Weighted Rankings

Now that we have a true three-party race, I'm noticing things about our voting system that I never noticed before. Depending on how one weights our ranked votes, it can dramatically change the outcome of the election. Here's an example.

There are three parties, called A, B, and C, that receive six ranked votes. The first three votes ranked A first, B second, and C third and so on as below:

In a standard election, the winner would be A (3 votes) followed by C (2 votes) and B (1 vote). However if the following weights are used, 1.0 for first place, 0.5 for second, and 0.0 for last, the following happens:

The winner of the election in this case is B, followed by A and C. What's fascinating is that this weighted method makes the loser of the previous the winner simply by giving value to the second-place ranking.

The question I have is, which system do you prefer? One that gives one group their first choice or one that gives everyone their second choice? Granted the above example is contrived to demonstrate an unusual outcome, it nonetheless shows the tradeoffs that can be made with weighted systems.

Right now, the system uses a compromise between these two examples. The weights are (n-1)/(n-1), (n-2)/n, (n-3)/(n+1), etc.

~Ulrika~

So in our currently running election,

1st preference = 1
2nd preference =???
3rd preference = 0

In our current election there are three parties, n = 3, with three places, p, so:

The score for an arbitrary place is:

~Ulrika~

you sayin' I'm fat?

Ok. My brain exploded but I think I have enough of it left to give a question/thought. hehe..

Since we came up with this voting method in order to vote on parties and not individuals, why would it matter if our second choice "person" won out over our 1st choice "person"?

Or are you saying this also affects the party vote preference too?

I believe so. However, the second place vote in our election gets a lower score than in Ulrika's hypothetical. Thus the current formula would give (in the imaginary 6 voter election)

A = 3 * 1.0 + 0 * 0.33 + 3 * 0 = 3.0
B = 1 * 1.0 + 5 * 0.33 + 0 * 0 = 2.66
C = 2 * 1.0 + 1 * 0.33 + 3 * 0 = 2.33

Giving the same order of finish (although closer margins) as a first preference only system.

I like it how it is. I mean, what's the use of ranking them if it would only "pay attention" to whatever you've ranked highest priority, just #1 and #1 alone?

In your above example, if 1 person thought "B" was the best option, and 5 thought it was the 2nd-best option... Then yes, it makes sense to me that it's getting a higher amount of votes, as oppose to A where 3 people give it the highest ranking and 3 give it the lowest.

IMHO the existing weighted system provides more fine-grained detail/focuses on capturing a more accurate snapshot of the public vote we're trying to measure.

I believe the Borda-count method will use the wieghts (n-1)/(n-1), (n-2)/(n-1), and (n-3)/(n-1)... corresponding to 1, 0.5, and 0, the "standard" election weights in Ulrika's example.

Since it is the method prescribed in the constitution, I think that is the method we should follow.

See Ulrika's post #3 in this thread

I guess my point is that the "standard" weighting is the one usd by a proper Borda-count ranking, and the "compromise" weighting is not. Thus, we should be using the standard weights:

S(p) = (n-p)/(n-1)

Bah! I transcribed the wrong chunk of code. Sorry, it was late. We are using the Borda method, which is linear.

In our current election there are three parties, n = 3, with three places, p, so:

The score for an arbitrary place is:

~Ulrika~