Al Gore is the most 'electable' at 98%, say the online odds markets

I don't like to bet (esp. with money), except for a rare lottery ticket or two and a rarer slot machine playing for fun.

But, I happened to check the online betting site Intrade for the 2008 election and found something very interesting. Link (click the "2008 US Election" link to the left).

They have the market prices for wins in the primary and the Presidency. For example, Hillary Clinton is trading at (bid):

2008.PRES.CLINTON(H) 47.6

Therefore, roughly speaking, the participants of Intrade odds market think that:
-- Hillary Clinton has a 71.6% chance of winning the nomination.
-- and she has a 47.6% chance of winning the Presidency, all told.

So, let's ask the question: what are HRC's chances of winning the general election if one assumes that she wins the nomination?

This quantity is a case of the notion of conditional probability. From the wiki:

Conditional probability is the probability (or chance) of some event A (taking place), given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the probability of A, given B".

In our case, if A is the event of candidate X winning the general, and B is event of X winning the primary, the P (A|B) is the probability/chance of X winning the Presidency given nomination, which can be considered to be a measure of "electability," arguably.

There is a related formal quantity called the joint probability:

Joint probability is the probability of two events in conjunction. That is, it is the probability of both events taking place. The joint probability of A and B is written P(A,B).

Since for a major party candidate, to win the Presidency, he/she needs to win both the primary and the general (3rd parties excepted), in our example, P(A,B) represents the probability of X winning the Presidency.

There is a simple formula that relates all of these quantities:

P(A|B) = P(A,B) / P(B)

Applied to our case, in other words: the chances of winning the general given the nomination = the ratio of winning the presidency over that of winning the primary.

Working out the numbers for HRC's case, we get:

the  chances of HRC winning the general given the nomination =
the ratio of the chances of winning the presidency over that of winning the primary =
47.6%/71.6% = 66%

Let's give the quantity "chances of winning the general given the nomination" a name: the Electability Index (and "Intrade Electability Index" when applied to numbers in a given snapshot at Intrade).

No, I don't like the word electability much, because it's been much abused in both the previous and the current presidential cycles. But, at least we have a solid definition of the word now!

I compiled the tables for 6 candidates from both of the major parties (plus Gore) from the snapshot of the Intrade market that I've found, and computed the Intrade Electability Indexes for each them. Here is how the numbers shake out:

As one can see, the table reveals some very interesting numbers.

On our side, the market currently thinks that Gore has the highest chance of winning the general election if nominated: 98%! Please see this diary by NYPopulist for some recent GE actual poll numbers for Gore.

Hillary and Obama follow at 66% and 60%, respectively. Edwards comes in (maybe surprisingly low or high depending on the person you ask) at 56% (I suspect that this number dropped for Edwards as a result of his decision to accept public funds.)

On the GOP side, Ron Paul tops the list at 55%, followed by McCain on our "Electability Index" scale here. I'm a bit surprised to see Huckabee's electability index come at a low 19%, but as one can see at the link, he tops the list of VP prospects for the GOP at 21% odds. Obama (21%) and Bayh (20%) top the VP list from our side.

Tags: 2008 elections, Al Gore, Intrade, Presidential Race (all tags)



Tips for inventing the 'Electability Index' :)

by NeuvoLiberal 2007-10-22 04:44AM | 0 recs
He's not going to run

the first week of november is the deadline to run in the NH primary,  I'm not sure about iowa.

If al doesn't register to run in the NH primary by the deadline will that convince you he's not running?

by TarHeel 2007-10-22 04:52AM | 0 recs
wah wah wah
why do you care? I think people can decide for themselves and it has very little to do with what you think or want. Actually Gore could miss the first few primaries and be just fine. If you were to let go and stop trying to control the situation (Gore people are not going to flock to Edwards, sorry) you would feel better.
by MollieBradford 2007-10-22 05:02AM | 0 recs
there are deadlines for states

at some point the number of remaining delegates will be less than needed to win and any candidate needs to be registered in those states.

for certain I would think Gore would need to be registered for the FEb. 5th states.

by TarHeel 2007-10-22 05:33AM | 0 recs
Re: wah wah wah

Many of the issues based Gore supporters are already backing Edwards, but that's beside the point. Gore's very unlikely to run and it's silly that some people are so obcessed over trying to get him to despite his statements that he doesn't plan to.

by Quinton 2007-10-22 06:30AM | 0 recs
Re: wah wah wah
Edwards is no Al Gore and you can find it as funny as you want. But he has never said he will not run. I am not going to argue with you about it, I just don't see why you all care so much. If you seriously think Gore supporters are going to flock to Edwards, you are fooling yourself.
by MollieBradford 2007-10-22 07:16AM | 0 recs
Re: wah wah wah

I agree with you here, Mollie.  A vast majority of current Gore supporters will be going over to Clinton instead of Edwards or Obama.

by georgep 2007-10-22 09:03AM | 0 recs
link please.

