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<div>Juho wrote (23 May 2010):<br>
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<snip><br>
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<pre>><i> 1. Rank the candidates. Truncation is allowed. Equal ranking is not
</i>><i> planned for (but we could come up with something).
</i>><i> 2. Label the candidates A, B, C, ... Z in descending order of first
</i>><i> preference count.
</i>><i> 3. Let the current leader be A.
</i>><i> 4. While the current leader has a majority pairwise loss to the very
</i>><i> next candidate, set the current leader to the latter candidate. (In
</i>><i> other words step 4 must be repeated until there is no loss or no other
</i>><i> candidates.)
</i>><i> 5. Elect the current leader.
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<pre>How about this example and LNH.
6: A>C
5: B>A
2: C>B
2: C
Candidate names indicate the order in first preferences. B beats A. C
beats B. C wins.
6: A>C
5: B>A
2: C>B
2: C>A
Two "C" voters have changed their vote to "C>A". B does not beat A. A
wins. The "C" voters were harmed when they included their later
preferences.
Juho
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<br>
Juho,<br>
<br>
The key word you missed in the definition is "majority". In both your
elections there are 15 ballots, so a <br>
"majority pairwise loss" requires a winning score of at least 8.
<pre>In both cases the FPP winner A wins, in the first because B's pairwise score against
A is 7, a pairwise win but not a "majority" pairwise win (and so of course not a
"majority pairwise loss" for A).
Chris Benham
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