Question

You are competing in a game with 2 other players with a 21-faces dice (labeled 1-21). All three of you gets to choose a number and then roll the dice. Whoever chose the number closest to the outcome wins. What is your strategy?

Answer 1

This is actually an interview question for IT jobs!

The game strategy changes if there can be communication or not.

If there is no communication then I would pick the expected value of the game, meaning the sum of each value times the probability of it showing up on the die (1/21) or:

`E(x) = 1*(1)/(21) +2*(1)/(21)+3*(1)/(21)+...+21*(1)/(21)`=
`(231)/(21) =11`

This mathematically minimizes the difference between my pick and the one that comes from the roll.

If there is communication, and a consensus can be reached, then the most reasonable answer would be the one that gives all of us the same possibility to win: 4, 11, 18, so each of us covers 7 numbers, and each of us would have 1/3 chances to win.

If there is communication, and a consensus cannot be reached, then I would have to calculate the difference between both their pics, if it is smaller than the expected value, and it inclines towards high numbers, for example: 10, 18, then i would pick one less than the smallest, meaning 9, because i would cover 1-9, the one picking 10 would cover 10-14 and 18 would cover 15-21. My range would be bigger. If it is smaller than the expected value, and it inclines towards low numbers, i would pick one over the highest, for example 4, 9. I would pick 10. If the difference is bigger than the expected value then i would calculate whether i would cover more picking a number in the middle of their range or to the side, depending on their pick. For example, if they pick 9, 21, I would cover more range picking 8 than 15.