True Vote Model Scenarios: What will it take for Romney to win?

Richard Charnin

Oct. 6, 2012

Update: This is the final Nov.5 projection: 2012 Presidential True Vote and Monte Carlo Simulation Forecast Model.

Click this link to the final 2012 forecast. It was exactly right: Obama had 51.6% (2-party) and 332 EV with a 99.6% win probability. But his True Vote was 55% with 380 EV. http://richardcharnin.wordpress.com/2012/11/07/4380/

The 2008 Election Model also predicted Obama’s recorded vote exactly at 365 EV and 52.9% with a 100% win probability. But his True Vote was 58.0% with 420 EV. http://www.richardcharnin.com/2008ElectionModel.htm

The 2012 True Vote Election Simulation Model consists of two components: a Monte Carlo Simulation, based on the latest state polls; and the True Vote Model, based on 2008 voter turnout in 2012 and corresponding vote shares of returning and new voters.

The state polls will be updated on Monday, Oct. 8. In the meantime, let’s consider an analysis that never appears in the mainstream media or by pollsters and election forecasting bloggers.

Obama is leading in the polls, but the race is *“tightening”*. Yes, once again, we have another media-induced *“horse race”*. This analysis does not consider pre-election polls. It is concerned with returning voter turnout and defection scenarios and uses the True Vote Model component.

The model shows that in order for Romney to win, he needs a 5% net turnout percentage advantage of returning McCain voters as compared to Obama voters. And he needs to capture nearly one out of five returning Obama voters (18% defection rate) while Obama wins just one in twenty (5% defection rate) returning McCain voters.

Obama won the 2008 state unadjusted exit poll aggregate by 58-40%. The True Vote Model indicated that he had the identical 58%. He won the National Exit Poll by a 61-37% margin. The 8% discrepancy is eight times the theoretical 1% margin of error. The probability of election fraud was 100% in 2008.

Given the above we will assume the following:

1.Obama had 58% in 2008, not the recorded 53%.

2.New voters will split 50/50 for Obama and Romney.

*Note that the 50/50 split in new voters is a conservative assumption. According to the adjusted 2008 Final National Exit Poll, which understated Obama’s unadjusted share by 8% in matching the recorded vote, Obama had 72% of new voters. The Democrats have consistently won a solid majority of new voters in every election since 1988. *

Let’s now calculate several alternative scenarios to see what it would take for Romney to win, given the above assumptions.

Scenario I: Equal returning Obama and McCain 95% voter turnout and zero net defection of Obama and McCain voters. In other words, Obama has 95% of returning Obama 2008 voters and 5% of returning McCain voters.

**Obama has 57.1% with 413 electoral votes.**

Scenario II: 90% of Obama 2008 voters turn out in 2012 and 95% of McCain voters turn out. As in Scenario I, there is zero net defection of returning voters.

**Obama has 55.9% with 388 electoral votes.**

Scenario III: 90% of Obama voters turn out, but he only wins 90% of them. Romney still wins 95% of returning McCain voters. He has a 5% turnout and defection rate advantage.

**Obama has 53.5% with 334 electoral votes.**

Obama wins all the above scenarios. So what will it take for Romney to win the election?

Scenario IV: The model shows that Romney needs 18% of Obama voters (nearly one in five), He needs a 5% turnout advantage and a 13% net defection advantage.

**Romney wins in a squeaker with 50.4% and 280 expected electoral votes. **

But if Obama voters turn out at the same 95% rate as McCain voters, Obama wins a squeaker with 50.5% and 275 expected electoral votes.

This analysis is predicated on an equal split in new voters. If history is a guide, new voters will be solidly Democratic, so Obama’s popular and electoral votes will be higher in each of the above scenarios.

That’s why the Republicans are trying to limit new Democratic registrations. GOP governors seek to limit early voting and impose strict voter ID requirements. But the GOP knows that disenfranchising 5 million voters won’t be enough for Romney to win. You see, there is this thing called the *red-shift*. Otherwise known as the *fraud factor*. Stay tuned.

Election Model Forecast; Post-election True Vote Model

2004 Election Model (2-party shares)

Kerry 51.8%, 337 EV (snapshot)

State exit poll aggregate: 51.7%, 337 EV

Recorded Vote: 48.3%, 255 EV

True Vote Model: 53.6%, 364 EV

2008 Election Model

Obama 53.1%, 365.3 EV (simulation mean);

Recorded: 52.9%, 365 EV

State exit poll aggregate: 58.0%, 420 EV

True Vote Model: 58.0%, 420 EV

2012 Election Model (2-party shares)

Obama 51.6%, 332 EV (Snapshot)

Recorded : 51.6%, 332 EV

True Vote 55.2%, 380 EV

helenofmarlowe

October 6, 2012 at 10:20 pm

Do you know of any good reason why we cannot, in the US, count paper votes by hand as they still do in Britain, Sweden, and a number of other countries? Certainly here are more voters in US but not more in the precincts. Why not have each precinct hand-count paper ballots?

Richard Charnin

October 7, 2012 at 2:12 am

Good question. Ask your congressman. But it won’t do any good. They like it just the way it is.