On 9/20/2012 9:56 AM, Michael Kleber wrote:
On Thu, Sep 20, 2012 at 11:30 AM, Allan Wechsler<acwacw@gmail.com> wrote:
If we relax the constraint that all the dice have the same number of sides, then the two-person problem can be solved with only three "faces". One person "rolls" a device that always yields 2; the other has a coin with 1 and 3 on its faces. Does the three-person problem really need 18 faces?
Great question. If it could be done with all dice having #faces in {1,2,3,6}, then it could be expressed as a 3d6 solution (with appropriately-clustered numbers), and if my computation of the eleven 3d6 solutions is right, then none of them fit the bill. But this doesn't rule out e.g. something using a d4.
--Michael
Or just one die with the faces labelled by the six permutations of Al, Bob, and Charlie (or whoever the three players are). Brent