[EM] Re: PR Condorcet algorithm implemented as experiment

[Andrew Myers]

David Gamble writes:
> Andrew, could you provide a worked example of your method with say 4 
> candidates for 2 seats.
> 
> Secondly are there any circumstances electing two candidates from four with 
> the vote set:
> 
> 34 A>B>C>D
> 23 B>C>D>A
> 22 C>D>B>A
> 21 D>B>C>A
> 
> in which A is not one of the two winners.
> 
> Under the two other Condorcet PR methods I'm familiar with ( Tideman's 
> cpo-stv and Sequential STV) if any candidate has a Droop quota of 1st preference 
> votes that candidate must win.
> 
> The reason I ask this is that when I voted in your ice cream PR poll ( for 3 
> winners) I gave my highest cardinal rating (999) to cherry chocolate chip. My 
> vote was the fourth vote cast in the poll and therefore represented a Droop 
> quota.
> 
> Cherry chocolate chip was not one of the three winning flavours.
> 
> The only way you get such a result using cpo-stv or Sequential STV  would be 
> if each of the 4 voters had given a different flavour as their first choice 
> and there was a 4 way tie which was resolved by selecting 3 of the 4 first 
> choice flavours at random.

David,

In the case of the election you refer to, you gave the two choices of chocolate
and chocolate chip both weights (997,998) nearly as high as cherry chocolate
chip. Because of the weights you assigned, the voting algorithm thinks you are
more satisfied by getting these two representatives instead of the single
higher ranked representative. If you would rather have cherry chocolate than
any other combination of choices, you should give it a weight larger than the
sum of the other choices. Here is the full breakdown:

These are the weights given by the four voters. I have truncated them
to only include choices (0-7) because that's enough to explain what is
going on. You were voter C, obviously (I trust your ice cream preferences
are not a deep secret).

A:  1, 4, 0,  0,  2,  3,  1,  1
B: 30,40,31, 35, 57, 78, 80, 70
C:  0, 0, 0,800,999,997,996,998
D: 25,13, 3,  7,  9, 15,  8, 11

Winning set: 1,5,7

The first choices of the various voters are 0,1,6, and 4, so it may be
interesting to compare the set 1,5,7 to 0,1,6, to 1,4,6, and to 0,1,4.
You might also be interested to see what happens with 4,5,7, in which
your number 1 choice is picked.

		f-pref (# candidates considered)
A weight	1	2	3
---------------------------------
1,5,7   	4	7	8
0,1,6		4	5	6
0,1,4		4	5	6
1,4,6		4	6	7
4,5,7		3	5	6

B weight	1	2	3
---------------------------------
1,5,7   	78	148	188
0,1,6		80	120	150
0,1,4		57	97	127
1,4,6		80	137	167
4,5,7		78	148	205

C weight	1	2	3
---------------------------------
1,5,7   	998	1995	1995
0,1,6		996	996	996
0,1,4		800	800	800
1,4,6		999	1995	1995
4,5,7		999	1996	2994

D weight	1	2	3
---------------------------------
1,5,7   	15	28	39
0,1,6		25	38	44
0,1,4		25	38	47
1,4,6		13	22	30
4,5,7		15	26	35


Now, consider 1,5,7 vs. 0,1,6.
A prefers 1,5,7 at 2 and 3
B prefers 0,1,6 at 1 but 1,5,7 at 2 and 3
C prefers 1,5,7 everywhere
D prefers 0,1,6 everywhere
A,B,C like 1,5,7 better when considering all 3 choices (and they get
to consider all 3 choices because there are 3 of them), so D 'loses'.
(On the other hand, he likes 1,5,7 more than a lot of other possible
results.)
1,5,7 wins.

1,5,7 vs. 0,1,4.
----------------
A prefers 1,5,7 at 2 and 3.
B prefers 1,5,7 everywhere
C prefers 1,5,7 everywhere
D prefers 0,1,4.
Again D 'loses' to A,B,C.
1,5,7 wins.

1,5,7 vs. 1,4,6.
----------------
A prefers 1,5,7 at 2 and 3.
B prefers 1,4,6 at 1, but 1,5,7 at 2 and 3.
C prefers 1,4,6 at 1, is indifferent at 2 and 3
D prefers 1,5,7 everywhere.
Here everyone is in agreement when all candidates are considered.
1,5,7 wins.

1,5,7 vs. 4,5,7.
----------------
A prefers 1,5,7 everywhere.
B prefers 4,5,7 at 3 but is indifferent at 1 and 2.
C prefers 4,5,7 eveywhere.
D prefers 1,5,7 at 2 and 3 but is indifferent at 1.
So A and D like 1,5,7 at 2, which is valid because A and D are half the voters.
B's preference for 4,5,7 only emerges at 3, so it is invalid.
So it looks like A and D support 1,5,7; B is indifferent; C supports 4,5,7.
1,5,7 wins.

Hope this helps (and that I didn't make too many mistakes.)
I think the result of the election is reasonable even if it
doesn't agree with CPO-STV. It looks like D is 'losing' a lot,
but consider that the selected choices are his second, third,
and fourth choice. C's second and fourth choices are
picked. A's first and second choices are picked. B's second
and third choices are picked. It might be fair to say that
this algorithm does relatively good job at finding consensus.

-- Andrew