# Slicing the Democratic Primary Pies with Three Leading Candidates

Slicing the Democratic Primary Pies with Three Leading Candidates.

Previously, we showed how the Democratic district vote would be split between two candidates, because only candidates with more than 15% of the vote were to be counted, and only two existed at that time.   Now, however, the Quinnipiac University poll shows three candidates with 15% or over.   So we analyze this particular case for 3, 4, 5, 6, 7, or 8 delegates per district.

To start with, the poll has not only Sanders beating Trump, with 53% to 40%, but so do the next five Democratic candidates.  In the Democratic Primary country-wide, we have Biden at 30%, Sanders at 19%, Warren at 15%, Buttigieg at 8%, Harris at 7%, and others 3% or lower.  13% are undecided or not available.  The accuracy was pm 3.5%.

The first Democratic rule is to drop everybody below 15%.  The three above 15% add up to 64%.  The other 36% may end up voting for the top 3 by the time they get to vote.  Anyway, for now, they are dropped.

The second Democratic rule is to rescale those above 15% to 100% by multiplying by 100/64 = 1.5625.

This gives Biden 46.875%, Sanders 29.685%, and Warren 23.4375%.

These will be used for all districts, regardless of the number of delegates.

The third rule is to multiply this times the number of delegates for the district, which is mostly 4 or 5, but also 3 or 6.

Biden has 2.34, Sanders has 1.48, and Warren has 1.17.

The fourth rule is to first look at the whole numbers, giving Biden 2 delegates, Sanders 1 delegate, and Warren 1 delegate.

There is still 1 delegate left.  This is given to the leading fractional number, which is the 0.48 to Sanders.

So we have Biden 2, Sanders 2, and Warren 1 for five total delegates.

Next, we do the case of 4 total delegates.

Multiplying by 4 gives Biden 1.875, Sanders 1.1874, and Warren 0.9375.  Truncating, Biden gets 1 and Sanders 1, with Warren 0.  The rest are to be awarded in order of the highest fractional remainder.  Since Warren wins the fraction, she almost has 1, so she gets 1.  Biden has the next highest fraction,  so he should get another delegate.  Thus, Biden has 2, Sanders 1, and Warren 1.

For 6 delegates, Biden has 2.8125, Sanders 1.7811, and Warren 1.406.  So Biden gets 2, Sanders 2, and Warren 1.  The sixth delegate goes to the largest fraction, which is Biden.  Finally, Biden has 3, Sanders 2, and Warren 1.

For 3 delegates, Biden has 1.406, Sanders 0.891, and Warren 0.703.  Clearly, Biden gets 1, and 2 are left over.  Sanders is highest, and Warren also close to 1.  Biden would round off lower to 1.  So Biden 1, Sanders 1, and Warren 1.  Equal.  Despite Biden actually having twice the vote of Warren.

For 7 delegates, Biden has 3.281, Sanders has 2.078, and Warren has 1.640.  The whole numbers give Biden 3, Sanders 2, and Warren 1, adding to 6.  Warren has the largest fraction, and gets the last one.  The final number is Biden 3, Sanders 2, and Warren 2.

For 8 delegates, Biden has 3.75, Sanders has 2.37, and Warren has 1.875.  The whole numbers give Biden 3, Sanders 2, and Warren 1.  There are two left over delegates, which by fractional ordering gives Warren 1 and Biden 1.  The end result is Biden 4, Sanders 2, and Warren 2.  These breakups get closer to the vote percentage as the number of delegates increases.

In California’s 53 House districts, there are 1 with 3 delegates, 16 with 4, 18 with 5, 13 with 6, 3 with 7, and 2 with 8.

We make a table of the results for B, S, and W delegates for districts with a given number of delegates:

Del.  B.    S.    W

3.      1.     1.     1

4.      2.     1.     1

5.      2.     2.     1

6.      3.     2.     1

7.      3.      2.    2

8.      4.      2.    2

Applying this to the number of districts for each number of delegates, gives a total number of delegates for

B.     S.     W

125.     89.    58

These total 272 delegates.

The percentage of delegates for each candidate is now compared to their relative percentages of the top three candidates, which are over 15%.

B has 0.460, S has 0.327, and W has 0.213.      Compare this to their vote ratios:

B with 0.469, S with 0.297, and W with 0.234.  The law of large numbers to the rescue.

Since the apportionment of delegates is really a round off to the nearest whole delegates, when you put enough together, it gives a convergent approximation.  It’s like pixelating a picture, still leaves an approximation of the picture.  Three about equal pixelizations of 4, 5, and 6 delegates are being merged here.