Joined: 4/20/2005
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In my old roleplaying groups, we once had a 3-sided dice. It was made out of stearin, and had 3 bent triangles mended together to form a dice. It actually worked fairly well. But to answer Warp's question, the answer is no. To quote Wikipedia: ""Rolling-pin style dice" (also called "rolling logs"[12]) are the only way to make dice with an odd number of flat faces." See Wikipedia page
Those 3 sided ones may seem like what I had in mind (it's hard to tell by those pictures), but the 5-sided ones don't have equal faces. it's 2 triangles + 3 rectangles.
The company claims they provide equal access to all faces though.
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Some guys I knew used to waste their time in this mmorpg crap. And the only kind of 3-faced die was this one: Kind of the same as the other one you posted, but shaped like a sphere.
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Bisqwit wrote:
edit: oh ok, I get the problem, it's about making a face end facing upwards... Let me think :)
edit#2: ok, I got it. I'm making a 3D for you.
Actually that's not exactly what I meant. I accept your answer as correct, according to the letter of my question (it does do what I asked), although not according to its spirit (I made some assumptions in my question which I didn't state explicitly), but I admit my fault: I should have been more precise. Ok, let me write more precisely what I meant:
Is it possible to construct a die with an odd number of faces so that each face is planar (in other words, the die is a polyhedron), each face has a number on it, and getting any of the numbers (by throwing the die normally) has equal probability?
(If it's not, why not?)
Is it possible to construct a die with an odd number of faces so that each face is planar (in other words, the die is a polyhedron), each face has a number on it, and getting any of the numbers (by throwing the die normally) has equal probability?
(If it's not, why not?)
I would still consider that not clear, because it dependent on the surface. For irregular shaped dice, it is possible to make them equal with probabilities by just tweaking some lengths, but this would be very difficult to do and prove (probably surface dependent due to bounce etc). For example consider a pyramid, you can just tweak the height until the square side has equal probabilities to the others (shorten to increase, heighten to decrease).
Geometry problem:
Is it possible to construct a die with an odd number of faces (and at least 3 of them) so that the probability of each face (when the die is thrown normally) is the same?
It is, for all odd numbers n>3, by constructing a right prism based on a regular polygon with n-2 sides of such relative dimensions that the spherical areas of the projections of each face onto the smallest sphere containing the prism are equal.
For all even numbers n>2, if 4|n the fairest way is to attach two right pyramids based on a regular polygon with n/2 sides with such relative dimensions that all vertices lie on a sphere (forming a dipyramid), otherwise the fairest way is to construct two conical collections of n/2 kites with one common vertex and two opposite vertices each shared with a neighbor and one unshared vertex with such relative dimensions that the unshared edges lie on a regular prism based on an n-sided polygon and all vertices lie on a sphere, and attach them (this is known as a trapezohedron).
The only exceptions occur when a regular or semi-regular polyhedron with congruent faces is possible: n=4, tetrahedron; n=12, dodecahedron,; n=20, icosahedron; n=24, deltoidal icositetrahedron; 30, rhombic triacontahedron. The cube and octahedron are not exceptions, because the former is in fact a triangular trapezohedron and the latter is a square dipyramid.
For the degenerate case n=3, use an elongated right triangular prism capped with tetrahedra, so that no normal from a triangular face intersects the center of mass; for n=2, use the attachment of two tetrahedra with a smaller tetrahedron removed from one vertex, at the larger tetrahedral side, with such dimensions that no normal from a quadrilateral face intersects the center of mass; for n=1, use a unistable polyhedron; and finally for n=0, use the empty set.
Interestingly, the limiting case differs based on the parity of n: The limiting case for 2n is a cone, while the limiting case for 2n+1 is a pin; it would normally be expected for the limit to be a sphere but it doesn't seem like it...
Also I read about this more than 3 years ago, and the problem had been solved for quite a while back then; I also actually bought the 5- and 7-sided dice mentioned in the article: http://www.maa.org/editorial/mathgames/mathgames_05_16_05.html
Joined: 1/16/2008
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Hmmm I think it'd be possible for any N>2 to make fair rolling-log dice with equally shaped&sized *landable* faces? (each sides of the log would be N triangles resulting in non-landable faces that form a point side)
for even N, the top side could be indicated with the result
for odd N, numbers could be written on/over edges, or the botton/landing face could be used as the result
edit: oh yeah, it's be a N-sided prism with N+1 sided pyramids attached
Joined: 11/30/2008
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I know DaTeL237 is correct.
