2004 Projected Presidential Electoral Votes as of 3/15/2004

TradeSports.com ^ | Monday, March 15, 2004 | Momaw Nadon

[painter]

Bush could lose FL and MO

Bush ISN"T going to lose in Missouri!!

[Wallaby]

The probability of Bush winning at least 270 electoral votes is above .6798.

[SuziQ]

No time to relax.

We will be able to relax on Wed. after the election, unless the Dem Vote Fraud Machine is working overtime this year.

[Political Junkie Too]

When I ran mine, it was lower. At first, I ran 200 iterations, then I bumped it up to 500. Also, I ran one using these probabilities, and another using Dales' probabilities (slightly different). I can't keep them straight now.

However, my expected value matches yours. I'll have to try it again over the weekend with more iterations and see why we're so different. One reason may be that I'm not using a Monte Carlo tool, I'm using home-grown randomization in MS Access. I'll try using some decision tools and see how they differ. You'll see from earlier posts that a 51-node decision tree is too much for my tools. I'll try running an analysis using my 9 supernode tree and see what comes out.

-PJ

[Wallaby]

In other words, Bush gets at least 270 electoral votes 67.98% of the time. The average sum of his electoral votes is 288.87.

[airborne]

If the Bush campain can put a major effort together to check to registration rosters in Philly and Pittsburgh to keep the vote fraud to a minimum, Bush has a chance. The Specter/Toomey primary results will be a good indicator of the mood of the Republican voters here in PA. I know many of my friends are unhappy with Specter, but Arlen has alot of clout , and he can deliver the bacon. I'm voting for Pat.

[Wallaby]

I'm running the thing in LISP using LispWorks. If you'd like to take a look at it, I could post the program.

[Political Junkie Too]

I'm going to run some decision analysis tools designed for S-curves and tornado diagrams.

Can't do it now, though. Gotta run. I'll try tonight.

-PJ

[Rome2000]

Kerry is such a douchebag that results like those below are entirely possible by November.

 

1980

85,496,851 Total Votes.  52.6% of Eligible Voters.

Candidate	Party	Popular Vote	Electoral College	Percent
Ronald Reagan (CA) George Bush (TX)	Republican	43,901,812	489	50.8%
Jimmy Carter (GA) Walter Mondale (MN)	Democrat	35,483,820	49	41.0%
John Anderson (IL)	Independent	5,719,437	0	6.6%
Ed Clark (CA)	Libertarian	921,299	0	1.1%

National Conventions

Republican: Detroit, MI 7/14/80
Democrat:  New York, NY 8/11/80

Ronald Reagan (R): 489  Jimmy Carter (D):  49

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1984
"It's morning again in America"

92,641,042 Total Votes.  53.1% of Eligible Voters.

Candidate	Party	Popular Vote	Electoral College	Percent
Ronald Reagan (CA) George Bush (TX)	Republican	54,455,000	525	58.8%
Walter Mondale (MN) Geraldine Ferraro (NY)	Democrat	37,577,000	13	40.5%
David Berland (CA)	Libertarian	228,314	0
Lyndon LaRouche (VA)	Independent	78,807	0

National Conventions

Republican:  Dallas, TX 8/20/84
Democrat:  San Francisco, CA 7/16/84

Ronald Reagan (R): 525  Walter Mondale (D):  13

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1988
 "Read My Lips, No New Taxes"

TOTAL: 91,594,136 Total Votes.  50.1% of Eligible Voters.

