I think I posted a discussion of this game in another Thread, perhaps in the Chit Chat Forum before there was a Maths Forum.

Consider three dice, each of which has the same number on opposite faces.
  • Die A has 1, 6, 8 (twice each).
  • Die B has 3, 5, 7 (twice each).
  • Die C has 2, 4, 9 (twice each).
Now consider playing a game using the above three dice. Each of three players uses one of the above three dice. For conveneince, call the players Albert, Bob, & Charlie using die A, B, & C respectively.

A play consists of each player rolling his die once. If the number rolled by Albert is higher than that rolled by Bob, Bob pays Albert $1.00, and vice versa. Similarly: Albert pays or gets paid by Charlie; And Bob pays or gets paid by Charlie.

Note that a single roll cannot result in a tie. From the point of view of one player (say Albert), there are three possibilites.
  • He loses to both Bob & Charlie and pays $2.00
  • He wins from both and gets paid $2.00
  • He wins from one and loses to the other player, breaking even.
I hope the above provides an adequate description of the game.

Now, calculate the following probablilites.
  • P(Albert winning from Bob).
  • P(Bob winning from Charlie)
  • P(Charlie winning from Albert)