Mathematically, Dems have 99.97% chance of winning the House

[Thorgrim]

Using Pollster.com numbers, there are 54 GOP seats within 6pts and the MoE...toss ups or lean Dems

The odds of getting 15 of 54 coin-flips to go your way are 99.97%
Binomial Distribution: Probability Calculator

[SpicyMcVoodoo]

Then why does tradesports have them at about an 80% chance of winning?

[Thorgrim]

Originally Posted by SpicyMcVoodoo

Then why does tradesports have them at about an 80% chance of winning?

They have GOP control chances at 17%, and thats only because people don't want to sell their shares who bought them back in the spring...they'd only get back a third of their money, better to let it ride and hope that they double it

Which bring me to an interesting point:

TS/IT have the investors putting GOP control at 17%

[SpicyMcVoodoo]

Originally Posted by Thorgrim

They have GOP control chances at 17%, and thats only because people don't want to sell their shares who bought them back in the spring...they'd only get back a third of their money, better to let it ride and hope that they double it

Which bring me to an interesting point:

TS/IT have the investors putting GOP control at 17%

Well that's pretty interesting if you're right...that it really is a 99% chance but because of whatever inefficiency, it's stuck at 80% chance. That's a hell of a money making opportunity if at the near end to every election, the prices are off by 20 points are more from their true value. Something tells me, though, if it were that simple, the difference would've been arbitraged away already. I mean, if it really is 99.9%, that's free money on the table for anyone who wants it.

[Thorgrim]

Originally Posted by SpicyMcVoodoo

Well that's pretty interesting if you're right...that it really is a 99% chance but because of whatever inefficiency, it's stuck at 80% chance. That's a hell of a money making opportunity if at the near end to every election, the prices are off by 20 points are more from their true value. Something tells me, though, if it were that simple, the difference would've been arbitraged away already. I mean, if it really is 99.9%, that's free money on the table for anyone who wants it.

It really is...the title on MSNBC last night, their main story, was "Question not if the Democrats will take the House, but by how much?" and every pundit I've seen, even blood-red conservatives, have said the exact same thing

Leaked information says that the GOP at every level has been telling its congressional members they're going to lose "at least 20 seats"

However, the "buy" for a 15 seat gain is 82.5...meaning for every 100 dollars you put down that you could lose EVERYTHING on, you only gain 18 bucks (at least I think thats how it would work)

Pretty much everyone thinks "ooo easy money" and if I had money I'd plunk it down there, but people think two things:
1) For risking 100 dollars, im only gaining 18, and risk is risk, no matter how small
2) There are a million things that could go wrong with electronic voting that could mess up the elections so badly i'd lose
3) I have a mystical fear of Karl Rove and the GOP GOTV machine

And actually, those aren't ridiculous reasons, #2 is why I am not taking out loans to buy up huge quanities of shares on these future markets

The electronic voting could be a disaster, and I'm not going to end up thousands of dollars into debt because of something I knew might come

[SpicyMcVoodoo]

Originally Posted by Thorgrim

Pretty much everyone thinks "ooo easy money" and if I had money I'd plunk it down there, but people think two things:
1) For risking 100 dollars, im only gaining 18, and risk is risk, no matter how small
2) There are a million things that could go wrong with electronic voting that could mess up the elections so badly i'd lose
3) I have a mystical fear of Karl Rove and the GOP GOTV machine

Well then maybe ~83% chance of winning is right. Maybe it's 99.9% right now per the polls, but once you factor in those unknown risks above, it's 83%. I mean, you have to factor in the risk of a terrorist attack before the election, Bin laden being caught, Saddam's guilty verdict impacting things, etc, etc...all of which are not incorporated into that 99.9% figure.

[Thorgrim]

Originally Posted by SpicyMcVoodoo

Well then maybe ~83% chance of winning is right. Maybe it's 99.9% right now per the polls, but once you factor in those unknown risks above, it's 83%. I mean, you have to factor in the risk of a terrorist attack before the election, Bin laden being caught, Saddam's guilty verdict impacting things, etc, etc...all of which are not incorporated into that 99.9% figure.

I really doubt Saddam's verdict is going to change any minds when they hear about it the day before the election, because
1) there will probably be more violence
2) there is an appeal process, so it isn't final
3) he won't actually be killed

Most importantly
4) It is such a small factor to Americans, and the virtually everyone has formed their opinions about whether or not they will turnout over the war and how they will vote by Nov 6th

[motivez]

What does the math say for the Senate?

