[r-t] # methods
Philip Saddleton
pabs at cantab.net
Sun Sep 19 14:00:26 UTC 2004
Chris Poole <poole at maths.ox.ac.uk> wrote at 12:14:27 on Fri, 17 Sep 2004
>Now that we've moved up from 6 bells, here's a question - what is an
>order of magnitude estimate for the number of (regular) methods where
>the treble, say, treble bobs, on stage N?
>
>(We know the number of possible place notations on each stage...)
If P(n) is the no of possible place notations at each stage (including
the identity), the number of possibilities with the treble in
12 is P(n-2)
23 is P(1).P(n-3)
34 is P(2).P(n-4)
etc.
so, the number of possible place notations up to and excluding the
half-lead is
P(n-2)^3 x P(1).P(n-3) x [P(2).P(n-4)]^3 x ...
= [P(2).P(4)...P(n-2)]^6 x [P(1).P(3)...P(n-3)]^2
There are n/2 possibilities at the half-lead, and 2 at the lead-end (for
a regular method).
The expected proportion of lead heads that are regular is 1 in
(n-3)(n-5)...3 (including those that come round in one lead or have a
short course).
Now as Richard has pointed out, P(n) is a Fibonacci term. It can be
written as
P(n) = [T^(n+1)-(1-T)^(n+1)]/sqrt(5)
where T is the golden ratio (1+sqrt(5))/2
The second term tends to zero for large n, and ignoring this we finish
up with an approximation (I think)
T^[(2n+3)(n-2)]/5^[2n-4] x n x [(n-4)/2]!.2^[(n-4)/2]/(n-3)!
This ignores the fact that some of these will be false in the plain
course (insignificant for large n), or have multiple blows in one place
(not insignificant).
--
Regards
Philip
http://www.saddleton.freeuk.com/
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