[r-t] Extents of minor

[Philip Saddleton]

Stephen Penney <stephen at ucalegon.com> wrote at 10:35:55 on Wed, 18 Aug 
2004
>Simple question: how many extents of minor are there?
>

A probabilistic approximation: if there are n rows and k place 
notations, there are n! permutations of the rows. The probability that a 
random pair of adjacent rows differ by a valid change is (k/(n-1)). If 
these probabilities were independent (which they are not), the number of 
extents is approx

n! x (k/(n-1)) ^ n

giving
Singles n=6, k=2: 2.9
Minimus n=24, k=4: 3.6 x 10^5
Doubles n=120, k=7: 10^51
Minor n=720, k=12: 10^466

This appears to be an overestimate for small numbers

Alternatively, there are k^n sequences of place notation. The 
probability that the ith row differs from the preceding i (including the 
initial rounds) is (n-i)/n. The nth row must be rounds, so the number of 
extents is approx

(k/n)^n x (n-1)!

giving
Singles: 0.16
Minimus: 5.4 x 10^3
Doubles: 10^49
Minor: 10^463

This appears to be an underestimate for small numbers

Maybe the answer is somewhere in between.

-- 
Regards
Philip
http://www.saddleton.freeuk.com/