Washblog

First, just for reference, before I launch into the actual comment, here's the existing language in the rules

Step 1: Tabulate the percentage of the vote that each presidential preference
_______ receives,
_______
Step 2: Retabulate the percentage of the vote, to three decimals, received by
_______ each presidential preference excluding the votes of presidential
_______ preference whose percentage in Step 1 falls below 15%,
_______
Step 3: Multiply the number of delegates to be allocated by the re-tabulated
_______ percentage received by each presidential preference.
_______
Step 4: Delegates shall be allocated to each presidential preference based on
_______ the whole numbers that result from the multiplication in Step 3.
_______
Step 5: Remaining delegates, if any, shall be awarded to the highest fractional
_______ remainders in Step 3. Ties shall be determined by lot.

and even though the replacement text below is longer, keep in mind that underneath "tabulate" is hidden a multitude of sins.

Subject: WSDCC Delegate Selection Plan public comment (III.F.6 algorithm)
From: Roger Crew <crew@cs.stanford.edu>
Date: Mon, 26 Mar 2007 15:20:32 -0700

I will strongly suggest a re-arrangement of the math in the delegate computation in section III.F.6 as follows:

On page 16 of the Delegate Selection Plan, replace lines 11-25 with

Step 1: Multiply the total number of votes cast for all presidential
_______ preferences by 15% to obtain the vote count threshold.
_______
_______ All preferences receiving fewer votes than this threshold
_______ are to be treated in Steps 2-5 as if they received zero votes.
_______
Step 2: Sum the total number of votes
_______ (i.e., excluding preferences zeroed in Step 1)
_______
Step 3: For each presidential preference, multiply the vote count
_______ by the overall number of delegates to be allocated.
_______
_______ Divide this product by the total in Step 2 to obtain an
_______ integer (whole number) quotient and a non-negative
_______ integer remainder
_______
Step 4: Each presidential preference with a nonzero quotient
_______ in Step 3 will be allocated that number of delegates
_______
Step 5: Remaining delegates, if any, shall be awarded to the
_______ highest remainders in Step 3. Ties in this ranking
_______ shall be resolved by lot.

Any mathematically equivalent procedure may be used so long as
it achieves this result.

In short, we change step 1 to use a vote count threshold rather than calculating individual percentages, we change Step 3 to do the multiplication first and use integer division rather than long division, and we have Step 5 rank integer remainders rather than decimal fractions.

Purpose

These changes address several issues:

1. The original rule is susceptible to consequential rounding errors and behaves inconsistently because of this (see the example below).
   
   In fact, that the original rule *could* be deemed ambiguous because there is no specification of whether "to 3 decimals" means "rounded to 3 decimals" or "truncated to 3 decimals" and this can change delegate allocations.
   
   By eliminating the use of long division and instead basing the specification on integer arithmetic, we remove all dependence on levels of precision and rounding. With the revised rule, there is no need to calculate any decimal places, ever.
   
   
2. The original rule contains unnecessary calculations
   
   There is no need to compute percentages for each candidate. Once you know what the overall 15% vote threshold is, a simple comparison of vote counts suffices to determine viability.
   
   
3. The revised rule makes the process more amenable to simplification in various special circumstances
   
   In particular, with this revision, vastly simplified methods (see below) become available for the 1,2,and 3-delegate caucuses which comprise over 40% of precincts statewide.

The revised rule otherwise follows the same principles as the original rule in enforcing the viability threshold and awarding delegates proportionately to the extent possible.

The original and the revised rules only differ in outcome where the former has roundoff errors in its application. Specifically, if the original Step 5 were to be modified so that decimal fractional remainders be taken as equal whenever they are sufficiently close together (i.e., absolute difference less than 0.001 times the number of delegates to allocate), then the resulting method would be equivalent to that of the revised rule. (I don't actually advocate that particular change to the rules because the revised rule above is so much simpler)

Example

Here is where rounding causes trouble in the original rule:

3 delegate precinct with 12 attendees. 3 candidates get 6, 4, and 2 votes, respectively.

