<div dir="auto"><div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Thu, Jul 6, 2023, 10:46 AM Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de">km_elmet@t-online.de</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 7/6/23 19:33, Forest Simmons wrote:<br>
<br>
> This produces the following example:<br>
> 2: A>B>C<br>
> 1: B>C>A (honest is B>A>C)<br>
> 2: C>A>B<br>
> <br>
> It's an A>B>C>A cycle: A has 2 first preferences out of 5, hence >1/3<br>
> first preferences. The probabilities of a ballot having a last<br>
> preference for<br>
> A is 1/5<br>
> B is 2/5<br>
> C is 3/5<br>
> <br>
> Since these events are mutually exclusive, the probabilities should not <br>
> surpass 100 percent.<br>
<br>
You're right. I had originally written an example with more voters and I <br>
forgot to update the numerators. The right numbers are:<br>
<br>
A: 1/5<br>
B: 2/5<br>
C: 2/5<br>
<br>
which can be seen by counting last preference votes.<br>
<br>
So the penalties should be:<br>
<br>
A: (C's probability): 2/5<br>
B: (A's probability): 1/5<br>
C: (B's probability): 2/5<br>
<br>
B>A=C<br>
<br>
Does that seem right?<br></blockquote></div></div><div dir="auto"><br></div><div dir="auto">Yes.</div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
-km<br></blockquote></div></div><div dir="auto"><br></div><div dir="auto">Let's compare with the MMPO analysis:</div><div dir="auto"><br></div><div dir="auto"><span style="font-family:sans-serif;font-size:12.8px">: 2 A>B</span></div><div dir="auto">1 B>C</div><div dir="auto">2 C>A<br></div><div dir="auto"><br></div><div dir="auto">A>B 4 to 1, B>C 3 to 2, C>A 3 to 2</div><div dir="auto"><br></div><div dir="auto">MPO's are 3, 4, and 3, respectively for A, B,and C.</div><div dir="auto"><br></div><div dir="auto">So A and C are tied for MinMPO. The tie is broken in favor of A, because its second highest MPO is only 1.</div><div dir="auto"><br></div><div dir="auto">B has the fewest losing votes against this MMPO candidate, so the sincere final is between A and C ... which is won by the sincere CW, namely A.</div><div dir="auto"><br></div><div dir="auto">The new method is based on the idea of finding the candidate X that needs the least supplematation to form a pair that covers the candidate set.</div><div dir="auto"><br></div><div dir="auto">If X is A, then Y=C is not yet covered because C beats A. The cost of supplementing with C is C's antifavorite count, which is 2.</div><div dir="auto"><br></div><div dir="auto">If X=B, we need a candidate (like A) that beats B to complete our covering. The cost of adding A is only 1, its antifavorite count.</div><div dir="auto"><br></div><div dir="auto">This is clearly the least costly ... so X=B is the candidate most cheaply built up to a covering. Therefore B (the burier in this example) is the method winner.</div><div dir="auto"><br></div><div dir="auto">The burier is rewarded (frown!)</div><div dir="auto"><br></div><div dir="auto">Now suppose that sincere had been</div><div dir="auto"><br></div><div dir="auto">2 A>C</div><div dir="auto">1 B>C</div><div dir="auto">2 C>A</div><div dir="auto"><br></div><div dir="auto">Then the MMPO method would have detected & restored C as the sincere CW in the A vs C sincere final ... while our new method would fave still elected B ... punishing the burier A by electing its sincere antifavorite.</div><div dir="auto"><br></div><div dir="auto">The equality in the B>A=C cycle confounds things a little with two equally weak links B>C and C>A.</div><div dir="auto"><br></div><div dir="auto">How about the monotonicity and Landau Efficiency of the new method?</div><div dir="auto"><br></div><div dir="auto">BTW in the formula for S(X) the sum of over all Y defeating X of RBAFP(Y) could be replaced by</div><div dir="auto">Max over all Y beating X of BottomCount(Y).</div><div dir="auto"><br></div><div dir="auto">This makes it easier to explain and compute ... no need of lottery/probability concepts, etc.</div><div dir="auto"><br></div><div dir="auto">-fws</div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
</blockquote></div></div></div>