[EM] Discounted Redundancy, Fibonacci, and Despair Minimization

[Forest Simmons]

Here's something fun for those who would like to see how these topics are
related.

Start with despair minimization:

If p1, p2, p3, ... pN are the respective probabilities that candidates 1,
2, 3, etc. would vote like you want them to on any random issue of
importance to you, then the product D = (1-p1)*(1-p2)* ... *(1-pN) is the
probability that all of them are going to vote opposite to your wishes.
The closer this quantity is to unity, the greater your despair, so let's
just refer to it as your probability of despair, or just "despair" for
short.

An overestimate of this despair product is given by the expression

D' = (1-(p1+p2+...+pN)/N)^N, which, for large N, can be approximated by

D'' = 1/(1+(p1+p2+...+pN)/N)^N  and   D''' = 1/e^(p1+p2+...+pN) .

If any of these primed expressions is small, so is the despair, at least
for large N.  In general, for positive distinct p's, the inequalities

   D < D' < D'' < D'''  hold true. 

All of the approximations are good for large N or for small p's.  Small
p's are the rule for voters with significant despair. 

Before getting carried away too far, let's take a second look at D' . We
get from D to D' by replacing each factor with the average of all the
factors. This makes sense because all of the winners have to cooperate to
get something done, so just because the first candidate thinks 100% like
you do, doesn't mean she's going to be able to make any headway for you
(unless other will cooperate). 
 
So even though D' is larger than D, it is actually a more realistic
measure of despair.

One advantage of using the expressions for D', D'', and D''' is that they
work perfectly well for votes at the extreme without completely
discounting the redundant layers, as would happen with D.

So I propose using D' or D'' in the case of small N, and D''' in the case
of large numbers (like 3 or four) seats to be filled.

Suppose, in the context of Approval Voting we wish to minimize despair.

Let k = k(B,C) be the number of candidates approved on ballot B for the
candidates in combination C.  Then for this ballot the respective despair
estimates D', D'', and D''' are  (1-k/N)^N, 1/(1+k/N)^N, and 1/e^k .

The "hope" estimate for this last one is  1 - D''' which is  1 - 1/e^k .

Maximizing hope is equivalent to minimizing despair, and fits into our
Discounted Redundant Layers of Representation format more easily. The
relative weight of the k_th layer relative to the first is calculated as

[(1 - 1/e^k) - (1 - 1/e^(k-1))]/[1 - 1/e^1] , which reduces to

1/e^(k-1) . 

Dividing all of the layer weights by the constant  e  doesn't change the
results of DRLR calculations, so we may as well give the k_th layer a
weight of  1/e^k , which is identical to the expression for D'''  .

For small N, use  1/(1+k/N)^N  for the weight of the k-th layer.

Note that the sequence  1/e, 1/e^2, 1/e^3, ... is a geometric sequence or
the form  r, r^2, r^3, etc. with  r = 1/e .

Another more conservative geometric sequence, i.e. with r closer to one,
is 1/g, 1/g^2, 1/g^3, ... where g is gamma, defined by (1 + 5^.5)/2 , the
golden mean. 

This sequence is asymptotically proportional to the sequence

1/1, 1/2, 1/3, 1/5, 1/8, ... 

the sequence of reciprocals of the Fibonacci sequence. 

Notice that this sequence starts out like the harmonic sequence, but the
denominators start to grow faster than the harmonic sequence. On the other
hand they grow much slower than the sequences for despair minimization.

Therefore, this reciprocal Fibonacci sequence yields a DRLR method that
lies somewhere in between  Proportional Representation  and  Despair
Minimization.

For those who believe that mere Proportional Representation doesn't give
enough voting power to representatives of minorities, but think Despair
minimization is too radical, this could be the golden compromise solution.

It would take an election with eleven winners for this Fibonacci method
to catch up with the proportional power method based on the sequence of
reciprocals of perfect squares, so for most practical purposes it is less
radical than that method.

Wasn't that fun?

Forest