jsiehler

This is a (perhaps) interesting little problem I haven't yet had time to tinker with substantially.
Suppose you draw a set of 3 numbers from {1,...,n} and record the maximum number.
Depending on n, you _may_ be able to partition the outcomes {3,...,n} into two disjoint subsets A and B, each with probability 1/2.

For example, with n=6, use (A,B)=({6},{3,4,5})

With n=8, use (A,B)=({3,5,8},{4,6,7}).

With n=7, there is no such partition.

n=9 and 10 are possible; n=11 is not.

For which n is / is there not such a partition?