20 equally likely events. An octahedron can give 2, 4, 8 equally likely events. Two cubic dice can give 9 equally likely events. What's the smallest combination of such marked regular polyhedral dice that will give 7 equally likely events?

Platonic solids, juicy_wishun. I know that there are other kinds of solids that will give you 1 in 7, the simplest being a double 7-sided pyramid. But that's trivial and totally uninteresting.

I'll expand the question, and ask for any scheme or procedure, using marked platonic solids, to develop 7 equally likely events. Obviously any product of number of sides of platonic solids won't ever have 7 as a factor. Be more imaginative.

acafrao341, a truncated cube indeed does have 14 sides, which can even be made to have equal areas. However, it's extremely difficult to show that all faces are equally likely to come out on top. or how to work out the size of the truncations for that to happen. This being a math problem, not a physics one, I'm asking for a solution using platonic solids.

To the idea of using a 1,2,3,4 tetrahedron and 1,2,3,1,2,3 cubic, we have 24 possible outcomes, of which there are 2,4,6,6,4,2 ways to add up to 2,3,4,5,6,7. So, that's not coming up with anything happening with 1 in 7 probability.

Bogg, best answer so far, a practical one. But mathematicians frequently aren't very practical, so I'm still looking for a "better" answer, even though "better" here could be debatable. Let's see what others have to say.

It's a puzzle, you know.

Answer 1

You can do it with a regular octahedron.... if you roll an 8, just ignore it and roll again. The expected number or rolls it will take is (1 + 1/8 + 1/64 + ... = 8/7)... though it could in theory take a lot of rolls if you keep rolling 8's.

By the way, you can generate a random integer between 1 and 7 with coin if you flip three times, let heads be 1 and tails be 0 and let the three flips correspond to a 3-digit binary number, and flip three more times if you get 3 tails. This takes 24/7 flips on average.

Answer 2

There is no way to get 7 equally likely events from rolling regular polyhedra.

That's because the number of events has to be some product of powers of 4, 6, 8, 12, and 20. That means that the probability of an set of events occurring has to be of the form:

A/2^n*3^n*5^n

Where A is an integer. In particular, there is no way to get a 7 in the denominator.

Answer 3

I think you cannot do it with any combination of regular polyhedra, as there is no combination of numbers of faces that is a multiple of 7. However, you can do it with a truncated cube (a uniform, or Archimedean, solid), which has 14 faces. Mark each face with 1 to 7; so, each number is on two different faces. Your rolls will then be 1 in 7.

In response to your response:

Would not a tetrahedron marked 1,2,3,4 together with a cube marked 1,2,3,1,2,3 do it? The sum of the upturned faces is 7, and each combination is equally likely.

Answer 4

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Answer 5

I may be misunderstanding the qeustion, but I think you are overcomplicating things.

If you want a ceratin number of equally likely events, all you need is one regular polyhedron with that many sides. They are hard to describe, but a regular polygon can be created with N sides for any N>1.