Candidate	Party	Popular Vote	Electoral College	Percent
George H.W. Bush (TX) James Danforth Quayle (IN)	Republican	48,886,097	426	53.4%
Michael Dukakis (MA) Lloyd Bentson (MN)	Democrat	41,809,074	111	45.6%
Lloyd Bentson (MN)** Michael Dukakis (MA)	Democrat		1
Ron Paul (TX)	Libertarian	431,616	0
Lenore Fulani (NY)	New Alliance	217,200	0
David Duke (LA)	Populist	46,910	0
Eugene McCarthy (MN)	Consumer	30,903	0

**One elector from West Virginia voted for Lloyd Bentsen for President and Michael Dukakis for Vice President.

National Conventions:

Republican:  New Orleans, LA 08/15/1988
Democrat:  Atlanta, GA 07/18/1988

George Bush (R): 426   Michael Dukakis (D): 111  Lloyd Bentson (D): 1

 

 

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[painter]

"Bob Taft has been a disaster to the GOP there. He has raised the taxes to the point that the SOS is getting signatures for a repeal."

Maybe that's why Teddy Roosevelt ran against his great grandfather in 1912.

[jaugust]

SO WHAT?!?

[dirtboy]

Pennsylvania is shaping up to be a key state for this election. I'd better get my fat a** signed up in the state Bush campaign.

[StevieB]

Bush may be gaining ground in Idaho, but it's going to get hard to make much more progress. Get's really difficult to move that 95 winning chance percentage any higher!

[Lonesome in Massachussets]

Yeah, I take it all back. When I calculated variance for a particular state I calculated it as:

VAR(state) = Nv(state)*p(state)*(1-p(state)),

where:

Nv(state) is the number of electoral college votes for a given state,

p(state) is probability that Bush carries that state.

It should be:>

VAR(state) = Nv(state)*Nv(state)*p(state)*(1-p(state)),

Then the expected number of votes for Bush is 288.88 votes, the variance 3027.5 and the standard deviation is 55.03. If one assumes that the sum of the votes is nearly normally distributed – central limit theorem and all that, then the error function can be used to approximated the probability that Bush will receive 270 or more votes. 288.88 – 270 = 18.88. 18.88/55.03 = 0.3431 standard deviations. The estimate of the probability of less than 270 votes is:

P(Bush Votes >=270) = .5+erf(.3431/sqrt(2))/2 = 0.6342

A simple exact analytic solution is available to calculate the sum of discrete independent random variables, as we are assuming in this model. (A binary tree would require 2^51 –1 nodes.) If you represent the outcomes in each state as a polynomial in x of the form:

P(state) = p(state)*x^(Nv(state)+ (1-p(state))

Then the probability distribution of the sum is represented by the product of all the polynomials. In this case it’s a polynomial in x to the 538. You can implement it very easily in matlab as follows:

% Nv(state) contains the number of electoral votes in each state
% p(state) is the assume probability that Bush will carry the state, don’t forget to divide by 100!
%
P = 1;
for k = 1:51%
P = conv(P, [1- p(k),zeros(1,Nv(k)-1),p(k)]);%
end
CDF = 1- conv(P, ones(size(P))); %
CDF = CDF(1:length(P)); %
%
%
%
%

P represents the discrete probability density of the resulting sum. P(n+1) = probability that Bush will receive exactly n votes, given this model. The average number of votes is the dot product of a vector

V = [0, 1, 2, 3… 538]

with P

V dot P = 288.88.

The variance equal to the

[V-288.88]^2 dot P = 1524.866

Standard Deviation = SQRT(VAR) = 39.0495

CDF(n+1) represents the discrete probability density of the resulting sum. CDF(n+1) = probability that Bush will receive at least n votes, given this model.

CDF(270+1) = Probability that Bush will win = 0.6701

I also did 1,000,000 mock elections, given these probabilities. GWB won 680,228 times, he got an average of 288.89 votes with a standard deviation of 39.06 votes, in good agreement with the analysis.

[Wallaby]

Why do you suppose there is still a discrepancy of one percentage point in the estimates for the probability of a Bush win by the analysis you gave and the simulations?

[Political Junkie Too]

I ran my decision analysis using Risk Detecitve. I made a 9-supernode tree, where the first eight nodes contain 5 states each, grouped by descending electoral vote count, and the 9th node taking the rest (11 states).

