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A few days ago,http://www.futilitycloset.com/published a short post based on the fourth problem of the 1987 Canadian Mathematical Olympiad (from on a problem from the 6th All Soviet Union Mathematical Competition in Voronezh, 1966). The problem is simple (as always). It is about water pistol duels (with an odd number of players)The answer is nice, an can be read on the blog.What puzzled me in this problem is the following: if we know, for sure, that at least one player won't get wet, we don't know exactly how many of them won't get wet (assuming that if they shoot at the closest, they hit him for sure) ? It is simple to run simulations, e.g. assuming that players are uniformly distributed over a square,NOTWET=function(n){x=runif(n)y=runif(n)(d=as.matrix(dist(cbind(x,y), method = "euclidean",upper=TRUE)))diag(d)=999999dmin=apply(d,2,which.min)notwet=n-length(table(dmin))return(notwet)}It is then rather simple to get the distribution of the number of player that did not get wet,N25=Vectorize(NOTWET)(n=rep(25,NSim))T=table(N25)plot(as.numeric(names(T)),T/NSim,type="b")The graph for different values for the total number of players is the following (based on 25,000 simulations)If we investigate further, say with 51 players, we have a distribution for the total number of players that did not get wet which looks exactly like the Gaussian distribution,NSim=25000N51=Vectorize(NOTWET)(n=rep(51,NSim))T=table(N51)plot(as.numeric(names(T)),T/NSim,type="b",col="blue")u=seq(0,51,by=.1)lines(u,dnorm(u,mean(N51),sd(N51)),col="red",lty=2)If anyone has an intuition (not to say a proof) for that, I'd be glad to hear it...