I'm a Gore supporter who donates to Obama's campaign who likes Dodd. You did see the poll where Al Gore did better than Hillary? No?

Gore beats Giuliani in Iowa
42%     Giuliani
51%     Gore
7%     Undecided rtEmail.aspx?g=2dbaa9be-dfa7-499e-9aa6-e 748067ea7ca

Gore can take Iowa 0601039&refer=columnist_hassett& sid=aQXnOvJRvTfs

by misscee 2007-10-22 05:33PM | 0 recs
Re: link please.

I fail to see what that has to do with my post?  

We have seen tons of evidence that tells us if Gore has x% support in a poll, then you remove Gore from the poll (either by re-asking the question or just using 2nd choices) Clinton ends up with a higher share of Gore's support than any other candidate.   So, logically, once Gore is categorically out of the picture (say, if he were to endorse someone for president) chances are good that Clinton gains the most from what we had seen previously.

by georgep 2007-10-22 10:48PM | 0 recs

my cut-off is Nov. 15, as by then a handful of state primaries close.

by NeuvoLiberal 2007-10-22 05:12AM | 0 recs

by TarHeel 2007-10-22 05:32AM | 0 recs
it's all a stat clas to me
one I would find hard to pass being a right brainer. But I like the implications. :)
by MollieBradford 2007-10-22 05:06AM | 0 recs

This reasoning is highly flawed.  The odds for  Gore to win the nomination are extremely low at 5.4% (update: sunk to 5.1%.)  His GE win odds are at an equally low 5.3%.  That means as of right now, for the GE there is a 94.7% chance he won't win the GE and a 94.6% chance he won't gain the nomination.

 Biden's nomination win odds are at a non-existent 0.3%, and his general win odds are at an even lower 0.2%, which in your mind means that he has a 67% chance to actually WIN the election, WERE he just nominated?    Balderdash.

BTW, Gore's primary numbers have just fallen to 5.1%.  Yet, his GE numbers are still at 5.3%.   Pulling out my trusted calculator that means that according to your logic he has now upped his probability of a GE win to a full 104%.  Imagine that.  A 6% swing upwards to even higher GE odds by seeing his PRIMARY odds fall by 15% within the span of a couple of hours.  Hey, if you wait a few more hours, Gore's primary numbers may fall even further, elevating his GE chances more.  You may get to 200% "electability" before the day is out.   :-)

But. thanks for giving us the opportunity to look at "real world" current Intrade trends for an update:

For the nomination:

Clinton 71%

Obama 11.2%

Gore 5.1%

Edwards 3.9%

Richardson 0.7%

Clinton is now just about 60% ahead of Obama's odds.

Edwards' odds are seen as astronomically low.  

If you look at the trends here, they are devastating for both Obama and Edwards.  The Intrade bettors have cut Edwards' already small chances from 7.5% down to half in the span of 10 days.  Obama's chances with Intrade bettors have taken a very dramatic nosedive from a high of almost 40% primary odds (just a tad below Clinton's) to the current 11.4%.


According to the Intrade betting community putting their money where their mouths are, Clinton's odds at winning the Democratic nomination over Obama are at higher than 6 to 1.  Her odds of winning the Democratic nomination over Edwards are at higher than 18 to 1.  

For the GE:

Odds to win the General Election:

Clinton 47.6%

Giuliani 16.8%

Romney 8.2%

Obama 6.8%

Gore 5.3%

Thompson 4.5%

Edwards 2.2%

Bettors are giving Clinton 3 to 1 odds to win the GE against Giuliani at this point.  Almost 6 to 1 odds over Romney. 10 to 1 odds over Thompson.   The other Democrats are not doing nearly as well as Clinton with Intrade bettors.  For example, Giuliani's odds to win the GE over Obama are nearly 3 to 1, and over Edwards almost 8 to 1.   Clinton is dwarfing everybody else when it comes to GE viability, with the nearest follower Giuliani's odds at winning the GE FAR FAR behind Clinton's odds (according to the betting community.)  

