These dice are used to play a gambling game: You pick the “best” of the three dice, then from the remaining two I pick the “second best.” We both roll and the high number wins. I can beat you, on average, even though you chose the better die. The surprise for nearly everyone is that there is no “best” die. Call the dice A, B, and C. If A beats B, and B beats C, it seems plausible that since A was better than B and B was better than C, A ought to be much better than C. Instead C beats A.
A) Kind of reminds me of a trick my friend Joe pulled on me where you take turns picking your best poker hand from the deck (“open hand draw poker” or some such). I said, but we’ll just both choose a royal flush and it’ll be a draw. He let me go first and it ended up just his way, though he did ask (and I laughed, demuring) if I wanted to draw any additional cards). Then he asked to pick first and chose four aces, let me pick an aceless straight flush, the. Reminded me it was draw poker, so he could go back into the deck and convert one of his aces into a royal flush. All the aces were thus consumed so I couldn’t improve on my straight flush and lost
B) really curious about these dice. Wonder how many such sets there are with 3 6-sided dice. And how many dice in the set before it becomes impossible to create this condition (first need to generalize beyond 3 dice...such that for every die there is at least one that beats it?)?
Note: stop reading the book of you want to figure out for yourself. He reveals a solution in short order. Avert thine eyes!
