Four strange dice
Four six-sided dice have these faces: Your opponent picks one die, then you pick one of the other three. You both roll, and the higher number wins (no ties are possible between different dice). Both of you choose as well as possible.
What is your probability of winning?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Compute the probability that A beats B, that B beats C, that C beats D and that D beats A.
Whatever die your opponent picks, is there always a die that beats it?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. These dice form a cycle: A beats B, B beats C, C beats D and D beats A, each with probability . So whatever your opponent picks, you can pick the die that beats it.
Step 1: the cycle.
- A beats B when A shows 4: .
- B beats C when C shows 2: .
- C beats D: C shows 6 (, wins always) or C shows 2 and D shows 1 (): .
- D beats A: D shows 5 (, wins always) or D shows 1 and A shows 0 (): .
Step 2: your reply. If your opponent takes B, take A; C, take B; D, take C; A, take D. You win with probability every time.
Step 3: can you do better? No die beats any other with more than here, and your opponent can pick any die, for example B, against which the best probability is (A beats B with ; C beats B with ; D beats B with ). So the answer is .
The lesson. “Beats” is not transitive. Expected values (A: , B: 3, C: , D: 3) do not decide which die wins head to head, and in a game like this the player who chooses second has the edge.