riddles on a line

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

In the Riddler of August 18, two riddles connected with the integer set Ð={2,3,…,10}:

1. Given a permutation of Ð, what is the probability that the most likely variation (up or down) occurs at each term?
2. Given three players choosing three different integers in Ð sequentially, and rewards in Ð allocated to the closest of the three players (with splits in case of equal distance), what is the reward for each given an optimal strategy?

For the first question, a simple code returns 0.17…

winofail<-function(permz){ 
  if (length(permz)>1){
    lenoperm=length(permz[-1])/2
    win=(permz[1]<permz[2])*(sum(permz[-1]>permz[1])>lenoperm)+
     (permz[1]>permz[2])*(sum(permz[-1]<permz[1])>lenoperm)+
      (runif(1)<.5)*(sum(permz[-1]>permz[1])==lenoperm)
    win=win&&winofail(permz[-1])
  }else win=TRUE
  return(win)}

and for the second question, a quick recursion produces 17, 18, 17 as the rewards

gainz<-function(seqz){
  difz=t(abs(outer(2:10,seqz,"-")))
  cloz=apply(difz,2,rank)
  return(apply((2:10)*((cloz==1)+.5*(cloz==1.5),1,sum))}

timeline<-function(prev){ 
  if (length(prev)==0){ 
   sol=timeline(10);bez=gainz(sol)[1] 
   for (i in 2:9){ 
    bol=timeline(i);comp=gainz(bol)[1] 
    if (comp>bez){
        bez=comp;sol=bol}}}
  if (length(prev)==1){
    bez=-10
    for (i in (2:10)[-(prev-1)]){
      bol=timeline(c(prev,i))
      comp=gainz(bol)[2]
      if (comp>bez){
        bez=comp;sol=bol}}}
  if (length(prev)==2){
    bez=-10
    for (i in (2:10)[-(prev-1)]){
      bol=c(prev,i)
      comp=gainz(bol)[3]
      if (comp>bez){
        bez=comp;sol=bol}}}
  return(sol)}

Thanks for visiting r-craft.org
This article is originally published at https://xianblog.wordpress.com
Please visit source website for post related comments.