# [Election-Methods] YN model - simple voting model in which range optimal, others not

Abd ul-Rahman Lomax abd at lomaxdesign.com
Mon Mar 24 19:44:15 PDT 2008

At 01:27 PM 3/24/2008, Abd ul-Rahman Lomax wrote:
>So we eliminate, for the individual voter, the exact match candidate.
>We now have a choice of candidates who agree with the voter on three
>out of four issues. The problem, you will note, isn't solved on that
>web page. We could make a nice neat assumption that somehow these
>issues have been arranged in a universal sequence of importance, so
>that agreement in the first position is more important than that in
>second, and so on. The effect is that the fourth issue is dropped,
>there are now eight unique positions, each held by two candidates;
>one of these, however, was in first rank, so the other one is now the
>second choice. And, by recursion, we can determine the vote of each
>voter. I haven't done it. Once upon a time, I would have. Life moves on.

I decided to construct the rankings. For each row in the table, if we
make the assumption above about issue importance (not the general
case, to be sure, but a reasonable example), we can determine the sincere vote.

the table (monospaced) shows the order of preference for each voter
profile. Each voter votes first preference for the candidate with the
exact same profile as the voter. Then the voters rank the candidates
according to two considerations: the voters always prefer a candidate
which agrees with them on more issues. The table shows, at the top,
the number of agreements (the first column total of 4 is not shown).
Then, within the set of candidates with the same number of
agreements, the issues are considered to be presented in order of
importance, which would necessarily be minor compared to the
preference between, say, a candidate with three agreements rather
than two. If a ranked ballot allowed equal preferences to be
expressed, then these columns with the same number at the top would
be equated, but then we'd need to deal with a tie-breaking method even more.

voter count
agree:  3    3    3    3    2    2    2    2    2    2     1    1
1    1   (0)
1st  2nd  3rd  4th  5th  6th  7th  8th  9th  10th 11th 12th 13th
14th 15th 16th
0 YYYY YYYN YYNY YNYY NYYY YYNN YNYN YNNY NYYN NYNY NNYY YNNN NYNN
NNYN NNNY NNNN
4 YYYN YYYY YYNN YNYN NYYN YYNY YNYY YNNN NYYY NYNN NNYN YNNY NYNY
NNYY NNNN NNNY
4 YYNY YYNN YYYY YNNY NYNY YYYN YNNN YNYY NYNN NYYY NNNY YNYN NYYN
NNNN NNYY NNYN
1 YYNN YYNY YYYN YNNN NYNN YYYY YNNY YNYN NYNY NYYN NNNN YNYY NYYY
NNNY NNYN NNYY
4 YNYY YNYN YNNY YYYY NNYY YNNN YYYN YYNY NNYN NNNY NYYY YYNN NNNN
NYYN NYNY NYNN
1 YNYN YNYY YNNN YYYN NNYN YNNY YYYY YYNN NNYY NNNN NYYN YYNY NNNY
NYYY NYNN NYNY
1 YNNY YNNN YNYY YYNY NNNY YNYN YYNN YYYY NNNN NNYY NYNY YYYN NNYN
NYNN NYYY NYYN
1 YNNN YNNY YNYN YYNN NNNN YNYY YYNY YYYN NNNY NNYN NYNN YYYY NNYY
NYNY NYYN NYYY
4 NYYY NYYN NYNY NNYY YYYY NYNN NNYN NNNY YYYN YYNY YNYY NNNN YYNN
YNYN YNNY YNNN
1 NYYN NYYY NYNN NNYN YYYN NYNY NNYY NNNN YYYY YYNN YNYN NNNY YYNY
YNYY YNNN YNNY
1 NYNY NYNN NYYY NNNY YYNY NYYN NNNN NNYY YYNN YYYY YNNY NNYN YYYN
YNNN YNYY YNYN
1 NYNN NYNY NYYN NNNN YYNN NYYY NNNY NNYN YYNY YYYN YNNN NNYY YYYY
YNNY YNYN YNYY
1 NNYY NNYN NNNY NYYY YNYY NNNN NYYN NYNY YNYN YNNY YYYY NYNN YNNN
YYYN YYNY YYNN
1 NNYN NNYY NNNN NYYN YNYN NNNY NYYY NYNN YNYY YNNN YYYN NYNY YNNY
YYYY YYNN YYNY
1 NNNY NNNN NNYY NYNY YNNY NNYN NYNN NYYY YNNN YNYY YYNY NYYN YNYN
YYNN YYYY YYYN
5 NNNN NNNY NNYN NYNN YNNN NNYY NYNY NYYN YNNY YNYN YYNN NYYY YNYY
YYNY YYYN YYYY

Is this the correct presentation of voter rankings, given the
assumptions? (I did cross-check it by looking at the binary patterns.)

Now, to try to use IRV with this, we immediately run into a rather
nasty problem. Massive ties. Most of the tie-breaking methods don't
work; I will eliminate the first candidate on the list of tied
candidates. So this is the way the rounds go:

YYYY 0 -
YYYN 4 5 5 5 5 5 5 5 5 5  5  5  -
YYNY 4 4 5 5 5 5 5 5 5 5  5  5 10 15 20 wins
YYNN 1 1 -
YNYY 4 4 4 5 5 5 5 5 5 5  5  5  5  -
YNYN 1 1 1 -
YNNY 1 1 1 1 -
YNNN 1 1 1 1 2 2 2 2 2 -
NYYY 4 4 4 4 4 5 5 5 5 5  5  5  5  5  -
NYYN 1 1 1 1 1 -
NYNY 1 1 1 1 1 1 -
NYNN 1 1 1 1 1 1 2 2 2 2  -
NNYY 1 1 1 1 1 1 1 -
NNYN 1 1 1 1 1 1 1 2 2 2  2 -
NNNY 1 1 1 1 1 1 1 1 -
NNNN 5 5 5 5 5 5 5 5 6 8 10 11 11 11 11

YYNY wins. IRV performs better than Plurality, which elects NNNN.
Now, this used a full set of rankings. One could look at this with,
say, RCV (three ranks), or using the five ranks that allow ranking
all the three or four issue matches.

I thought it would be interesting to do this with Bucklin. Allow to
proceed until majority or to the fifth round (which is the last round
in which voters are voting for candidates they match on three issues
out of four.) In Duluth Bucklin, perhaps, these votes would be third
round votes, which allowed multiple candidates to be chosen. Notice
that if every voter votes "strategically," they get NNNN. Prisoner's
dilemma, isn't that?

YYYY 0 4 4 4 4 = 16, majority
YYYN 4 0 1 1 1 =  7
YYNY 4 1 0 1 1 =  7
YYNN 1 4 4 1 1 = 11
YNYY 4 1 1 0 1 =  7
YNYN 1 4 1 4 1 = 11
YNNY 1 1 4 4 1 = 11
YNNN 1 1 1 1 5 =  9
NYYY 4 1 1 1 0 =  7
NYYN 1 1 1 1 4 =  8
NYNY 1 1 4 1 4 = 11
NYNN 1 1 1 5 1 =  9
NNYY 1 1 1 4 4 = 11
NNYN 1 1 5 1 1 =  9
NNNY 1 5 1 1 1 =  9
NNNN 5 1 1 1 1 =  9

Bucklin does detect the issue-majority winner, the candidate who
agrees with the majority on all four issues.

These studies could be replete with errors. Handle with care. The
full ranking table is probably correct, I would not bet much on the
IRV counting. Remind me not to count an IRV election with many candidates.

If someone wants to create the Condorcet matrix....