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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 $\frac{1}{2n+1}$.
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    $\begingroup$ Note that if you generalize to $P_{2n}$, die with even $n$ do have a "top" face (for example, the 8-sided octahedron, a.k.a. $P_8$). You can modify $P_{2n}$ with odd $n$ to get a top face by rotating one of the two halves by half a face; such shapes are called trapezohedrons and are how most real d10s are constructed. $\endgroup$ – 2012rcampion Mar 23 '16 at 19:46
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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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  • $\begingroup$ Fair enough... When I posted my question, I did not really know how to define a fair dice. I accept your mathematical answer. If somebody comes up with a better solution that takes physics and mechanics into account, I may reconsider and accept the other solution. $\endgroup$ – Alexis Feb 16 '15 at 16:31
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Let $n$ be an arbitrary positive integer.

Start with a right cylinder with cross section a regular $n$-gon. To each of the bases, attach a pyramid (with regular $n$-gon base). The resulting polyhedron has $n$ rectangular faces and $2n$ triangular faces. If the cylinder and pyramids are tall enough, it will be impossible for the die to land on the triangular faces (the center of gravity would not be over those faces). So, although this die has $3n$ sides (so it doesn't exactly answer your question), it is only possible to land on $n$ of them, and by symmetry all $n$ are equally likely.

This die looks a bit like a pencil that's sharpened at both ends.enter image description here

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  • $\begingroup$ Ohh, that's pretty. $\endgroup$ – No. 7892142 Feb 16 '15 at 11:29
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    $\begingroup$ By continuity, you can also show that a die exists that will land on any of the $3n$ sides. As you say, in your picture the die can only land on the long faces. If we make another with very short "long" faces, it will (almost) never land on them. Somewhere in between there is a $3n$ side die that will land on each face equally. $\endgroup$ – Ross Millikan Dec 20 '15 at 16:29
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    $\begingroup$ @RossMillikan Since it quite strongly depends on how the dice is thrown, and even on things like air pressure, dew point, local gravity and table surface, such dice can hardly ever be considered fair. $\endgroup$ – yo' Mar 23 '16 at 21:40
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I can loosely prove that such dice exist for n > 1, though actually constructing such a die might be trickier.

Consider a pyramid whose base is a regular 2n-sided polygon, with a height h. This shape has 2n+1 faces.

As h approaches 0, it becomes more and more likely that the die lands on its base, asymptotically approaching 1/2 probability, as this shape approaches a flat "coin". As h approaches infinity, it becomes less and less likely that the die lands on its base, asymptotically approaching 0 probability. I claim that the probability of landing on the base is continuous, and so there must be some value for the height such that the probability of landing on the base is 1/(2n+1), as 0 < 1/(2n+1) < 1/2. The rest of the faces then total (2n)/(2n+1) probability, and by symmetry must each also have a 1/(2n+1) probability of coming up.

A similar construction is also possible with a (2n-1)-polygon-prism instead of a (2n)-polygon-pyramid.

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  • $\begingroup$ Not completely satisfactory because one face is singled out. The probability can change depending on how you throw the die, but I don't think you can do better. $\endgroup$ – Florian F Feb 15 '15 at 23:11
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    $\begingroup$ Probably even less satisfactory because the probability of landing on the 'bottom' face is likely to be affected by factors such as the elasticity of the surface onto which it is tossed and/or the amount of angular momentum given to the die when it is thrown. $\endgroup$ – Penguino Feb 16 '15 at 0:30
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For practicality, you would just take a 2n die and treat opposing sides as equal on the lower half. For example, a 6 sided die could be used as a three sided die with two of each possibility, 1 and 6 are 1, 2 and 5 are 2, and 3 and 4 are 3. Die faces are properly labeled so that opposing sides total to the same amount (7 on a 6 sided die) and if you're using a D20 (opposing sides total 21) to roll for a D10 since you lost it under the couch or the roommate's cat ate it, you would use the rolled number if 10 or less or subtract it from 21 if 11 or more. Or you could alternately just relabel the sides before using it.

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  • $\begingroup$ 1 and 6 are 1, 2 and 5 are 2, and 3 and 4 are 3. Eww, just halve the top face and round up. $\endgroup$ – Jacob Raihle Dec 19 '15 at 22:00
  • $\begingroup$ @JacobRaihle From my experience, that much more confusing. What I do to obtain a 3-sided dice is that I erase (or colour in black or fill or whatever) some of the dots, namely changing 6 to 1 (keeping one of the middle dots), 5 to 3 (keeping 3 dots in a line) and 4 to 2 (keeping two diagonal dots). This works wonders. $\endgroup$ – yo' Mar 23 '16 at 21:32
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Geometrically, it is the TOP face that is the value, and a corresponding BOTTOM face that rests on a surface (a table). A die consists of numbered faces each of the same size. Since each side must have a corresponding face on the opposite sides, there is no circumstance where an odd number would not end with a vertex on top (unless somehow the die rested on its edge.) Instead, use the suggestion above and use an even number of faces that are labeled the same, i.e the top and bottom always are the same label. In this fashion you can make "fair dice" for any value >2

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A similar solution to the one provided by Julien Rosen is when your sides are mathematically speaking not sides, because your dice is not a polytope. Rather, it has $n$ smooth surfaces as sides, with $n$ smooth curves as edges and $2$ vertices, and it looks a bit like a rugby ball or a straight banana. This is a fair dice that can (in theory) be made with any number of faces.

enter image description here

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    $\begingroup$ The question defines dice as being polyhedrons. $\endgroup$ – f'' Mar 23 '16 at 23:05
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Make an n-1 sided (even) dice as per your description and flatten the top and bottom. Label the top and bottom with "n". By continuity, there is a cutoff where landing on the ends matches the probability of landing on the sides. This would be hard to agree this die is fair.

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