At Large Strategic Voting in Wilmington City Council Primaries
In the upcoming Democratic Primary for the Wilmington City Council At-Large seats, being second favorite might be best. There are four at-large seats on City Council with no more than three at-large members allowed to be from any one party. For decades this has meant 3 Democrats and one Republican due to the more than 4-1 Democratic registration edge (32,343 Democrats to 6947 Republicans) in the city. The Republicans have only two candidates for three ballot slots on the November ballot,so they don’t have a primary.
The Democrats have two incumbents seeking re-election, Charles “Bud” Freel and Loretta Walsh.The third current At-Large council member, Theo Gregory, is running for City Council President. Facing the two incumbents are 6 challengers in the Democratic Primary: Darius Brown, Dwight Davis, Samuel Guy, Gary Hutt,Vincent White and Justen Wright.
In the primary each voter has the option of voting for as many as three candidates,but some people only vote for one or two. In an 8 way race, being the second choice for the supporters of other rivals may be a path to victory. The numbers cited below indicate that the average votes cast per voter in the last 5 city wide primary was between 1.91 and 2.21.
This means that while some voters vote may vote only once ,many voters make two or three choices .
1988 -10,236 voted in mayor’s race , 20,043 voted in the city council at-large,
votes cast per voter in the at-large race 1.96
1992-10,773 voted in mayor’s race, 23,607 voted in the city council at-large,
votes cast per voter in the at-large race 2.19
1996-10,549 voted in mayor’s race, 23,329 voted in the city council at-large,
votes cast per voter in the at-large race 2.21
2000-9353 voted in the mayor’s race , 17,835 voted in the city council at-large,
votes cast per voter in the at-large race 1.91
2004-total Wilmington voter turnout 7,558 (no mayor’s primary),15,915 voted in city council at-large,
votes cast per voter in the at-large race 2.11
The case can be made that the winner could make it with “one choice only voters’ and that is mathematically possible,but not likely. While we can not determine how many people actually vote for only one candidate, we can make some calculations by process of elimination which show that winners tend to garner support from supporters of their rivals.
In 1988 Loretta Walsh and Theo Gregory were the top two finishers in a field of five and got a combined 10,040 votes. There were 10,236 votes cast in the mayoral primary that year,so if Walsh and Gregory both had only single choice voters that would leave only 236 voters times 3 votes per person to be shared by the other three candidates. However, 3 x 236= 708. 708 plus 10,040=17048 potential votes cast. Since 20, 043 votes were actually cast in the at-large race, we can be fairly certain that Walsh and Gregory’s combined total included many second choice votes.
1992,2000 and 2004 present even clearer evidence.
1992 Walsh and Gregory got a combined 10,796 votes in a six way race,but only 10 ,773 votes were cast in the mayor’s race. Unless 23 people voted in the at-large race and skipped the mayor’s race it is guaranteed they got second choice votes.
In 2000 Theo Gregory and Bud Freel were the top two vote getters in a four way race with a combined 9801 votes. Since only 9353 votes were cast in the mayor’s primary, unless over 450 voters skipped the mayor’s race and voted in the at-large race they also mathematically had to get some second choice voters.
In 2004 there was no mayor’s race,but 7,559 city Democrats voted. 7588 of them voted for one of the two top vote getters in a 5 way race, Theo Gregory and Loretta Walsh, indicating they again got second choice votes since the most possible single choice votes they could have gotten would have been 7,559.
Because in 1996 there was 14 way primary for 3 ballot slots after two opened up when Walsh and fellow city council member, Bob Poppiti, both ran for mayor, this type of math is not as easy to establish. Gregory and Freel were the top two vote getters, but garnered a combined 6, 775 votes. 10, 549 votes were cast in the mayoral primary. While it is likely they each got their share of second choice votes, it is not so easy to quantify as when there is a smaller ballot. 10,549 minus 6,775= 3774 which is how many voters would be left over if all of Gregory & Freel’s votes were from single choice voters. 3,774 x 3 votes per person makes 11,322. 11,322 plus 6,775= 18,097. Since 23,329 votes were cast in the at-large race, we can tell that Freel and Gregory must have gotten some second choice votes because 23,329 minus 6,775=16,554. If all of their votes were single choice only votes there would not be 18,097 voters left to cast the votes that were actually cast.
Sometimes the best way to predict the future is to look at the past. While the position expressed in this essay may have been discussed previously I had never seen it quantified. I had heard that candidates ran on the plan of getting their own supporters to vote for them only and it appears that might not be enough.
There are 8 candidates in this year’s at large field, so the math may be closer to the 1996 scenario than the other four elections discussed,but I am betting that whoever wins is going to rack up their share of second choice support from voters that other candidates convince to go to the voting booth.