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With four current candidates ("M", "G", "Y" and "W") and three choices (1st, 2nd and 3rd) there are 24 ways to vote all three, 12 ways to vote 1st and 2nd and skip 3rd, and four ways to vote for only 1st.

I'll ignore voting only for 2nd and/or 3rd and skipping first and/or second.

What are the guidelines for the voting strategy here?

  1. Are there any benefits for voting for any less than all three choices (1st, 2nd and 3rd)?
  2. Does the voting robot let us get away with checking any less than all three?

Python

from itertools import permutations
for n in range(3, 0, -1):
    options = list(permutations('MGYW', n))
    options = [''.join(option) for option in options]
    print(n, len(options), options)
    print('')

results in:

(3, 24, ['MGY', 'MGW', 'MYG', 'MYW', 'MWG', 'MWY', 'GMY', 'GMW', 'GYM', 'GYW', 'GWM', 'GWY', 'YMG', 'YMW', 'YGM', 'YGW', 'YWM', 'YWG', 'WMG', 'WMY', 'WGM', 'WGY', 'WYM', 'WYG'])

(2, 12, ['MG', 'MY', 'MW', 'GM', 'GY', 'GW', 'YM', 'YG', 'YW', 'WM', 'WG', 'WY'])

(1, 4, ['M', 'G', 'Y', 'W'])
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1 Answer 1

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TL;DR

Are there any benefits for voting for any less than all three choices (1st, 2nd and 3rd)?

No.

Does the voting robot let us get away with checking any less than all three?

Yes.


The simplest way to understand it is it's more or less working like ranked choice voting.

Let's say we have 100 voters and candidates A, B and C. Our threshold for winning is 51 votes. The first round breaks down like so

  • A - 35 votes
  • B - 32 votes
  • C - 33 votes

Nobody hits the threshold so we eliminate the lowest candidate, which is B. Now we look at the second choice for people who voted for B (assuming all B voters had a second pick)

  • A - 10 votes
  • C - 22 votes

That means our second round ends up as

  • A - 45 votes
  • C - 55 votes

C wins, crossing the 51 vote threshold.

Now, if you didn't pick a second choice and voted for B, your vote falls into the exhausted pool (doesn't directly benefit any candidate).

You can see how rounds play out with more candidates by looking at the Stack Overflow 2019 vote results. I should note that I oversimplified the process a bit so it's understandable. In larger elections, where all votes are exhausted and there's still no winner, SE uses Meek STV to pick a winner (if you thought orbital mechanics were complicated...). It's unlikely STV will pick a winner for Space.SE (1 seat, fewer candidates and smaller vote pool)

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  • $\begingroup$ So even in the hypothetical case that there were three candidates and three votes allowed and there was one specific candidate that I was loath to help win, I would be better off or at least exactly equally well off voting for them 3rd versus only voting for 1 and 2 and not voting 3rd at all? $\endgroup$
    – uhoh
    Commented Feb 26, 2020 at 3:46
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    $\begingroup$ If there's only one spot for grabs, then your third vote would never count in a 3-way race because we eliminate candidates until we reach (N + 1) candidates (where N is the spots available). If there's no winner after that, Meek STV is used to break the deadlock. SO Election 2018 went that far, if you want an example $\endgroup$
    – Machavity
    Commented Feb 26, 2020 at 13:20

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