[EM] IRV proponents figure out how to make IRV precinct-summable

Raph Frank raphfrk at gmail.com
Tue Mar 24 07:28:58 PDT 2009

```On Tue, Mar 24, 2009 at 2:30 AM, Dave Ketchum <davek at clarityconnect.com> wrote:
> Let's try it slowly for IRV, assuming multiple districts to avoid shortcut
> temptations:
>
> 1  Count ala Plurality.  If leader has a majority, that is winner.
>
> 2 Sum vote counts, starting with weakest count and ending before doing the
> next candidate that would make a majority.  None of those counted could win,
> so mark them all as losers and go back to step 1.

That isn't true.

The rule is actually that you can eliminate the weakest N candidates
in one step, if the sum of their votes is less than the (N+1)th
weakest candidates.  The procedure is then to find the largest
possible N.

40: A
25: B
15: C>E>B
9: D>E>B
7: E>B
4: F>E>B

Round 1:
A: 40
B: 25
C: 15
D: 9
E: 7
F: 4

According to your rules, eliminate F+E+D+C.  Eliminating B as well
would cause a majority of votes, so B is safe.

Round 2:
A: 40
B: 60

B wins

However, with full IRV, the results are

eliminat F

Round 2
A: 40
B: 25
C: 15
D: 9
E: 7+4 = 11
F: -

Eliminate D

Round 3
A: 40
B: 25
C: 15
D: -
E: 9+11 = 20
F: -

Eliminate C

Round 3
A: 40
B: 25
C: -
D: -
E: 20+15 = 35
F: -

Eliminate B

Round 4

A: 40
B: -
C: -
D: -
E: 35+25 = 60
F: -

E wins

So, the result is different.

>> Less formally, the method is summable if you can "count in precincts" to
>> produce managable data chunks that can then be combined to get the result
>> for all precincts or districts involved, no matter the size of each
>> district.
>>
> Not clear how this helps.  You have to get the totals for round 1 to decide
> how to proceed - matters not how many chunks.

I guess IRV is "summable in precincts, subject to central office instructions".

An election can be verified by checking all the precinct sums/counts
and that the central instructions were correct, assuming that the
precinct sums were also correct.

```