If you were to construct an odd sided die with say, 10 001 faces, you would expect the prisim to become extremely narrow, to compensate for the small projected size of the side faces of that particular prism. As the number of faces approaches infinity, the end faces area aproaches 0. The end faces of the prism would indeed have the same area on a spherical projection, but it would seem clear to me, that a toothpick is not going to land on it's point, but on it's edge.
If we were to say that the toothpick has 10 001 faces, then we would say that this prism indeed has an appearance similar to a pin at this point. In reality, we would never expect a pin to land on it's point, but theoretically, let's consider the times when the prisim does land on it's point, however unlikely that may be. The prisim would need virtually zero rotation in order to land on it's point without tipping. and with a dice roll, the probability of this occuring would be expected to be below 2/10 001 because the act of rolling a dice practically means (almost by definition) that significant rotational force is applied. (however, we cannot discount rotation on the axis perpendicular to the end faces)
However, I don't have any proof at the moment. This was purely made by intuition. correct me if I'm wrong.
I also think it is possible to create fair prisms. If the above is incorrect, then assume that the end faces are infinitely small. Then there is literally zero chance of the object landing on it's end faces. However, make the side faces smaller and smaller in comparison to the end faces, and there is zero chance that the die landing on a side face. Somewhere in the middle, there is a happy medium where the probability of the die landing on any face is equal.
If you read further down the article, you would have found that there were indeed errors with the spherical projection model, however, to manage with the errors, there is another model, called the energy state model, which take into account the energy required to topple a die onto a more stable face. I quote:
Before the first bounce, the [spherical projection] Model can be used to determine which face of the die has current influence ... the amount of energy possessed by the die decreases geometrically with each bounce . In order for the model to work, some nth bounce must relate to an identity matrix, relating to a state where there isn’t enough energy left to shift the die from any face to any other face.
The spherical projection model does not take into account the shape of the dies, and therefore can be inaccurate. I would say that given a sphere divided into n portions with equal area, the die would be fair, but a shape that produces the same spherical projection will not necessarily be as fair.
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andymac wrote:
I also think it is possible to create fair prisms.
the article arflech linked has an explanation why dice with unequally shaped faces will never be fair... basically, it'll be influenced by the elasticity of the die and surface it lands on (e.g. a flexible surface might slightly favor one face while a less elastic surface might favor another face)
arflech wrote:
ok, well at the very least (having read the clarification by Warp) that it is not possible for 3 faces, because there is no polyhedron with fewer than 4 faces
this turns out to be true because it is possible to construct the dual of any polyhedron, in which each face is mapped to a vertex, each vertex to a face, and each edge to an edge, and the dual is also a polyhedron...but the dual of a 3-sided polyhedron would be a triangle, because three vertices determine at most one polygonal face
also I wasn't sure to what degree physical considerations would be needed, so for most cases I just was like "if you randomly rotated it, there would be an equal probability of any face being the one landed on if the polyhedron were placed gently on a horizontal surface at that rotation"
Finally, in that pic, the manufacturer of the 5-sided die made those customizations, but it is possible to make a triangular prism into a fair die.
Joined: 1/16/2008
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I just realized that if you simply want a fair die with 1 (whats the point), 2 or 3 values.... then you could just use a regular 6-sided one and re-purpose the values :) (e.g. 1&4=1, 2&5=2, 3&6=3)
then you'd have equal probability for all values AND for all faces :) it's just that face-count isnt same as value-count
You can also roll the die twice, then subtract 1 from the first roll, multiply the result with 6, and then add the second roll.
See? There's no need for 36-sided dice at all!
If you want a 30-sided one, just reroll if the first throw gets you a 6!
I've been told you can use similar magic to replace 216-sided ones as well, but I guess it's just rumors..