There is that GE electability almost everybody sees, right before your eyes, folks.  Remember that Intrade has a very good track record when it comes to its predictive qualities.  The reason behind that is that Intrade bettors are generally about the best informed people out there, as they generally do a lot of research and look at all angles before placing a high-Dollar bet.

by georgep 2007-10-22 05:56AM | 0 recs

In my GE odds section right below Clinton the Giuliani odds did not show up.  They are currently traded at 16.8%

by georgep 2007-10-22 05:59AM | 0 recs
Gore is not in the race. That's why his odds

of winning show a low number because it turns out, people track the developments and since Gore is not in the race, his odds of becoming the nominee won't be high (but they're still higher than Edwards' odds, eventhough JRE is a candidate and Gore isn't (at least yet)).

by NeuvoLiberal 2007-10-22 06:23AM | 0 recs
Re: Gore is not in the race. That's why his odds

But don't you see the flawed logic that as Gore's primary chances fall further that in your model his GE electability quotient goes up?  Say, his primary betting falls to 2.6% as a result of some of his primary odds money going to Clinton.  Yet, his GE quote stays the same.  Then, as a result of bettors losing faith in Gore for the primaries his GE "electability" rises to 200%.   I trust that you can see how flawed that line of reasoning is.  

by georgep 2007-10-22 06:34AM | 0 recs
No, it doesn't.

That's because the chances of winning the Presidency drop as well.

"Say, his primary betting falls to 2.6% as a result of some of his primary odds money going to Clinton.  Yet, his GE quote stays the same."

You can't assume that "his GE quote stays the same" because the GE quote is for winning the Presidency and assuming Gore would run as a Democrat if he does, then his GE quotes should go up or down adjusting to the changes in the situation (if those betting are seeing the same factors driving his nomination chances numbers go up or down).

Say, Gore decides to enter the race. Then, his nomination quote could go up to 40% (readjusting others' numbers in the process) and Presidency quote could both go up to 35%, eg.

In other words, both the numbers are elastic and are changeable with events.

by NeuvoLiberal 2007-10-22 06:59AM | 0 recs
Re: No, it doesn't.

No they really don't, or I should say not to the extent necessary to make a cogent argument.  The two metrics are independent from each other.  You can see that displayed as of right now, as I laid out.   Gore's primary odds dropped by 15% (0.3%) over the last few hours, while his GE odds stayed the same.   It had the effect that a drop in primary odds furthered his GE "electability," which is preposterous on its face and shows the inherent flaw in this argumentation.  

Dodd is at a devastatingly low 0.1% for the primaries.  If his GE chances were also pegged at a devastatingly low 0.1% according to your logic his GE "electability" would be at 100%, besting every candidate, Democrat or Republican as for "electability."  

  This type of reasoning makes no sense at all.

by georgep 2007-10-22 07:08AM | 0 recs
Re: No, it doesn't.

We should look at Gore's numbers at the end of the day. Since he is close to 100% on this scale, we can't look at instantaneous shifts.

The problem with Richardson/Dodd/Biden's numbers is rounding. 0.2 could be coming form 0.151, and 0.1 could be coming from 0.149 when rounded. Only numbers about 1% should be considered usable.

Essentially on our side, that means only HRC, Obama, Gore and Edwards.

by NeuvoLiberal 2007-10-22 07:24AM | 0 recs
'Only numbers above 1%'

by NeuvoLiberal 2007-10-22 07:29AM | 0 recs
Re: Al Gore is the most 'electable' at 98%, say th

I just  asked my wife this morning what my odds are of getting the democratic nomination.
"A trillion to one"

Then I asked her what my odds were of becoming president.
"A trillion to one"

There you have it!  According to this opinion poll, my electibility index is 100%!!

