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)}

(but the analytic solution exhibits a cool Pascal triangular relation!) and for the second question, a quick recursion or dynamic programming produces 20, 19, 15 as the rewards (and 5, 9, 8 as the locations)

gainz<-function(seqz){
  difz=t(abs(outer(2:10,seqz,"-")))
  cloz=apply(difz,2,rank)
  return(apply(rbind(2:10,2:10,2:10)*
   ((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)}

After reading the solution on the Riddler, I realised I had misunderstood the line as starting at 2 when it was actually starting from 1. Correcting for this leads to the same 5, 9, 8 locations of picks, with rewards of 21, 19, 15.