<div dir="ltr">Yes, by that definition the Maximal Lottery fails.<br><br>The intuition for why it's a much smaller issue here is that when the voter adds their ballot, they are transferring probability from their top-ranked candidate to their second-ranked candidate. However, their third, 4th, etc. ranked candidates also lose probability in the winning lottery to their 2nd, which cancels out, such that the voter is still happier with the result than they would have been if they had not voted.<br><br>Another intuition is that the algorithm is switches from being forced to compromise between several irreconcilable alternatives by randomizing and over to being able to choose a single "consensus pick".</div><br><div class="gmail_quote gmail_quote_container"><div dir="ltr" class="gmail_attr">On Wed, Jun 25, 2025 at 11:09 AM Daniel Kirslis <<a href="mailto:dankirslis@gmail.com">dankirslis@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr">I have always seen the participation criterion defined as 'Adding votes that rank candidate A over B will not make it more likely for candidate B to win over candidate A'. The proof that Markus shared appears to establish that any Condorcet method will fail this criterion. It is not dependent on how the lotteries are defined over the Smith set, because it hinges on the fact that when a Condorcet winner exists, their probability of winning is 100%. It constructs an election with a cyclic tie between 4 candidates. Then, it shows that the tie can be broken to create a Condorcet winner by adding votes that rank that Condorcet winner second, thus moving those votes' first place candidate from a non-zero probability of winning to a zero probability of winning. </div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Jun 25, 2025 at 12:57 PM Closed Limelike Curves via Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.electorama.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div><div dir="auto">Markus—different generalizations/definitions of no-show (equivalent in the deterministic case) yield different results in when you allow lotteries. I'd have to double-check which is satisfied for Maximal Lotteries, but the most common are either:</div></div><div dir="auto">1. Turning out to vote will always yield a better lottery than not turning out, or</div><div dir="auto">2. Turning out to vote will probably improve the outcome for you, i.e. if you do a random draw from the winning lottery if you do vs. don't turn out to vote, you will prefer the random draw from the one where you turn out more often than vice-versa.</div><div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, Jun 24, 2025 at 5:27 AM Markus Schulze via Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.electorama.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">Hallo,<br>
<br>
it has been proven by Moulin that the Condorcet<br>
criterion and the participation criterion are<br>
incompatible:<br>
<br>
Herve Moulin, "Condorcet's principle implies<br>
the no show paradox", Journal of Economic Theory,<br>
volume 45, number 1, pages 53-64, 1988,<br>
DOI: 10.1016/0022-0531(88)90253-0<br>
<br>
Here is a short version of Moulin's proof:<br>
<br>
<a href="http://lists.electorama.com/pipermail/election-methods-electorama.com/2003-October/011042.html" rel="noreferrer" target="_blank">http://lists.electorama.com/pipermail/election-methods-electorama.com/2003-October/011042.html</a><br>
<br>
Markus Schulze<br>
<br>
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