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I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.

\left( \sum_{i=1}^m x_i/m + \sum_{i=m+1}^{2m} x_i/m \right) / 2 = \sum_{i=1}^{2m} x_i /(2m)

I'd add that sampling is still very useful in contexts when the partition function is known or efficient to calculate. What's generally interesting is the general character of the …