# [EM] A Question about Pages 22-29/58 of Chapter 12 of Fobes' 'Ending the Hidden Unfairness of U.S. Elections

VoteFair ElectionMethods at VoteFair.org
Thu Sep 17 11:44:23 PDT 2015

```On 9/13/2015 12:22 PM, steve bosworth wrote:
> ...
> [...] please [...] explain why the following simpler set of
> calculations would not also always allow us to discover the most
> popular sequence:
>
> Firstly, find the grand total of preferences given by the 100 voters to
> each of all the candidates (4 in this example) over each of the other
> candidates (3 in this example). The result is:
> Elliot 200
> Selden 180
> Meredith 90
> Roland 80
>
> At least in this case, the same sequence is produced:
> Elliot 1st, Selden 2nd, Meredith 3rd, and Roland 4th.

Yes, sometimes -- in SOME situations -- a simpler calculation (such as
this one) can identify the same winner and even the same ranking.

However, typically a shortcut fails to provide fair results in ALL
situations.

Consider that plurality (first-past-the-post) voting is a shortcut that
mistakenly assumes that the candidate with the most first-choice votes
is always the most popular.  This shortcut does not work if the
candidate with the most first-choice votes does not ALSO get a majority
of (more than half) the votes.

In a similar way, instant-runoff voting is a shortcut that does work in
some situations.  It is based on the shortcut of (mistakenly) assuming
that the candidate with the fewest first-choice votes is least popular.

Also in a similar way, your APR method will work in some situations.
Yet it too puts too much emphasis on first-choice votes without
considering secondary preferences (or somewhat-equal preferences for
those who prefer approval voting).

In contrast to methods that work SOME of the time, full fairness
requires that a method must produce fair results either ALL the time
(which is mathematically impossible if all fairness criteria are
considered), or MOST of the time.  In the best methods, unfairness
(according to any fairness criteria) is rare.  To get this level of
fairness, the voting method must look beyond the first-choice counts.

Back before computers became available, mathematical shortcuts were
often useful in some situations.  Now, both in terms of calculation time
and the work of coding software, it's easier to do full calculations
using a fully-fair algorithm, compared to writing code that handles both
the shortcut and all the needed validity checking and related decision
handling (to handle the cases where the validity checks fail).

I hope this information helps not only you/Steve, but also helps some
other participants in this forum.

Richard Fobes

BTW, page numbers in an ebook reader do not match the page numbers in a
different ebook reader, and do not match the page numbers in a printed
edition.

On 9/13/2015 12:22 PM, steve bosworth wrote:
> Hi Richard,
>
> Unfortunately, my other obligations have not allowed me to reply to your
> answers to our 20^th VoteFair/APR dialogue until today.However, before I
> do so, I would very much appreciate it if you would clarify a question
> that I have about your book, ‘Ending the Hidden Unfairness of U.S.
> Elections’.With regard to pages 22 to 29/58 of Chapter 12, I finally
> understand how you arrived at the ‘score’ of 400 for the winning
> sequence i.e. Elliot(E)>Selden(S)>Meredith(M)>Roland(R):
>
> You started by adding up the number of the 100 voters who had preferred
> each of the three candidates to the left of R, over R, i.e. 60+70+70 =
> 200.You then added the number who preferred those to the left of M, over
> M, i.e. 70+70=140.Next you added the number who preferred E over S, i.e.
> 60.The sum of these 3 totals is 400, i.e. the largest score for any one
> of all the possible sequences.
>
> However, please also explain why the following simpler set of
> calculations would not also always allow us to discover the most popular
> sequence:
>
> Firstly, find the grand total of preferences given by the 100 voters to
> each of all the candidates (4 in this example) over each of the other
> candidates (3 in this example). The result is:
> Elliot 200
> Selden 180
> Meredith 90
> Roland 80
>
> At least in this case, the same sequence is produced: Elliot 1st, Selden
> 2nd, Meredith 3rd, and Roland 4th.
>
> Why do we also have to calculate the score for each possible sequence?
>
> What do you think?
>
> Steve
```