Consider the case where n=5. If n is 0, 1, 4 or 5 then this strategy will work. By simple binomial calculations, this will occur with probability 0.375. In the remaining 2 cases (each with prob 0.3125), the strategy only works when all three people in the majority guess there hat correctly at random. For n=5, this occurs with probability 1/8. Using Law of Total Probability to combine these we get P(Success) = 0.45.

The strategy only works more than half the time for n > 9, and for n=41 the success rate is 0.75. (I confirmed this with my own simulations).

Still, an interesting article.

]]>You need to understand the system your are looking at and the different metrics for measuring it. Science matters too. ]]>

Anyway, nice distro!

]]>my code:

test.func = function(x){

for (i in 1:length(x)) {

tt[[i]] <- x[i]; tt[[i]] = Line(tt[[i]]); tt[[i]] = Lines(list(tt[[i]] ), 'i')

tt1 = SpatialLines(list(tt[[i]]))

}

return(tt1)

}

Assuming the brackets are there to denote precedence, the first for loop should be:

for(t in (T-1):1){

print(“yes”)

…

}

That should print 9 “yes”. Hope this helps.

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