Using the original method, we get

______ votes _______ % ___________ step3 _ delegates
Cand_A:   6   /12=   0.500   x3=   1.500   1
Cand_B:   4   /12=   0.333   x3=   0.999   1
Cand_C:   2   /12=   0.167   x3=   0.501   1

But, if "to three decimals" is interpreted as meaning "truncated to three decimals" instead of "rounded to 3 decimals" (or if someone simply makes a mistake in rounding), we get:

______ votes _______ % ___________ step3 _ delegates
Cand_A:   6   /12=   0.500   x3=   1.500   2
Cand_B:   4   /12=   0.333   x3=   0.999   1
Cand_C:   2   /12=   0.166   x3=   0.498   0

There is also a sense in which both of these outcomes are wrong.

Consider that in this situation there are 4 voters per delegate, thus, removing Cand_B and her 4 voters from the picture should not, in principle, change anything about the allocations to the other two candidates. And yet, if we do that, we'll have a situation with 8 attendees for which the original method gives

______ votes _______ % ___________ step3 _ delegates
Cand_A:   6   /8=    0.750   x2=   1.500   1 + 1/2
Cand_C:   2   /8=    0.250   x2=   0.500   0 + 1/2

"+ 1/2" meaning there is a tie to be resolved by lot between the extra two voters for A (i.e., beyond the 4 needed for A's first delegate) and the two voters for C, and, since weighting individual voters differently is to be frowned upon, resolving this with a coin toss is the only right thing to do. But if so, then for consistency, this is what we should be doing in the previous 12-attendee situation as well.

Meanwhile, using the revised rule

______ votes __________________ step3 ___ delegates
Cand_A:   6   x3=   18   /12=   1,rem=6   1 + 1/2
Cand_B:   4   x3=   12   /12=   1,rem=0   1
Cand_C:   2   x3=    6   /12=   0,rem=6   0 + 1/2

That is, there's a tie between the two remainders of 6 and there's no longer any ambiguity.

In the 8-attendee-2-delegate situation, we get

_______ votes _________________ step3 ___ delegates
Cand_A:   6   x2=   12    /8=   1,rem=4   1 + 1/2
Cand_C:   2   x2=    4    /8=   0,rem=4   0 + 1/2

and these cases are now handled consistently.

Equivalent Methods for 1-3 Delegate Caucuses

The following simplified methods can be proven equivalent to the revised rule in the case of caucuses with 1 to 3 delegates.

Note that the caucus envelope can include a worksheet tailored to the specific number of delegates for that caucus, at which point the precinct chair and participants need not concern themselves with what the general rule actually is. With the right worksheets, everything can be done with subtraction and table-lookup, and the process goes that much faster.

See http://41dems.org/2008/rfc/pcaucus_proposal/wksht.pdf for examples.

Again, this will simplify life for over 40% of the precinct caucuses.

These methods all replace steps 3-5. As per Step 1, we treat preferences under 15% as having zero votes, and "total votes" refers to the Step 2 total.

1 Delegate Caucus

Award the delegate to the preference with the most votes. Resolve ties by lot.

2 Delegate Caucus

Find the two highest preferences (resolve ties by lot) and take the difference of their vote counts.

• If this is less than half the total votes
  each of the top two preferences gets a delegate.
• If this is greater than half,
  the top preference gets both delegates.
• If exactly half, resolve the second delegate by lot.

3 Delegate Caucus

Find the three highest preferences (resolve ties by lot).

If either of the following two comparisons comes out equal, flip a coin to decide whether it holds.

• Take the difference of the two highest.
  If this is more than two-thirds of the total votes the highest preference gets all 3 delegates
• Take the difference of the highest and the third highest.
  If this is less than one-third of the total votes each of the top three preferences gets 1 delegate.

If neither comparsion holds,
the highest preference gets 2 delegates and the second highest gets 1.

Further Tweaks

Note that under the revised rule, the multiplication in the new Step 1 can be replaced with a simple table lookup: