Condorcet Exploded

The Condorcet explanation that Alex gave was perhaps a little dense, so I’m going to do my best to make this easier to explain.

The trick to Condorcet is the fact that the ballot that you fill out is not actually a ranked ballot, but rather a much larger ballot. A ballot that would look like this:

A –
B –
C –
D –

would actually break down to a bunch of different Conditions:

A>B –           B>A –
A>C –           C>A –
A>D –           C>A –
B>C –            C>B –
B>D –            D>B –
C>D –            D>C –

like this. Now, let’s consider a sample ballot:

A – 3
B – 1
C – 4
D – 2

which expands to:

A>B –                          B>A – x
A>C – x                       C>A –
A>D –                         D>A – x
B>C – x                       C>B –
B>D – x                       D>B –
C>D –                          D>C – x

where each of the Conditions from the ballot is tallied.

Say we start over with a new set of results:


which would break down into this:

A>B – xx                     B>A – x
A>C – xxx                   C>A –
A>D – xx                     D>A – x
B>C – xxx                   C>B –
B>D – x                        D>B – xx
C>D – x                       D>C – xx

Now, you subtract the lower of either of the two in each row from the greater of the two, leaving a margin of victory for each condition. In the first row, for example, B>A is subtracted from A>B, because the latter got more votes. This would then be expressed as the winning condition, with the margin beside it. Our example would look like this, with winning contitions bolded:

A>B – xx         minus   B>A – x           =          A>B – x
A>C – xxx       minus   C>A –              =          A>C – xxx
A>D – xx         minus   D>A – x           =          A>D – x
B>C – xxx       minus   C>B –              =          B>C – xxx
B>D – x           from    D>B – xx         =          D>B – x
C>D – x           from    D>C – xx         =          D>C – x

The next step in the process in the ranking of all the pairs:

A>C – 3
B>C – 3
A>B – 1
A>D – 1
D>B – 1
D>C – 1

Now, start developing the final ranking by creating a system which fulfills the requirements of the pairs, in order from most votes to least votes. Note that all conditions that are tied must be considered at the same time (ie. they may not conflict).

The Threes
A>C gives us A>C
B>C gives us (A,B)>C
The Ones
A>B gives us A>B>C
A>D gives us A>((B>C),D)
D>B gives us A>D>B>C
D>C gives us A>D>B>C

We now have a full ranking. If we had a greater number of these, we might eventually run into a condition that conflicts with the established order to that point. Condorcet ignores any lower ranked Condition that conflicts with a higher ranked Condition.

Now, our election is going to have higher voter turnout than 3 (hopefully), so assuming about about a 9% voter turnout, we could have an election like this.

A>C – 3942
B>C – 3210
A>B – 2523
A>D – 2221
D>B – 2182
D>C – 2090

This would give us the same result as before, but without the need to consider conditions simultaneously.

So, in summary, Condorcet works like this:

  1. Everybody votes.
  2. Ranked ballots are expanded into votes for Conditions.
  3. Opposite conditions are subtracted from one another.
  4. Conditions are ranked in order of most votes received.
  5. A ranking is constructed based applying Conditions in the order of votes received.
  6. Somebody wins.

5 Responses to “Condorcet Exploded”

  1. 1 radicalbeer
    January 27, 2009 at 5:03 pm

    And in the oh-my-god rare case no one wins, a random ballot is added!

  2. 2 radicalbeer
    January 27, 2009 at 5:06 pm

    Also, if you want to vote strategically, you still can, but it’s much riskier and requires a ton of more information than FPP. Personally, I’ll be voting honestly. Good luck!



  3. 3 Darren
    January 27, 2009 at 7:21 pm

    An even simpler explanation (I hope):

    On your ballot, you rank the candidates. Ties are allowed and anyone you don’t rank is last, so marking an X next to one name is fine. It just provides less information on what you think of the other candidates.

    Your preference order is translated into pairs. If you ranked candidate F 3rd and candidate B 7th, you liked candidate F better than candidate B, and F wins a point against B.

    Instead of counting how many first-place votes a candidate got, you count how preferred candidates are relative to each other, by tallying the results of these pairs. If 1700 people liked F better than B and 1500 liked B better than F, F wins that pair by 200.

    These wins are combined, starting with the most lopsided race, ultimately ordering the candidates by preference. This means you’ve identified the most-preferred candidate. They win the election. F may have beaten B by 200, but F beat A by 500, so that’s considered first, and candidate C beat A, B and F, each by about 20, so C’s more preferred and should win over A, B and F.

    While it makes no difference in the F vs B pair whether they were 1st and 7th or 5th and 6th, the candidate you rank first gets points in more pairs, and this comes out when the pairs are ranked. In almost every AMS race, one candidate will get enough first place votes that the choice of vote-counting system is irrelevant. In ranked pairs, such a candidate easily wins in all pairs and ends up on top of the heap more or less immediately. Where the system gets interesting is in a very tight and polarized race, where supporters of each side hate the other so much they’d rather see a keg of beer elected than that loser. They may well get what they wished for.

    The full, detailed technical explanation is available for anyone who wants it, such as the elections people or Matt. The average voter doesn’t need to know the wording or gory details, any more than they need to know the exact legal definition of “second-degree murder” or “usury”. For most, “you rank the candidates and the most preferred one wins” is enough.

  4. 4 radicalbeer
    January 28, 2009 at 9:53 am

    If the people want a Keg to rule the land, then we support the decision of the people.

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