asymmetric information

This article is originally published at https://xianblog.wordpress.com

The Riddler of 16 October had the following puzzle:

Take a real number θ uniformly distributed over (0,100). Among three players, the winner is whoever guessed the closest price without going over θ. In the event all guesses exceeded θ, the contestant with the lowest (and therefore closest) guess is declared the winner. The second player knows the first player’s guess and the third player knows both other guesses. What is the optimal guess for the first player, assuming all players maximise their probability of winning?

Looking at the optimal solution z for the third player leads to six possible choices, depending on the connection between the other guesses, x and y. Which translates in the R code

topz=function(x,y){
  if((2*y>=x)&(y>=1-x))  z=y-.001
  if(max(4*y,1+y)<=2*x)  z=y+.001
  if((2*x<=1+y)&(x<=1-y))z=x+.001
  z}
  
third=function(x,y) ifelse(y<x,topz(x,y),topz(y,x))

For there, the optimal choice y for the second player follows and happens on a boundary of one of the six regions, which itself returns that the optimal choice for the first player is x=2/3, leading to equal chances of winning (although there is some uncertainty on the boundaries). It is thus feasible to beat the asymmetric information. The picture above was my attempt at representing the probabilities of gain for all three players, some of the six regions being clearly visible, with first axis being x and second being y [and z is one of x⁻,x⁺,y⁻,y⁺]. The R code is too pedestrian to be reproduced!

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This article is originally published at https://xianblog.wordpress.com
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