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The Riddler of today has a secretary problem, where one measures sequentially N random variables until one deems the current variable to be the largest of the whole sample. The classical secretary problem has a counter-intuitive solution where one first measures N/e random variables without taking any decision and then and only then picks the first next outcome larger than the largest in the first group. The added information in the current riddle is that the distribution of those iid random variables is set to be uniform on {1,…,M}, which begs for a modification in the algorithm. As for instance when observing M on the current draw.

The approach I devised is clearly suboptimal, as I decided to pick the currently observed value if the (conditional) probability it is the largest is larger than the probability subsequent draws. This translates into the following R code:

M=100 #maximum value
N=10  #total number of draws
hyprob=function(m){
# m is sequence of draws so far
n=length(m);mmax=max(m)
if ((m[n]<mmax)||(mmax-n<N-n)){prob=0
}else{
prob=prod(sort((1:mmax)[-m],dec=TRUE)
[1:(N-n)]/((M-n):(M-N+1))}
return(prob)}

decision=function(draz=sample(1:M,N)){
i=0
keepgoin=TRUE
while ((keepgoin)&(i<N)){
i=i+1
keepgoin=(hyprob(draz[1:i])<0.5)}
return(c(i,draz[i],(draz[i]<max(draz))))}


which produces a winning rate of around 62% when N=10 and M=100, hence much better than the expected performances of the secretary algorithm, with a winning frequency of 1/e.

Filed under: Kids, R Tagged: mathematical puzzle, R, secretary problem, stopping rule, The Riddler

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