Write-in "Markjay" for democratic nominee if you really want to take back the Whitehouse!!!!

by markjay 2007-10-22 07:16AM | 0 recs
Re: Al Gore is the most 'electable' at 98%, say th

Markjay for president, the only candidate who shows a 100% GE electability quotient.   :-)  

by georgep 2007-10-22 09:00AM | 0 recs
So what? NuevoLiberal is still right

Let's say you were Jesus. We would all agree the chances of Jesus being our presidential nominee next year is infinitesimally small.  For the sake of argument, let's say it's 1 in a trillion (i.e. it's likely to happen once in every one trillion Democratic presidential primary seasons).  For the sake of argument, let's imagine that if Jesus were our nominee, it would be an unbeatable ticket, no matter who got picked to run as Jesus's VP, and no matter who the Republicans nominated.  But mathematically the chances of Jesus being elected cannot be higher than the chances of Jesus being nominated, since we're assuming Jesus is running on a party ticket. So the chances of Jesus being elected - assuming a Jesus ticket is unbeatable (100%) - would be precisely 1 in a trillion.  Just because it is an extremely tiny number, doesn't mean that Jesus doesn't indeed have the highest electibility of any potential Democratic candidate.

by Rob in Vermont 2007-10-22 07:31PM | 0 recs
Wrong math

Your math is wrong buddy.

The joint probability is the product of Event A and Event B, which means you have to multiply, not divide.

Therefore, the probability that said candidate can win both the nomination (Event A) and the General election (Event B) is as follows:

Hillary Clinton: 71.6% x 47.6% = 34.08%
Barack Obama: 11.4% x 6.8% = 0.78%
Al Gore: 5.4% x 5.3% = 0.29%
John Edwards: 3.9% x 2.2% = 0.09%

Richardson, Biden and Dodd are close to 0%.

Rudy Giuliani: 43.5% x 16.8% = 7.31%
Mitt Romney: 25.8% x 8.2% = 2.12%

All other Republican candidates get less than 1%.

So you had good intentions, but you flopped.

by RJEvans 2007-10-22 10:22AM | 0 recs
Re: Wrong math

I'm sorry, conditional probability, which means, Event B is dependent of Event A.

by RJEvans 2007-10-22 10:36AM | 0 recs
My math is fine

You're misreading the two columns.

They stand for:
-- odds of winning the nomination
-- odds of winning the Presidency

by NeuvoLiberal 2007-10-22 05:36PM | 0 recs
Re: My math is fine

How did I misread it? Did you read my comments?

by RJEvans 2007-10-22 07:02PM | 0 recs
Re: My math is fine

NuevoLiberal is correct.  The first hurdle is the nomination. Let's call your probability of winning the nomination P(n).  The next hurdle is the general election. Let's call your chances of winning that P(e). In order to become president, you must do both.  So the probability of winning the presidency, call it P(w), is:

P(w) = P(n) * P(e)

Your chances of beating your fellow Democrats is P(n).  Your chances of beating your Republican opponent - i.e. your electibility - is P(e).

P(e) = P(w) / P(n)

So NuevoLiberal plugs in P(w) and P(n) to solve for P(e).  Division is the proper operation here, not multiplication.

by Rob in Vermont 2007-10-22 07:48PM | 0 recs
Re: My math is fine

Well if you are using your formula, his numbers are still wrong:

Ex. Hillary Clinton

P(34.08%) = P(71.6%) * P(47.6%)

P(47.6%) = P(34.08%) / P(71.6%)

Which does not make any sense because P(e) is the same as the probability of winning the general election alone.

What he wants to do is find the probability of winning both the nomination and the general election because the general election is dependent on the nomination.

Your first equation is the one I used.

So in general, something is wrong.

by RJEvans 2007-10-22 08:38PM | 0 recs
No, nothing is wrong

My first equation and my second equation are the same equation. I've just isolated P(e) in order to solve for it.  Here, I'll show it step by step, algebraically:

p(w) = p(n) * p(e)

That part you agree with, right?  OK, now divide both sides of the equation by p(n):

p(w)/p(n) = (p(n)*p(e)) / p(n)

Since p(n) appears in both the numerator and demominator on the right, you can cancel it out:

p(w)/p(n) = p(e)

Now just flip it around:

p(e) = p(w) / p(n)

So to calculate Sen. Clinton's electibility, p(e), i.e. the chances she will beat the Republican opponent if she is nominated, we plug in her probability of winning the presidency, p(w), and her chances of winning the nomination, p(n).  According to NeuvoLiberal:

Hillary Clinton has a 71.6% chance of winning the nomination...and she has a 47.6% chance of winning the Presidency

So plugging in these numbers:

p(e) = .476 / .716 = .665

The problem you're having is you keep trying to plug in 34.1, which is an incorrect number, since you got it by applying an incorrect operation.

by Rob in Vermont 2007-10-22 09:05PM | 0 recs
Is our children learning math?

by Rob in Vermont 2007-10-22 01:55PM | 0 recs


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