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On 1/30/11 2:39 PM, Paul Kislanko wrote:
<blockquote cite="mid:66C06D5741FC45FD816CBD5334FB6F7A@jkis01"
type="cite">
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<div dir="ltr" align="left"><span class="493153419-30012011"><font
color="#0000ff" size="2">Strike my previous reply... Didn't
notice that #6 pairwise beat #1, but pairwise lost to #2-#5.</font></span></div>
<div dir="ltr" align="left"><span class="493153419-30012011"></span> </div>
<div dir="ltr" align="left"><span class="493153419-30012011"><font
color="#0000ff" size="2">Here's a case where I'd actually
like to see instead of the pairwise matrix the matrix that
shows counts of votes for #1, #2, ... #5. In particular,
which is the Bucklin winner?</font></span></div>
<div dir="ltr" align="left"><span class="493153419-30012011"></span> </div>
<div dir="ltr" align="left"><span class="493153419-30012011"><font
color="#0000ff" size="2">#6 loses or ties with every
alternative except #1.</font></span></div>
</blockquote>
I've attached the ballots. Note that there were actually 15
candidates in this election; I just showed the action for the top 6
in my earlier mail. Each ballot is one row, and position x shows
the ranking that the voter assigned to candidate x. The listing uses
the unranked numbers for the candidates, so the top 6 candidates are
not candidates 1-6. There is also one more ballot in the listing
below than for the matrix I sent earlier, but the same relatively
interesting situation still pertains.<br>
<br>
-- Andrew
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