1
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ I don't know. Have you proved it? It seems highly unlikely. If B rolls 10, then he has won unless A gets a 10,11 or 12. If he rolls again he could easily worsen his position. But I have not done the calculations. $\endgroup$
    – almagest
    Commented Sep 13, 2014 at 20:52
  • $\begingroup$ You have not stated that A's die is 1-12 (on the faces) and B's 1-20 and both are fair dice. Obviously there is no hope of approaching this question if we know nothing about the dice. Or to be more accurate, the reasoning will be totally different. $\endgroup$
    – almagest
    Commented Sep 13, 2014 at 21:05
  • $\begingroup$ @almagest Fair dice should be assumed unless the problem states otherwise. $\endgroup$
    – MJD
    Commented Sep 13, 2014 at 21:20
  • $\begingroup$ @MJD. Thanks. Where do I find that? Is there a list of standard assumptions somewhere? Or are you just saying assume the most plausible way of making the question make sense? $\endgroup$
    – almagest
    Commented Sep 13, 2014 at 21:21
  • 2
    $\begingroup$ I don't know if there's some master book of background assumptions somewhere. But I am hereby informing you that when people talk about dice, they mean fair dice unless they say otherwise. $\endgroup$
    – MJD
    Commented Sep 13, 2014 at 21:25

2 Answers 2

Reset to default
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @BrianWong. If this was not enough, let me know and I will add some more detail. But I am out for the next 5 hours or so (now 7:20am London, UK). $\endgroup$
    – almagest
    Commented Sep 14, 2014 at 6:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .