- #1

Master1022

- 611

- 117

- 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.

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.

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.