A d30 against a d20 with a reroll
Alice rolls a fair 30-sided die and Bob a fair 20-sided die, at the same time. Without seeing Alice's roll, Bob may reroll his die once and must then keep the second result. Ties go to Bob: Alice wins only with a strictly larger number.
Bob plays to maximise his chance of winning. What is the probability that Alice wins?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
With final number , how likely is Bob to win? Notice the form of this function.
Bob wins with probability , which is linear in . So he should simply maximise his expected number.
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Bob wins when Alice's roll is at most , which has probability because . That is linear in , so maximising the chance of winning is the same as maximising the expected value of his final number.
Step 1: Bob's rule. A reroll is worth on average. So Bob keeps 11 to 20 and rerolls 1 to 10.
Step 2: his expected number.
Step 3. Bob wins with probability , so Alice wins with probability
When this breaks. If Bob had a 40-sided die, his winning chance would be , no longer linear: a 35 is no better than a 30, and the rule “maximise the expected number” is no longer right.