[EM] counts of possible rank-order votes

[robert bristow-johnson]

On Feb 3, 2010, at 4:31 PM, Warren Smith wrote:

> as RBJ (and I earlier, and probably many others earlier) proved,
> the number of possible rank-order votes
> among C candidates is
> C!
> if rank-equality forbidden and all candidates must be ranked.

actually Warren, i never attempted to prove that.  and i didn't get  
at all how you came to say that until Kristofer told us that you were  
assuming "Australia rules" that requires every candidate to be ranked  
(a rule that didn't occur to me since that's not how IRV works  
here).  but given that rule and the no-equal-rank rule, i can quickly  
see that C! is correct.

the thing that i didn't expect, and was so dubious of, was that this  
closed form:

      C
     SUM{ C!/n! }  =  floor(e C!)
     n=0

is *exactly* correct.  *that* i did not expect, but was later able to  
see that it's true.


> ...The underlying reason for that is basically that the unranked
> pseudolevel can be regarded as just another ranking level...

well, i think that was the earlier argument i was having with Kathy  
and Abd ul.  maybe not, maybe it's something different.

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."