But plug in

(( R & R ) | R ) & R & R

into the OP code and it tends towards 5, which was how I was interpreting the expression:

( R & R | R & R ) & R

My mistake. Perl evaluates ( R & R | R & R ) & R as ( (R & R) | (R & R) ) & R and that, according to my hypothesis, results in 7 bits set on average.