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2004 Projected Presidential Electoral Votes as of 6/21/2004
TradeSports.com ^ | Monday, June 21, 2004 | Momaw Nadon

Posted on 06/21/2004 3:20:15 PM PDT by Momaw Nadon

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To: Momaw Nadon; All
I'm wondering if pollsters are falling into the same trap old generals fall into, fighting the last war. My state, New Jersey, is in play at this point with about 4 months until Election Day. It shouldn't be.
Is it possible the "Battleground States" could have shifted? Could California in play for President Bush as a result of Gov. Swarzeneger(? spell).
It's important to view this as dynamic model. Am I playing the part of Captain Obvious? Maybe, but thinking outside the box in necessary to shake up the status quo. It keeps the mind fresh.
21 posted on 06/21/2004 9:28:38 PM PDT by olde north church (Never give a liberal an even break!)
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To: Political Junkie Too
The outcome in Ohio is not based on the outcome in Texas (no "So goes Maine, so goes the nation?"), which is my understanding of how dependent events works.

Ah, perhaps you do not understand what is meant by "independent events". Events are statistically independent if, and only if, the probability of both/all events occurring is equal to the product of the probabilities of each event occurring.

The notion of statistical independence has little to do with causality, and does not respect normal principles of transitivity. It is possible for events A and B to be independent, and for B and C to be independent, and likewise A and C, and yet for the three events not to be mutually independent of each other. For an example of this, suppose I have two dice, one red and one white, and I roll them. Consider the following three events:

  1. The red die shows an odd number.
  2. The white die shows an odd number.
  3. The sum of the two dice is an odd number.
Any two of these three events will be independent because the probability that e.g. the red die will show an odd number and the sum will also be odd, is 1/4. On the other hand, the three events are not mutually independent because the probability of all three events occurring is zero.
22 posted on 06/21/2004 9:35:28 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: supercat

But won't that dependence be reflected in the polls or the "markets" also?


23 posted on 06/21/2004 9:39:21 PM PDT by Doctor Stochastic (Vegetabilisch = chaotisch is der Charakter der Modernen. - Friedrich Schlegel)
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To: supercat
But your example creates a mutually exclusive scenario, i.e., the sum of two odd numbers is also an odd number. There is no such similar mutually exclusive scenario in the way that states vote.

-PJ

24 posted on 06/21/2004 9:44:39 PM PDT by Political Junkie Too (It's not safe yet to vote Democrat.)
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To: Doctor Stochastic
But won't that dependence be reflected in the polls or the "markets" also?

I'm sure it is reflected in the markets. I would be surprised if the wagered probability of Bush winning matched and tracked the conditional probability one would compute assumuming all state probabilities are accurate and independent.

25 posted on 06/21/2004 10:16:06 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: Political Junkie Too
But your example creates a mutually exclusive scenario, i.e., the sum of two odd numbers is also an odd number. There is no such similar mutually exclusive scenario in the way that states vote.

While it is theoretically possible that any of the 3*2^51 combinations of electoral votes could occur (the 3x being a consequence of Maine having six possibilites rather than 2: RR+R RD+R RD+D DR+R DR+D DD+D), the actual probabilities of certain combinations occurring do not necessarily track well with what one would combute by multiplying their individual probabilities. For example, what is the probability of all of the following states voting for Bush: Alabama, Alaska, Idaho, Indiana, Kansas, Kentucky, Mississippi, Montana, Nebraska, North Dakota, Oklahoma, South Carolina, South Dakota, Tennessee, Texas, Utah, Virginia, and Wyoming? I for one would consider that much better than an even-money wager, and yet if the state probabilities are to believed, the probability that all those states will voting for Bush would have to be about 40% for the probabilities to be independent.

26 posted on 06/21/2004 10:28:35 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: supercat
Isn't that why you do a decision tree and cumulative probability distribution to arrive at the expected value of the total electoral college vote? Those extremely unlikely combinations would result in the less than P05 or greater than P95 possibilities. Wouldn't the probability of the 270 electoral vote result become the probability of winning?

-PJ

27 posted on 06/21/2004 10:34:20 PM PDT by Political Junkie Too (It's not safe yet to vote Democrat.)
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To: Momaw Nadon
Date Prob. Bush Win Mean EVs Std. Dev.
01/21 96.8% 341.5 41.1
01/26 95.5% 334.8 40.6
02/02 92.2% 323.8 39.7
02/09 83.0% 307.8 40.3
02/16 78.4% 300.4 39.4
02/23 76.2% 298.2 39.6
03/01 74.5% 295.9 39.3
03/08 68.0% 289.2 39.8
03/15 68.0% 288.8 39.0
03/22 68.5% 289.3 38.8
03/29 69.4% 290.1 38.8
04/05 71.2% 292.3 39.1
04/12 70.4% 290.6 38.1
04/19 68.6% 288.1 36.7
04/26 64.9% 284.5 36.3
05/03 66.3% 285.7 36.3
05/10 65.6% 285.3 36.8
05/17 65.2% 284.8 36.6
05/24 60.0% 280.3 36.9
05/31 61.1% 281.2 36.8
06/07 60.5% 280.6 36.5
06/14 65.0% 285.0 36.6
06/21 63.9% 284.0 36.8

28 posted on 06/22/2004 7:47:16 AM PDT by jdege
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To: jdege

Thanks jdege!


29 posted on 06/22/2004 12:49:10 PM PDT by Momaw Nadon (Goals for 2004: Re-elect President Bush, over 60 Republicans in the Senate, and a Republican House.)
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To: Political Junkie Too
Isn't that why you do a decision tree and cumulative probability distribution to arrive at the expected value of the total electoral college vote? Those extremely unlikely combinations would result in the less than P05 or greater than P95 possibilities. Wouldn't the probability of the 270 electoral vote result become the probability of winning?

I started my paragraph with one thought and rather confusingly munged it with another. Let me ask you a few questions: What would you consider to be the probability that Bush wins all of the following states: Alabama, Alaska, Idaho, Indiana, Kansas, Kentucky, Mississippi, Montana, Nebraska, North Dakota, Oklahoma, South Carolina, South Dakota, Tennessee, Texas, Utah, Virginia, and Wyoming?

30 posted on 06/22/2004 5:30:11 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: supercat
Well, by multiplying the probabilities of just those states, I get 31.9% for all of them going to Bush. It only takes one state out of the 18 in the list to go the other way for the ALL proposition to fail.

Your point is that you'd bet better than even money that all states would go to Bush, meaning that the probability is better than the 32% that the numbers indicate. Are you thinking that because of an inherent dependency in the character of the states, for instance, that New England as a block would go for Kerry, that the South as a block would go for Bush, that the urbanized states as a block would go for Kerry, etc? Is that where you find your dependency? I'm thinking that that dependency is encoded into the individual state probabilities already.

To rephrase your question, out of 100 elections, how many times would you expect all of the states in your list to vote for Bush?

-PJ

31 posted on 06/22/2004 6:10:14 PM PDT by Political Junkie Too (It's not safe yet to vote Democrat.)
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To: Political Junkie Too
To rephrase your question, out of 100 elections, how many times would you expect all of the states in your list to vote for Bush?

Most of them--probably about 60% or so. For Bush to lose any of those states, something would have to go severely wrong for him campaign. While I wouldn't consider such an event impossible, I wouldn't regard it as being even a 50% probability.

I would posit that my simplified model of the election would regard each state as having a "Bush favorability requirement" and a random dither value, and I would regard the nation as having a "national Bush favorability rating". If the national Bush favorability rating exceeds the sum of a state's favorability requirement and its dither value multiplied by a random number -1 to +1, the state will vote for Bush; otherwise the state will vote for Kerry.

If a state like Nebraska votes for Kerry, it will most likely be because Bush messed up badly. A mistake large enough to cost Bush Nebraska would also likely cost him Virginia and Tennessee as well.

32 posted on 06/22/2004 7:17:35 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: Political Junkie Too
Let me ask you this question: which of the following events do you consider more likely [accepting that neither is very likely]
33 posted on 06/22/2004 8:18:54 PM PDT by supercat (Why is it that the more "gun safety" laws are passed, the less safe my guns seem?)
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To: supercat
Neither is very likely, given the probabilities.

The Georgia Loses scenario has a 0.005% probability of occuring, while the other scenario has a 0.003% probabiliy of occuring. That makes the Georgia Loses scenario more likely.

My gut instinct tells me that it is more likely to lose one state in a slate than it is to win all states in a slate, or paramutual betting would never survive as an industry.

-PJ

34 posted on 06/22/2004 11:50:01 PM PDT by Political Junkie Too (It's not safe yet to vote Democrat.)
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To: Political Junkie Too; supercat

I disagree. It would take a major surge in Bush's favor for him to have a prayer of winning Rhode Island. Such a surge would be broad-based and national, so much that he cannot possibly lose conservative pro-Bush Georgia.

In both scenarios, we must premise our evaluation on the assumption that Bush wins Hawaii (CERTAINLY KERRY in my estimation but not confirmed by polls), New Jersey (LIKELY KERRY), and Rhode Island (CERTAINLY KERRY).

So which is more likely? Bush wins Delaware (CERTAINLY KERRY based upon 2000 election results but unconfirmed by polling and previously a bellwether), and Washington (POSSIBLY KERRY, a swing state).

OR Bush loses Georgia (LIKELY BUSH--and almost certinly Bush unless Kerry takes Miller as veep).

The latter situation is more likely (provided that Delaware is as certainly for Kerry as we think it is), but only if we ignore our premise. But if we consider the premise, then we give up our "statistically independent" assumption used to derive the probabilities. That assumption presupposes that the state races for the electors for president occur independently of each other. But they don't.

If Kerry cannot hold Hawaii and Rhode Island, he probably will have collapsed entirely, and hence he certainly won't win Georgia. If on the other hand, Bush manages to persuade the liberal voters in Hawaii and Rhode Island to choose him over Bush, he definitely won't have trouble in Georgia, likely will win Washington, and may take Delaware.

Of course, I'm basing my prediction on the assumption of a two-way race, wherein minor-party candidates take only a small portion of the vote in the named states. If enough Kerry voters in Rhode Island, for example, defect to Nader either to throw the state to Bush or to give Nader an outright electoral victory, then a liberal shift to throw Georgia to Kerry (or Nader) might be rational to contemplate.


35 posted on 06/23/2004 8:04:53 PM PDT by dufekin (John F. Kerry. Irrational, improvident, backward, seditious.)
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To: dufekin
I agree that the state elections aren't independent in the sense that they all take place within the same environmental influences (the same media reports, the same news, the same economy, etc.) My point is that the impacts of that are encoded into the probabilities already.

It almost sounds like you are saying that the fact that a slate of states leans towards Kerry or Bush is the dependency in and of itself, given that if a news event sways one leaning state it is likely to sway them all. My point is that those states independently chose to lean that way in the first place, and it is almost coincidental that a slate of states has the same common leaning as its common bond. I doubt that there was some unmodeled dependency that caused the states, and only those states, to lean in the same direction, and therefore lean in another direction based on some event, together as a dependent block.

-PJ

36 posted on 06/23/2004 10:00:39 PM PDT by Political Junkie Too (It's not safe yet to vote Democrat.)
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