Variables/Condorcet Factorial math

classic Classic list List threaded Threaded
1 message Options
Reply | Threaded
Open this post in threaded view
|

Variables/Condorcet Factorial math

Craig Carey-2
A. Variables
The recent postings show many of the math problems in elections.
Namely, there is a low to high variable in
(a) the number of voters,
(b) the number of candidates,
(c) the number of winners, and
(d) the number of rankings (in ranking methods) made by each voter (truncated
ballots).
For each of them, there is also an odd or even variable to consider
(especially for tie possibilities).
An election method that apparently works with one low number variable may
have defects with higher number variables (especially with 3 and 4 and
perhaps 5 and 6 but probably not with 7 or more). The only sure test for a
method is to do all the combinations of variables at low numbers to test the
tie breakers in the method and see if such tie breakers do or do not produce
strange results. If strange results are produced, then the tie breaker is
obviously defective.

B. Condorcet Factorial math

The general math for a total of T candidates and W (1 or more) winners in the
Condorcet method is as follows.
The first winner is one of T candidates. The second winner is one of T- 1
candidates. The third candidate is one of T- 2 candidates. And so forth.
Because of the use of combinations, the factors must be multipled together---
T x (T- 1) x (T - 2), etc.
With the Condorcet method, there is in effect a test loser to be tested
against the test winner(s). The first choice votes cast for each of the other
remaining losers are transferred to an additional choice (if any) who is a
test winner or the test loser.
Thus, the winner combination is multiplied by T - W (i.e. each test loser
coming in rotation from the group of all losers).
Factorial means a number x (number -1) x (number -2) x (number -3), etc down
to 2 x 1.
Thus, 4 factorial means 4 x 3 x 2 x 1 or 24. The word factorial is
represented by the ! symbol.
 
Thus, there are (T!/(T-W)!) x (T-W)/2 combinations which becomes T!/((T-W-1)!
x 2).

Note- there is a 2 in the denominator because if a given candidate is a test
winner or the test loser the math is duplicated.

Example 10 candidates, 3 winners and thus 7 losers
10 ! / (6 ! x 2) equals (10 x 9 x 8 x 7)/2 or 2,520 combinations to be tested
of 3 possible test winners versus 1 possible test loser of 7 possible losers.


In other words, in each test combination the voters for each of the other 6
test losers have their votes transferred to 1 of the 3 possible test winners
or the possible test loser.

In the single winner case, W equals 1, so the formula becomes  (T X (T-1))/2
combinations of pairs of candidates to be tested (i.e. head to head pairs in
the Condorcet method).
For the simple 1 winner case the number of combinations can be thought as an
expanding corner of a checkerboard (or a right angle pyramid).
The formula becomes ((T x T) - T)/2 .
The T x T is the full checkerboard, the minus T is a diagonal (where a
candidate is paired against himself/herself) and the checkerboard is divided
by 2.
Thus, on the standard 8 by 8 checkerboard with 8 candidates and 1 winner,
there would be ((8 x 8) - 8)/2 or 28 combinations (which equals (8 x 7)/2).