Link Thread

[Nate the Great]

If you need a random 1/2 event, you flip a coin. For a random 1/4 event, use a tetrahedral die, 1/6 use a normal die, etc., but what if you need a random 1/3 event? How thick does a coin have to be where landing on its edge is as likely as either end face?

Matt Parker of standupmaths (and Numberphile, etc.) decides to find out.

You need a cylinder where the diameter is its height times the square root of three.

Of course the more practical solution is simply assigning 1 and 6 on a normal die to the first outcome, 2 and 5 to the second outcome, and 3 and 4 to the third outcome.

FYI, if you don't care about statistical distributions I'd bail out after ten minutes or so.