[EM] Exact spatial model probabilities?

[Daniel Carrera]

On Thu, Jan 27, 2022 at 4:20 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> > I'm a bit confused by the notation. If p(A>B>C) is the probability that
> > a random voter picks A>B>C then what is p(A>B>C) * p(A>B>C) * p(B>C>A)?
> >
> > I also don't get what you mean by "e_s = (A>B>C, A>B>C, B>C>A)" being an
> > election.
>
> The former is just a statement of statistical independence. Suppose that
> we roll three ordinary dice, call them x, y, and z. Then the probability
> of the result being all ones is p(x=1) * p(y=1) * p(z=1) due to
> independence - the different outcomes don't constrain each other.
>


Ok. The part that wasn't obvious to me was that "e_s = (A>B>C, A>B>C,
B>C>A)" was the election with exactly those ballots in that order. You
might want to instead define an election so that it's only about the
combination of ballots and not the permutation; so if you shuffle the
ballots it's the same election.

e_s = {2: A>B>C, 1:B>C>A}

=> p(e_s) = Choose(3,2)*p(A>B>C)*P(B>C>A)



> (Again on a side note: since it doesn't matter in what order the voters
> vote, you could save some computations by using a multinomial - e.g. if
> you find out that method M is susceptible to manipulation with election
> (x, y, z), you automatically know it's susceptible to manipulation with
> election (x, z, y) too. But it would make the code more complex.)
>


One way to avoid the combinatorial part is to define the election so that
the ballots are sorted in some canonical order. For example, sort the
possible ballots alphabetically:

b1 = ABC
b2 = ACB
b3 = BAC
b4 = BCA
b5 = CAB
b6 = CBA

Let nk be the number of voters that cast the ballot bk. Then:

p(e) = Prod_k p(bk)^nk

Cheers,
--
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
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