Back in Newport Beach, the die was being cast. I passed Warren’s test when I told him if the dice are numbered as A = (3, 3, 3, 3, 3, 3), B = (6, 5, 2, 2, 2, 2) and C = (4, 4, 4, 4, 1, 1) then calculations show that, on average, A beats B two-thirds of the time, B beats C five-ninths of the time, and C beats A two-thirds of the time. Other sets of nontransitive dice are possible as well. I have entertained people by marking a set of three dice like this and letting my opponent pick his die first. After trying all three dice in turn and losing each time, people are typically stumped.
Did you know you can buy non-transitive dice? https://www.amazon.com/Non-Transitive-Dice/dp/B00F4I8IW0/
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Brian
