Rolling a Long String of Ones

A die with k sides, numbered from 1 to k, has an equal probability (1/k) of landing with any one of these numbers up whenever it is thrown. On average, how many times do you have to throw the die to generate n sequential outcomes of the number 1?

Which takes more throws on average?

  • k and n?
  • k − 1 and n + 1?

Know the answer? Send your solution to ar@casact.org.

Infinitely Many Equal Pieces

The solution below is based on the solution submitted by Eamonn Long, FCAS, who correctly recognized that the answer hinges on the truth of the axiom of choice. In practice this axiom is usually assumed to be true. However, there are skeptics, and Long demonstrates both points of view.

The Question: “Given any finite positive real number A, can you define a set S, in the two-dimensional plane, with area A and a partition of S into infinitely many sequentially numbered subsets S 1, S 2…. such that any two of these subsets are isometric? Show it or prove it to be impossible. Isometric in this case will specifically mean two subsets related by a one-to-one mapping that only involves translation and/or rotation in the plane. Partition means that the subsets are pairwise disjointed and that their union equals S. Can you generalize your answer to a higher dimension?”

Remark: The Question could be viewed as appearing to relate to the second of Kant’s antinomies. I will give an answer arguing it can be possible, and I will give an answer arguing it can be impossible.

Proof that it is possible (sketch, assuming axiom of choice is true)

The Area A is a red herring; scaling in finite dimensions can reduce us to the case of the unit hypersphere.

It is convenient to use the complex number plane to handle the situation in two dimensions. Let S be the set of all eiΘ where Θ is real and i is the square root of −1.

Define x equivalent to y if and only if there are real numbers a and b such that x = eia and y = eib where ab is an integer. This equivalence relation defines a partition in the usual way.

Let A be the set formed (using the axiom of choice) of exactly one element of each equivalence class. For a given integer z, let Az be the rotation of A defined by Az in the set of all ei(Θ+z) where eiΘ is in A.

S is the union of all the Az sets. Since the integers are countable, the number of Az sets is countable. The Az sets are pairwise disjointed, since the angle between any two points in different sets is not an integer. However, one Az set can be rotated by such a non-integer angle onto another Az set. So, the Az sets are an example solution in two dimensions.

To generalize from the circumference to the whole circle, join rays from the origin to the edge and associate points on the circumference with all those on the rays joining from the origin (excluding the origin itself).

To generalize to higher dimensions, using n-spheres and n-balls (excluding the origin), just rotate in any two dimensions, as just shown, leaving the coordinates in the other dimensions fixed.

Proof by contradiction that it is not possible (assuming the axiom of choice is false)

In the Solovay model (see https://people.math.wisc.edu/~awmille1/old/m873-03/solovay.pdf), every set of real numbers is Lebesgue measurable.

Given that we are asked to consider a finite area being split into infinitely many isometric subsets (which, being Lebesgue measurable, must have the same measure), we must conclude that this is impossible since the isometric subsets can neither have a finite nor zero measure for the objective to be achieved.

Therefore, it is not possible in this model of set theory.

Bob Conger also submitted a solution.