Probability: Dice Game between 20- and 12-sided dice with re-rolls

[Master1022]

Homework Statement

One person has a 12 sided die and the other has a 20 sided die. They each get two rolls and they can each chose to stop rolling on either one of the rolls, taking the number on that roll. Whoever has the higher number wins, with the tie going to the person with the 12 sided die. What is the probability that the person with the 20 sided die wins this game? Assume the players cannot see the others' roll.

Relevant Equations

Probability

Hi,

I was reading around and found this problem. I have seen some discussion about the solution (but nothing verified) with some disagreement.

Problem: One person has a 12 sided die and the other has a 20 sided die. They each get two rolls and they can each chose to stop rolling on either one of the rolls, taking the number on that roll. Whoever has the higher number wins, with the tie going to the person with the 12 sided die. What is the probability that the person with the 20 sided die wins this game? Assume the players cannot see the others' roll. (note this is asked as an interview question, so resources available are limited)

My question: does the re-roll matter in terms of calculating the answer? Some people seem to think that: "Then, part of the trick is realizing that the second roll doesn't matter. Whatever the strategy is for the second roll, both parties will use it and therefore, their chances of winnings are the same. "

If that is the case, I can understand a way to get to the solution. Let $W$ represent the event that the person with the 20-sided die wins the game. Then,

$$ P(W) = P(W|\text{20-side die} \leq 12)\cdot P(\text{20-side die} \leq 12) + P(W|\text{20-side die} \geq 13)\cdot P(13 \leq \text{20-side die} \leq 20) $$

where $P(W|\text{20-side die} \geq 13) = 1$ and $P(W|\text{20-side die} \leq 12) = \frac{\frac{144 - 12}{2}}{144} = \frac{11}{24}$ and these can lead to the answer. However, I don't get that assumption about the second roll.

Any help would be greatly appreciated.

 

[PeroK]

I think you should be able to do a lot more analysis than this. I don't see the argument about the second roll being irrelevant. Consider, for example, a 20-sided die against a two-sided die. The second roll is of little use to 2-sided die, but much more use to the 20-sided die, who can practically guarantee a win by re-rolling whenever he gets a 1 or 2.

So, that argument doesn't hold up in my view.

Next, the odds should be easy to calculate for one roll.

Finally, each player must choose a number below which they will re-roll. The odds will vary according to these two decisions, so I imagine we are looking for an optimum strategy for both players.

You could calculate (or simulate using Python) all the possible strategies for both players and see what you get.

 

[haruspex]

"Then, part of the trick is realizing that the second roll doesn't matter. Whatever the strategy is for the second roll, both parties will use it and therefore, their chances of winnings are the same. "

The better trick is not to trust opinions of random commentators online.