# riddles on a line

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

*Given a permutation of Ð, what is the probability that the most likely variation (up or down) occurs at each term?**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.

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