I have recently bought a fair $10$-sided dice for generating equally distributed random numbers from the range $1,2,\ldots,10$. Let me describe this dice in some more detail. Consider a regular 5-gon $ABCDE$ in 3-dimensional space that is situated in the $xy$-plane so that its center lies in the origin. Consider a further point $F$ on the $z$-axis somewhere above the origin, and a point $G$ that results by reflecting $F$ across the origin. If one connects every corner of the 5-gon $ABCDE$ by two edges to $F$ and $G$, this will yield a polyhedron $P_{10}$ with 10 congruent triangular faces. My fair 10-sided dice is essentially $P_{10}$ with the sides labeled by the numbers 1,2,...,10. Throw it up into the air, look how it lands, and take the side on which it lands as your random number. (With a cube one always takes the top side, but $P_{10}$ has no top side). By the symmetry of $P_{10}$ it is obvious that all $10$ numbers are equally probable.
If I modify this construction and replace the regular 5-gon by a regular $n$-gon, I shall get a fair $2n$-sided dice with an even number of sides.
My question is whether there also exist fair $(2n+1)$-sided dice with an odd number of sides. By a fair dice I mean
- a $(2n+1)$-sided polyhedron with $2n+1$ congruent sides,
- so that the probability of landing on each of its sides is precisely $\frac{1}{2n+1}$.
Answer
A partial solution
You asked for the existence of a $(2n+1)$-sided polytope/die, so that all $2n+1$ sides are congruent and so that the probability of landing on each of these sides is precisely $1/(2n+1)$.
I do not see how to mathematically model the property that the probability of landing on each of the sides is precisely $1/(2n+1)$. This seems to need some assumptions from physics and/or mechanics. Instead, I propose a purely mathematical formulation of fairness:
Definition of fairness: A polytope is a fair die, if for any two sides there exists a symmetry of the polytope that maps one side into the other one.
This fairness definition works for the standard six-sided die (which is highly symmetric) and also for the 10-sided polytope that you describe in your puzzle. Note that the definition trivially implies congruence of any two faces.
I will prove below that for this definition of fairness, there does not exist a fair die with an odd number of sides.
Non-existence proof for an odd number of sides
(1) Suppose for the sake of contradiction that there exists a $(2n+1)$-sided polytope that is fair. Let $s$ denote the number of edges of every side. The polytope has altogether $2n+1$ sides, and every side has $s$ edges. Hence the total number of edges is $(2n+1)s/2$, as every edge is counted twice.
The polytope has $e=(2n+1)s/2$ edges and $s\ge3$ is an even integer.
(2) Consider a side of the polytope. The side has $s$ vertices, and we let $d_1\le d_2\le\cdots\le d_s$ denote the number of edges that are incident to these $s$ vertices. A vertex with $d_i$ incident edges is a vertex of $d_i$ different sides. Hence, the polytope contains exactly $(2n+1)/d_i$ vertices of this particular type. We conclude:
The polytope has $v=(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s})$ vertices.
(3) Next we use Euler's polyhedral formula that says that a polytope with $f$ sides (faces), $e$ edges, and $v$ vertices must satisfy $v+f-e=2$. By plugging in $f=2n+1$ and the expressions derived in (1) and (2), we get
$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2})=2 ~~~~~~(*)$
Since $d_i\ge3$ for all $i$, we furthermore derive
$(2n+1)(s\frac{1}{3}+1-\frac{s}{2}) ~\ge~ (2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2}) ~=~ 2 $
Since $2n+1$ is positive, also the value $1-s/6$ in the other bracket must be positive; this implies $s\le5$. Since $s\ge3$ is an even integer, we arrive at the following fact:
$s=4$
(4) Since $s=4$, the equation $(*)$ now simplifies to
$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4}-1)=2 ~~~~~~(**)$
If $d_1\ge4$, then the sum of the four reciprocals in $(**)$ would be at most $1$; a contradiction. We conclude $d_1=3$, and $(**)$ further simplifies to
$\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4} ~=~ \frac{2}{3}+\frac{2}{2n+1} ~~~~~~(***)$
This only leaves four possible cases for $d_2$ and $d_3$ (in all other cases, the sum of the three reciprocals would be at most $2/3$):
- (a) $d_2=3$ and $d_3=3$
- (b) $d_2=3$ and $d_3=4$
- (c) $d_2=3$ and $d_3=5$
- (d) $d_2=4$ and $d_3=4$
(5) It remains to do the case work.
In case (a), equation $(***)$ turns into $d_4=(2n+1)/2$; this is impossible, as $d_4$ would not be integer.
In case (b), equation $(***)$ turns into $d_4=12(2n+1)/(2n+25)$. Then the odd number $2n+25$ must divide $3(2n+1)=6n+3$. Since $6n+75=3(2n+25)$, also the difference $6n+75-(6n+3)=72$ must be a multiple of the odd number $2n+25$; a contradiction.
In case (c), equation $(***)$ turns into $d_4=15(2n+1)/(4n+32)$. Then the odd number $15(2n+1)$ must be a multiple of the even number $4n+32$; another contradiction.
In case (d), equation $(***)$ turns into $d_4=6(2n+1)/(2n+13)$. Then the odd number $2n+13$ must divide $3(2n+1)=6n+3$. Since $6n+39=3(2n+13)$, also the difference $6n+39-(6n+3)=36$ must be a multiple of the odd number $2n+13$; the final contradiction.
(6) As all possible cases have ended up in a contradiction, we conclude that there is no fair die with an odd number of sides.
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