[Thorgrim]

Originally Posted by motivez

What does the math say for the Senate?

the math i am using rests on massive amounts of coin flips, the senate races don't have coin-flips like that

VA, MO and maybe MT could be "coin flips" but honestly its more like

A) The wave is small and most tossups remain R
B) The wave is basically large, and VA, MO and MT all flip
C) The wave is huge, and add AZ and TN to that list

B gives a 51 seat majority, C gives a 53 seat majority (no lieberthreats)

All the math in the world (currently available) can't tell you what kind of wave this is going to be...

In 1994 almost no one called a GOP takeover, but did acknowledge a "wave" the few that did, thought it would be a 40+ seat takeover...huge, but barely making it...it was a 54+ seat blowout that ended up +60 when the dust settled

[Ardentfrost]

Nice use of statistics I thought about that a while back, but was too lazy to look up real numbers

[Phantom]

I guess that's a fair stat if you assume all those races are perfect toss-ups, but that's quite an enormous assumption.

[Ardentfrost]

Originally Posted by Phantom

I guess that's a fair stat if you assume all those races are perfect toss-ups, but that's quite an enormous assumption.

Well, he's obviously basing it on the polls. That's what every news agency bases their predictions on, so really that's the best guess you can muster.

[nbiggershaft]

Thats assuming the races are independent trials. Chances are the percent chance for some races to go republican are highly or atleast somewhat related to the chances of other races going republican. So I doubt you can really directly multiply them.

[Ardentfrost]

Originally Posted by nbiggershaft

Thats assuming the races are independent trials. Chances are the percent chance for some races to go republican are highly or atleast somewhat related to the chances of other races going republican. So I doubt you can really directly multiply them.

they're independant events. The outcome in state A has nothing to do with the outcome in state B. That's why the binomial distribution works. If memory serves, binomial distribution is defined as n independent Bernoulli (only 2 choices) trials each with a constant probabilty of success finds the probability of success in a fixed number of trials.

ie. X ~ Bin(n,p) where n is the number of trials and p is the probability.

So he's saying that there are 54 (that's n) seats that are within 6 points (poll numbers, we just have to accept them because it's all we have) of going from GOP to DEM (that is, all the same, all iid) and he's rounding to say 6 points isn't enough of a spread to ensure victory one way or the other, therefore they're tossups, or understood to be 50% chance of going one way or the other (that's p).

So you have X ~ Bin(54, 50%). E(X)=n*p=27 (expected value), but that's not what he's really looking for... he says dems NEED 15. So the pdf if P(X=x)=(n choose x)p*(1-p)^(n-x). All you do is plug that in from 0-15, take 1-answer and you have the total probability that at least 15 of these tossup seats will change hands. But since the expected value is 27, that's where there's a 50% chance it'll be higher (>=) and a 50% chance it'll be lower (<=), so you can see how the chance that it'll be lower than 15 decreases quickly.

Actually, I just realized I can complete the math...

X ~ Bin(n,p) can be approximated by the following:
P(X<=x) = Iota([x+.5-n*p]/SQRT[np(1-p)])
P(X>=15) = 1 - Iota([15 + .5 - 54 * .5]/SQRT[54 * .5 (1 - .5)])
P(X>=15) = 1 - Iota(-11.5/SQRT[13.5])
P(X>=15) = 1 - Iota(-3.1299)
--Iota is a table lookup, so I gotta get my stats book here, one sec--
P(X>=15) = 1 - .0009
P(X>=15) = 99.91%

My tables only go to 2 sigfig, so I did -3.13 for Iota, but since -3.11 through -3.13 are all .0009, I'd say it's not affected.

[Phantom]

Originally Posted by Ardentfrost

Well, he's obviously basing it on the polls. That's what every news agency bases their predictions on, so really that's the best guess you can muster.

that's all good, but the error is in assuming that a race within the MOE is a perfect toss-up. I don't think anyone thinks a race where one candidate is up 4 pts, although within the MOE, is a perfect toss-up. If you put faith in the polls (which one must do for any of this to be relevant), there's a much greater chance the candidate up by 4 will win the race.

[Thorgrim]

Originally Posted by Phantom

that's all good, but the error is in assuming that a race within the MOE is a perfect toss-up. I don't think anyone thinks a race where one candidate is up 4 pts, although within the MOE, is a perfect toss-up. If you put faith in the polls (which one must do for any of this to be relevant), there's a much greater chance the candidate up by 4 will win the race.

Not neccesarily, if an incumbent is up 46-42...that's 12% undecided...now who do you think they'd break for in a democratic year?

[Phantom]

Originally Posted by Thorgrim

Not neccesarily, if an incumbent is up 46-42...that's 12% undecided...now who do you think they'd break for in a democratic year?

it's really irrelvant whether we're talking dem or rep, the idea is that those races are not perfect tossups. For your math to work, there would have to be exactly two candidates in each race with exactly 50% chance of winning. I think that assumption is so far from reality that the results are meaningless.

Everyone is free to disagree though

[SpicyMcVoodoo]

Originally Posted by Phantom

I guess that's a fair stat if you assume all those races are perfect toss-ups, but that's quite an enormous assumption.

I think the problem with using polls to come up with probability is that polls don't account for the probability of future events. They just tell you what would happen if the election were the day of the poll. The only predictor which attempts to account for future events and present polls are futures markets.

[nbiggershaft]

Originally Posted by Ardentfrost

they're independant events. The outcome in state A has nothing to do with the outcome in state B. That's why the binomial distribution works. If memory serves, binomial distribution is defined as n independent Bernoulli (only 2 choices) trials each with a constant probabilty of success finds the probability of success in a fixed number of trials.

ie. X ~ Bin(n,p) where n is the number of trials and p is the probability.

So he's saying that there are 54 (that's n) seats that are within 6 points (poll numbers, we just have to accept them because it's all we have) of going from GOP to DEM (that is, all the same, all iid) and he's rounding to say 6 points isn't enough of a spread to ensure victory one way or the other, therefore they're tossups, or understood to be 50% chance of going one way or the other (that's p).

So you have X ~ Bin(54, 50%). E(X)=n*p=27 (expected value), but that's not what he's really looking for... he says dems NEED 15. So the pdf if P(X=x)=(n choose x)p*(1-p)^(n-x). All you do is plug that in from 0-15, take 1-answer and you have the total probability that at least 15 of these tossup seats will change hands. But since the expected value is 27, that's where there's a 50% chance it'll be higher (>=) and a 50% chance it'll be lower (<=), so you can see how the chance that it'll be lower than 15 decreases quickly.

Actually, I just realized I can complete the math...

X ~ Bin(n,p) can be approximated by the following:
P(X<=x) = Iota([x+.5-n*p]/SQRT[np(1-p)])
P(X>=15) = 1 - Iota([15 + .5 - 54 * .5]/SQRT[54 * .5 (1 - .5)])
P(X>=15) = 1 - Iota(-11.5/SQRT[13.5])
P(X>=15) = 1 - Iota(-3.1299)
--Iota is a table lookup, so I gotta get my stats book here, one sec--
P(X>=15) = 1 - .0009
P(X>=15) = 99.91%

My tables only go to 2 sigfig, so I did -3.13 for Iota, but since -3.11 through -3.13 are all .0009, I'd say it's not affected.

I know the math, but I doubt the favor towards republicans and democrats in one state is independent of the favor towards republicans in another. Or rather to put it better I doubt the favor towards the parties in the individual states are independent of the favor nationally. If for some reason on election day there is a 10 percent chance to love republicans. Or nationwide there is a 10 percent chance that republicans will show up at the polls in greater numbers than expected, then the states will sway one way or another together, not independently.

Example, say factors for a democratic senator are X, Y, and Z. X and Y are national issues, Z is a state issue. To simplify the math, we can say all the deviation on national issues between states can make Y a state issue to account for X and Y. Also we assume that a candidate must be agreed upon on all 3 conditions to be voted for to simplify.

So for state A for a republican senator lets say you have a local Z of .75 chance for approval a Y of .5 chance for approval and a X of .3 chance for approval.

For state B you have .74, .5, and .3 as well.

So be independent trials to elect both senators you would have a probability of of .012656.
If you assume the national issue X will sway one way or another for both you have a new probability that both will be elected .0421875 a 3 percent difference.