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The Riddler of this week had a quick riddle: if one does want to avoid parallel parking a car over a six spot street, either the first spot is available or two consecutive spots are free. What is the probability this happens with 4 other cars already parked (at random)?

While a derivation by combinatorics easily returns 9/15 as the probability to fail to park, a straightforward R code does as well

l=0 for(t in 1:1e6){ k=sort(sample(0:5,4)) l=l+1*(!!k[1]|3%in%diff(k)|!k[4]%%3)}

since

> round(choose(6,2)*F/1e6) [1] 10

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