| I don't quite know where to begin with a solution, so I'll articulate the problem.
"A right cylinder of uniform density has a radius, *r*, and a height, or thickness, *h*. If *r* is sufficiently larger than *h*, as with a coin, and the object is tossed about, the probability of it coming to rest on a base are very high. If *r* is relatively small, as with a caber, it's likely to come to rest on its lateral face.
If *r* is 1, what is the value of *h* such that, if the object is tossed and allowed to come to rest on a face, the probability of it resting on its lateral face are 0.333?"
Now, I didn't say "land on its lateral face" because I understand that an object could land on one face and then fall / roll into another, if that second face drops the object's center of gravity, and if there's sufficient momentum to raise the center of gravity due to the rolling itself.
The only value for *h* that doesn't cause any rolling at all would be *h* = 2, where the side-view of the cylinder is resembles a square. But that seems too large a value. The probability of such an object landing on its lateral face seems much higher than 0.333.
As i say, I'm stumped. |

physicsguyAlso:

http://www.google.com/search?hl=en&c2coff=1&q=%22Probability+of+a+Tossed+Coin+Landin

serpentgodI guess the answer depends on what you're really looking for - a solution to a math problem, or a solution to the "true" d3. If it's the former, then the PDF is quite thorough.

If it's the latter, though, I think applied physics make the notion of a cylindrical d3 pretty much moot. The math in the PDF works fine in a vacuum, but as a page late in the document shows, the materials of the die, the surface you're throwing the die onto, and even HOW you're throwing the die can have an impact on the probability. The probability of a die landing on faces of two different shapes is unlikely to be the same under all conditions.

Whenever I've seen a die that was an irregularly-shaped polygon, the only faces which -counted- were still the same shape as each other. For example, I've seen cylindrical dice - but a Cd6 would have a hexagonal cross-section, and the ends wouldn't count.

The only way to get a "pure" d3 - short of making it a Cd3, like a prism - would be to make it a three-sided regular polyhedron. That's a topological challenge, though, not a geometric one.

serpentgodcouldhave a three-sided object, but any variation would basically resemble a sharpened prism.ratofthelabAnyway, a quick google for "3 sided coin" points to this article which suggests the maths would give h = 0.707r, while their practical 10,000 roll test seemed to suggest that h = r was the right option.

In terms of D3's the easiest way to do it in "a game sense" would be to roll a D6, divide by 2 and in the case of a fraction, round up to the next integer value (ie, 1 or 2 => 1; 3 or 4 => 2; 5 or 6 => 3) Or if you're really dead set on a fair D3, I believe some dice manufacturers make prism-like dice with a rounded edge too (another quick search result here)

(And oops, originally posted that as a comment to your mathematics community post ;).. figured I'd move it here where I originally intended to post it.. apologies)

flamingophoenixakitromAs it turns out something as simple as the "bounciness" of the surface you're rolling on, makes a big difference,

ratofthelab