I essentially take the states in groups of 5 at a time, run the analysis on the group, and then simulate the cumulative probability distribution with its 10/50/90 numbers each getting 25%/50%/25% probabilities. I then run the 9 nodes through an analysis which resulted in 19,683 scenarios. I'd like to try running each states as its own node, but that may create too many endpoints to resolve. I will try using Supertree later to see if that helps.

I take back what I said earlier about the 270 vote probability. My analysis now shows a 32% probability of Bush getting lower than 270 votes. My expected value of electoral votes is 290.75, with P10=242, P50=287, P90=346.

The statistical stuff (which I don't really understand) is: EV=290.75, Standard Deviation=39.37242, Variance=1550.1875, Coef. of Skewness=0.27, Coef. of Kurtosis=8893.

-PJ

[Lonesome in Massachussets]

Beats me. Not worth looking into more deeply.

[Wallaby]

I suspect that your slightly lower probability of a Bush win (67% instead of 68%) is due to an error of requiring more than 270 votes, rather than at least 270 votes. When I ran the simulation using that incorrectly higher requirement for victory, my results matched your analysis for the probability.

[Lonesome in Massachussets]

Wallaby is absolutely correct.

The way I calculate CDF, CDF(n) is equal to the probability that Bush equal or exceed n. So it is consistent with the analysis. BTW, while we shouldn't take this too seriously, R.W. Hamming, Professor at the Naval War College, made the point that the purpose of scientific computations was insight, not answers. (Of course poor analysis or simulation lead to misleading insights.)

One interesting insight is that there is an approximately 1-percent (0.0098) chance that the race will end in an electoral college tie 269-269. I believe the House has decided the election only once before, selecting Thomas Jefferson over Aaron Burr in 1800. The House votes by "delegation" each state getting one vote. Since the delegations are overwhelmingly Republican, this sort of means we should likely award the election to W in case of tie.

If the election is close it is likely that Kerry will have the majority of the popular vote since Democrats are more concentrated in a few populous states which they win by large margins. It'll be SO much fun to hear the NYT and CNN squeal.

[Political Junkie Too]

the point that the purpose of scientific computations was insight, not answers. (Of course poor analysis or simulation lead to misleading insights.)

Last night, I reran my analysis several time. Even Supertree couldn't handle a 51-node tree. I was able to break down the tree into two 20-node trees and one 11-node tree, grouped by descending electoral vote count. It was a time-consuming process, and the results weren't that different from Monte Carlo sampling, so I'll stick to Monte Carlo from now on.

I modified my MS Access program to run 10,000 samples (I'll run more once I get a faster computer). The prior stats were based on the tradesports.com probabilities from this thread, where Bush's probability of reaching 270 EVs is about 67%. I also ran the simulation using Dales' predictions (on my implied probability scale documented in his thread Electoral College Breakdown 2004, March 17th Update). Dales' probabilities are not as Bush-optimistic as tradesports.com, so the probability of Bush reaching 270 in that simulation was about 59%. I will run new simulations as Dales updates his assessments.

Anyway, using these tradesports.com probabilities, I looked at the states that Bush lost that he should have won (where his probability was >= .5) for simulated elections where he failed to reach 270 electoral votes. What I found was the following:

Bush lost Florida (27 EV) in 60.68% of the losing elections.
Bush lost Ohio (20 EV) in 48.2% of the losing elections.
Bush lost Missouri (11 EV) in 44.74% of the losing elections.
Bush lost Arizona (10 EV) in 33.89% of the losing elections.

I didn't look at combinations yet, to see how often he loses both Florida and Ohio (and perhaps Missouri). However, this seems to be the conventional wisdom without the sampling -- that Bush has to win either Florida or Ohio to win.

I also haven't looked at states that Bush won where he should have lost, in elections that he won. I wonder what that would show?

-PJ