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How about using a coin and a 6 sided die for 7 possibilites? You roll the die and you flip the coin. Then add 1 if the coin gives heads, or 0 if it's tails. Is that fair for all values from 1 to 7?
Flip 3 coins of different types and assign each type as a digit of a three digit binary number. Treat heads as one and tails as zero. Re-flip if you get all zeroes :P
Joined: 1/16/2008
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Find a number N such that
N > 0
cross_sum(N) % 2006 = 0
cross_sum(N+1) % 2006 = 0
where cross_sum(x) means the sum of the digits (in decimal notation)
e.g. cross_sum(38421) = 3+8+4+2+1 = 18
the difference between cross_sum(N) and cross_sum(N+1) is important, it has to be a multiple of 2006.
If N doesn't end with a 9 it's:
cross_sum(N+1) = cross_sum(N) + 1
If N ends with a 9 (but the second-last digit is not 9), it's:
cross_sum(N+1) = cross_sum(N) - 9 + 1
consequently, if the last i digits are 9 (but the i+1'th isn't), we have
cross_sum(N+1) = cross_sum(N) - 9*i + 1
so we need a number i with
(9*i - 1) % 2006 = 0, for example 223*9 - 1 = 2006
Our number ends with 223 digits of 9. The cross sum of those is 2007 but 2007 % 2006 = 1. We just need to preface more digits with a cross sum of 2005. The lowest number satisfying all our properties has these digits (from left to right):
N = 1x 8, 221x9, 1x 8, 223x 9, cross_sum(N) = 4012 = 2*2006
N+1 = 1x 8, 222x9, 223x 0, cross_sum(N+1) = 2006
done.
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
259999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
should be the shortest one.
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
259999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
should be the shortest one.
You could turn "25" into "7" (that's the number I got)
Thanks, DaTeL237. This problem was fun. ^^
Assume we have a rectangular grid consisting of m vertical lines and n horizontal lines. All the vertical lines are of equal length and arranged horizontally, at equal distances from each other. Likewise all the horizontal lines are of equal length and arranged vertically, at equal distances from each other. The length of the vertical lines is the distance between the outermost horizontal lines, and vice-versa. The two sets of lines is superimposed so that the four outermost lines coincide at their endpoints, and thus they form a large rectangle (with all the other lines inside it). No line goes outside of this rectangle.
In other words, the grid contains m x n line intersections (and consequently there are (m-1) x (n-1) small empty rectangles inside the grid). For example, a standard Go board has a 19 x 19 grid (19 vertical lines and 19 horizontal lines, totaling 361 intersections).
(While the idea is quite simple, I tried to be as unambiguous as I could above, which is why the description became somewhat lengthy.)
Let's define the smallest possible grid to be 2 x 2 (because it's the smallest that can be formed with lines of non-zero length).
Such a grid forms many rectangles. A 2 x 2, rather obviously, forms only one rectangle. However, a 3 x 2 forms 3 rectangles (the two small rectangles and a third one, which is formed by the outermost lines). A 4 x 2 grid forms 6 rectangles (3 small rectangles, 2 medium-sized and 1 encompassing the whole grid). Likewise a 3 x 3 grid forms 9 rectangles and a 4 x 4 grid forms 36 rectangles. And so on.
So the task is simple: Write a function f(m,n) which tells how many rectangles can be found in an m x n grid. (Explain how you came up with the function).
(Edit: Corrected the number of rectangles in a 4x4 grid.)
Joined: 4/20/2005
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A rectangle of size a x b can be placed in m-a horiztonal positions and n-b vertical positions (this is easy to see if you draw a picture of the grid). So a total of (m-a)*(n-b) total ways to place it in the grid. In order to fit in the grid, a must be an integer between 1 and m-1. In the same way, b must be an integer between 1 and n-1. If we sum over all possible combinations of a and b we get:
Is this right? With this formula I get 36 rectangles on a 4*4 grid while you get 30, but after drawing up the grid myself, I see that you can indeed put 36 and not 30 rectangles on a 4*4 grid, so my formula might actually work.
Is this right? With this formula I get 36 rectangles on a 4*4 grid while you get 30, but after drawing up the grid myself, I see that you can indeed put 36 and not 30 rectangles on a 4*4 grid, so my formula might actually work.
