How Big will Barrett Score in Dane and Milwaukee Counties?
Holding 2010 totals constant for the state’s other 70 counties, Barrett would need to carry 75.1 percent of the vote in Wisconsin’s two most populated counties to beat Walker on Tuesday (up from 64.1 percent in 2010)
The Wisconsin Government Accountability Board estimated last week that turnout in the state’s gubernatorial recall election Tuesday could hit an astounding 60 to 65 percent, from somewhere between 2.6 and 2.8 million voters.
Turnout in 2010 was only 49.7 percent, or 2.1 million voters.
Underdog Democratic challenger Tom Barrett will obviously have to improve his 2010 numbers across the state’s 72 counties in order to score an upset victory over Republican Governor Scott Walker.
No doubt part of his recipe for success is a particularly large turnout in the state’s most populated counties, Dane and Milwaukee, where Barrett tallied his second (68.0 percent) and third (61.6 percent) best showings in the state nearly two years ago respectively.
(Barrett also won 77.9 percent of the vote in sparsely populated Menomonee County and 61.6 percent in Ashland County).
Presuming, for the moment, that turnout patterns increase equally across all counties in the state in the recall race, and presuming that the Barrett/Walker county splits hold constant from 2010 in the other 70 counties, how big a share of the vote does Barrett need to snag in Dane and Milwaukee Counties to win the election?
In 2010, more than a quarter of the votes cast in the gubernatorial race came from Dane and Milwaukee Counties: 561,290 out of 2,160,832 total votes, or 26.0 percent.
Barrett carried these two counties by a 64.1 percent to 35.1 percent margin, netting 162,781 votes on Walker.
Walker meanwhile won the rest of the state by 287,419 votes, for a final victory tally of 124,638 votes.
Estimating voter turnout on Tuesday in the middle of the Government Accountability Board’s projections at 2.7 million, that would mean 701,612 votes will be cast in Dane and Milwaukee Counties.
Looking at the Barrett-Walker splits from 2010 across the state’s 72 counties with the new turnout model, Barrett is operating at a deficit of 153,900 votes that he needs to make up, or 1.409 million to 1.256 million statewide.
To make up those 153,900 votes in Dane and Milwaukee Counties, Barrett would have to increase his vote share from 64.1 to 75.1 percent (526,683 votes out of 701,612 cast), or a 17.2 percent increase, while Walker drops to 24.1 percent (169,316 votes).
Note: The model also presumes independent candidate Hari Trivedi captures approximately the same total as 2010’s three third party candidates for governor.
Of course, the truth is, Walker does not need 75.1 percent of the vote in these two counties.
For if he increases his vote tally to even 68 or 69 percent in Dane and Milwaukee Counties he will undoubtedly be increasing his tally from 2010 in many of the state’s other 70 counties – whether he won them or not in the last cycle.
Then again, it is also possible that turnout in either or both of these two counties could theoretically be much higher (or lower) than the rest of the state vis-à-vis the 2010 race.
So, while the numbers above do not reflect a predictive model, they nonetheless capture a rough gauge of the baseline from which Barrett will need to improve in order to have any chance at victory Tuesday evening.
The converse is also true: if Governor Walker improves on his 2010 best performances in Washington (75.0 percent) and Waukesha (71.5 percent) Counties, then his chances of building upon his 2010 margin of victory also likely increase.
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“…truth is, ‘Walker’ does not need 75.1%…” Sure he does, just nowhere near as much as BARRETT (also, whoever the Ds nominate as their presidential standard bearer must expand her or his electoral appeal well beyond the progressive Dane and proletariat Milwaukee, something Gore succeeded in 2000 and HRC failed in 2016).