Later...
-> Anyway, I'll make the change (REMing that line) and also increase T
-> to a larger value, to show up smaller variations in frequency. I bet
-> this will show that your routine still produces different frequencies
-> for the outcomes!
Did that. It turned out that the outcomes acb, bac and bca were each
about 25% more frequent than each of the other three. So I suppose these
more-frequent outcomes must each have five of the 27 branches of the
tree leading to them, giving each of them a 5/27 probabilty of
occurrence, while the other three each have four branches, and 4/27
probabilities. Of course, 3*5/27 + 3*4/27 comes to 27/27.
                              dow
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