Qua locus

Keeping both eyes on the long game.

A hyper-rational race to 2nd place?

This blog has been split in three and all content moved to the new blogs – Qua Locus LifeQua Locus Tech and Qua Locus Puzzle – comments are disabled here.

Five hyper-rationalists have been forced to pick a number between 1 and 100. The individual (or individuals) who pick the 2nd highest number is allowed to live, all others are executed by some capricious demon. What number do the hyper-rationalists pick and how many survive?

NB: Me and some friends spent 10 minutes discussing this and could not agree on a solution, is there one?

Written by CGF

February 10, 02013 at 18:25

Posted in Puzzles

Tagged with

4 Responses

Subscribe to comments with RSS.

  1. I don’t understand why they would not all choose 100, because they will all survive. Perhaps the clause should be that if there is no second highest, then all of them die.

    Akshat Rathi

    February 10, 02013 at 18:30

    • I agree that if all pick 100 then they are all 1st equal and all of them would die. Therefore, no one would ever pick 100, so everyone would pick 99, and they would all die, and so forth all the way down to 1, whereupon they all die. So, since they know this, how does this change their picks.

      Thought from Alex (http://alexflint.weebly.com/) – maybe there are no Nash equilibrium here so the answer is indeterminate?

      Christo Fogelberg

      February 10, 02013 at 18:36

      • Hmmm…. under the rule that if they all pick the same number then they all die, then I think the Nash equilibria are actually (?,1,1,1,1), (1,?,1,1,1), (1,1,?,1,1), (1,1,1,?,1), (1,1,1,1,?). That is, any situation where 4 players choose 1 (and the remaining player picks anything) is a Nash equilibrium. In this situation the four players that choose 1 have nothing to gain by changing their decision (assuming the other players’ choices are fixed), and the fifth player will die no matter what he chooses (assuming the other players’ choices are fixed), so he has no reason to change his decision either.

        Alex Flint

        March 10, 02013 at 21:10

      • Hmmm, interesting analysis!

        It seems to me that if a HR values his own life at x or more of his peers’ (where x > 4) then any permutation of (1, 1, 1, 1, ?) would be a Nash equilibrium. Since each HR knows this though, and since each HR also knows that whoever guesses ‘?’ will die, I think the only Nash equilibrium that would occur in practice would be (1, 1, 1, 1, 1). I.e. all 5 HR would always die.

        However, what if a HR valued his own life less than 4x? In that case he can’t guarantee that he can save his peers, but by guessing 100 he maximises the chance that one or more of them could live, since however many guess the closest value to 100 will also live. Of course, they all think this, and they all know that they all think this, so in this case they would fatalistically all go to their death guessing 100.

        Finally, what if a HR valued his own life at exactly 4x a peer’s life? In that case he is indifferent to his own survival versus the survival of all of his peers and he knows that his peers feel the same way. If he could influence a single peer to sacrifice themselves he would. One way would be for the HR to collude and draw straws, with all but the short straw guessing 1. In that case, if each HR is truly indifferent and if changing your mind takes energy, 4 would survive and 1 would perish. I’m not sure what would happen if the HR could not communicate though. Randomness?

        Christo Fogelberg

        March 11, 02013 at 21:41

Comments are closed.

%d bloggers like this: