Probability that 20 sided die beats 12 sided die with reroll

Question

Alice rolls a 12 sided die (the faces labeled 1 through 12) and Bob rolls a 20 sided die (the faces labeled 1 through 20). After seeing their roll (but not the other person's roll), each person can choose to roll again. Bob wins if his final roll is strictly greater than Alice's final roll. What is the probability that Bob wins?

I'm assuming that Alice's optimal strategy is to reroll if she gets a roll that is less than 7, and Bob's optimal strategy is to reroll if he gets less than 11. Is this correct?

Answer 1

HINTS

[more hints added, assuming downvote means my first attempt does not make it easy enough]

I am assuming that A's die is 1-12 and B's is 1-20 (not stated in the question as I write this).

An elementary point to get started. If your die has 12 sides, you have a half-chance of getting 1-6. So you might think it was best to reroll iff you got that. If you follow that strategy, you have a half chance of getting 7-12 on the first roll, and a half chance of rerolling. So you have $\frac{1}{24}$ for each of 1-6, and $\frac{3}{24}$ for each of 7-12, expected value 8.

Now consider B's strategy. If B gets 13 or better he is certain to win. If he leaves a 9 he expects to win 50% of the time. So suppose he rolls a 10 (which if he followed the analogous strategy to A would mean a reroll). It is easy to check that A's chance of leaving a 10 or better (on her assumed strategy) is $\frac{9}{24}$. In other words, if B does not reroll, he expects to win $\frac{15}{24}$ of the time. So should he reroll? It is easy to check that he should not.

But how do we avoid a kind of infinite regress? How can either A or B decide whether to make a second roll without knowing the other's strategy? Do we need some kind of minimax approach?

Note the asymmetry. With a single roll, the game would clearly favour B.

Answer 2

What I will do to solve that question :

Consider the 13 cases for Alice and 21 cases for Bob (from reroll always, to never reroll, through reroll on 1, reroll on 2 or less, reroll on 3 or less...) and compute the probability of winning for each cases (use a computer program for the $21\times13$ cases).

Then I'll compute the nash equilibrium and the minmax. If they are unique and the same, you have your best moves for Alice and Bob, if not, then it depends of what's in the head of the other player.