[EM] Maximal Lotteries

[Closed Limelike Curves]

Maximizes the minimal expected margin of victory in a competition against
any other lottery. Essentially extends minimax by allowing for breaking
ties using randomness.


The important thing is there is always a lottery that has, on average,
majority support when compared to any other lottery.

100% agree this is the "objectively correct" generalization of Condorcet to
races with cycles, to the extent that such a thing is possible.

On Tue, Jun 24, 2025 at 8:39 AM Toby Pereira via Election-Methods <
election-methods at lists.electorama.com> wrote:

> What is maximal about the lotteries?
>
> Toby
>
> On Tuesday 24 June 2025 at 13:16:27 BST, Daniel Kirslis via
> Election-Methods <election-methods at lists.electorama.com> wrote:
>
>
> Here you go bud:
> https://en.wikipedia.org/wiki/Condorcet_winner_criterion
> http://en.wikipedia.org/wiki/Maximal_lotteries
>
> On Mon, Jun 23, 2025 at 8:24 PM Chris Benham via Election-Methods <
> election-methods at lists.electorama.com> wrote:
>
> I don't know what the "maximal lotteries method" is, and I guess that is
> true of other members of this list. But just going by its name I doubt
> that it would appeal to me.
>
> Is the Condorcet "winner principle" something different from the
> Condorcet criterion?   Because that is a binary pass-or-fail thing.
>
> Chris Benham
>
>
> On 24/06/2025 7:44 am, Daniel Kirslis via Election-Methods wrote:
> > For those of you who believe in the Condorcet winner criterion, is
> > there anyone who doesn't agree that the maximal lotteries method is
> > the theoretically soundest Condorcet method?
> >
> > Amongst the Condorcet methods, it seems to me that maximal lotteries
> > is clearly the best, at least in principle (that is to say, if we
> > ignore more practical concerns about ease of administration and
> > popular understanding). All deterministic Condorcet methods fail the
> > participation criterion. Therefore, a non-deterministic method is the
> > way to go, and the question becomes: "How shall we assign
> > probabilities amongst the Smith set?" I cannot imagine a more elegant
> > and fair-minded way of doing so than the maximal lotteries method.
> >
> > Is there anyone out there who understands the maximal lotteries method
> > but still thinks that there exists another method that better
> > satisfies the Condorcet winner principle? If so, why?
> >
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