Reply #46: FINALLY, the estimated total miscount is divided up and allocated to the 5 [View All]

[AirAmFan]

candidates (including the Disqualified candidate).

First, it is split into two shares, using the proportions of major-candidate votes Kerry and Bush were awarded at the cluster. We have denoted these proportions as

m(K) = n(K) / (n(K) + n(W)) and (1-m(K)) = n(W) / (n(K) + n(W)) .

Then each major candidate's share is divided into c(c-1) pieces (where c is the number of clusters), and the pieces are allocated according to the values of the "number of wrong-machine ballot positions" variables wmbp_K? and wmbp_W? .

Continuing with the example of cluster 171, the estimate of the total miscount is 249.057. 91.9 percent of this total, or 228.883, goes to Kerry ballot miscounts. The other 8.1 percent, or 20.174, goes to Bush miscounts.

Because there are six wrong-machine ballot positions in any 3-precinct cluster, divide the Kerry miscounts into 6 pieces, each of size 38.147. Divide the Bush miscounts into 6 pieces of size 3.362 apiece.

Since wmbp_KD = 2 (see the end of post #44), 76.294 extra votes for Kerry are pulled from the pile of spoiled ballots, and in particular that portion of the pile attributable to votes for the Disqualified candidate. 38.147 votes are transferred to Kerry from Badnarik, and the same number from Peroutka. The other 76.294 come from Bush's vote, since wmbp_KW = 2 for the configuration of ballot orders at cluster 171.

Similarly, Bush gets 3.362 votes from each of Disqualified and Peroutka, and 6.724 from each of Bednarik and Kerry.

These Kerry votes awarded to Bush offset 6.724 of the 228.883 already awarded to Kerry above, so that kerry's total increases by the difference between these numbers, or 222.159 votes. Bush's total falls by 76.294 - 20.174 = 56.120. So correcting the "caterpillar" miscount in cluster 171 would reduce Bush's "lead" by 222.159 + 56.120 = 278 votes.