As Oleg and Titus imply, one idea is to change the test so that it allows for things that differ by a small amount to be treated as roughly equal. Another thing that sometimes works is to refine your answer by another application of mean. This works in some cases but doesn't change the fundamental fact that two theoretically identical things may be slightly different when represented by double precision on a computer. Anyway, here's what I had in mind:
>> x = repmat(0.86,1e5,1);
>> fprintf('%25.18g\n',x(1))
0.85999999999999999
>> m = mean(x); fprintf('%25.18g\n',m)
0.86000000000030574
>> m = m + mean(x-m); fprintf('%25.18g\n',m)
0.85999999999999999
