lew (04/11/83)
This is regarding the problem I posed of the distribution of possible
scores obtained by the taking the sum of absolute differences between
the rank of each of 15 items and its "correct" rank.
First, there are 57 (not 56) possible scores (0 thru 112 by 2's).
Second, I think I found the number of ways to score 112. It is 15*7!*7!.
This is obtained by reducing 2*8!*7! - 7!*7!. This is obtained by
observing that 112 is scored if and only if:
{15,...,8} <-> {8,...,1} (draw a picture)
This gives one chance in 3432 of getting 112 by random guessing, which
jives with 8 scores of 112 out of 30000 tries that I got with a Monte
Carlo run.
Stay tuned for further progress.
Lew Mammel, Jr. ihuxr!lew