Competition problem — Combinatorics

Library / /1320 of 1394

Combinatorics Difficulty 6.1 National Olympiad Prove it United States

Problem:

Consider a three-person game involving the following three types of fair six-sided dice.
- Dice of type $A$ have faces labelled $2,2,4,4,9,9$ .
- Dice of type $B$ have faces labelled $1,1,6,6,8,8$ .
- Dice of type $C$ have faces labelled $3,3,5,5,7,7$ .
All three players simultaneously choose a die (more than one person can choose the same type of die, and the players don't know one another's choices) and roll it. Then the score of a player $P$ is the number of players whose roll is less than $P$ 's roll (and hence is either $0$ , $1$ , or $2$ ). Assuming all three players play optimally, what is the expected score of a particular player?

Solution

Solution:

Short version: third player doesn't matter; against $1$ opponent, by symmetry, you'd both play the same strategy. Type $A$ beats $B$ , $B$ beats $C$ , and $C$ beats $A$ all with probability $5/9$ . It can be determined that choosing each die with probability $1/3$ is the best strategy. Then, whatever you pick, there is a $1/3$ of dominating, a $1/3$ chance of getting dominated, and a $1/3$ chance of picking the same die (which gives a $1/3 \cdot 2/3 + 1/3 \cdot 1/3 = 1/3$ chance of rolling a higher number). Fix your selection; then the expected payout is then $1/3 \cdot 5/9 + 1/3 \cdot 4/9 + 1/3 \cdot 1/3 = 1/3 + 1/9 = 4/9$ . Against $2$ players, your EV is just $E(p1) + E(p2) = 2E(p1) = 8/9$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.