# Re: Bush/Kerry optimal wagers

From: John Conover <john@email.johncon.com>
Subject: Re: Bush/Kerry optimal wagers
Date: 8 Oct 2004 01:03:10 -0000

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The new Reuters/Zogby poll, (they are about the most reputable,) was
just released:

and Bush has a 46% to 44% advantage over Kerry in the popular vote as
of today, with a +/- 3% statistical estimate in the sampling accuracy.

Using that data, and considering the election is 25 days away, (or the
distribution on election day would be about the sqrt (25) = 5%,) Bush
would have a 2 / 5 = 0.4 standard deviation of winning, or about 66%,
(note that the chances of Bush winning increased since September 21,
when he had a 3% advantage over Kerry, and only a 2% today.)

Including the +/- 3% uncertainty of the sampling error, (and using a
Cauchy distribution for a worst case-not so formal-estimate,) the
"standard deviation" would be +/- 8%, Bush would still have a 2 / 8 =
1 / 4 standard deviation, (we really should use the median and
interquartiles,) or about a 60% chance of winning. (Considering the
distributions to be a Cauchy distribution-we add them linearly since
Cauchy distributions are highly-actually the maximum-leptokurtotic,
with a fractal dimension of 1.)

Or, if we consider the distributions to be Gaussian/Normal, (i.e., we
add the root-mean-square of the distributions root mean square; it has
a fractal dimension of 2,) then the distribution on election day would
be sqrt (3^2 + 5^2) = 5.8, or about 6%. Then Bush's chance of winning
would be 2 / 6 = 1 / 3 standard distribution, or about 63%.

So, Bush has between a 60 and 63 percent chance of winning, (these are
the two limits-between no leptokurtosis and maximum leptokurtosis,)
probably laying closer to 63%, including the uncertainty of the
sampling error, and making no assumptions about leptokurtosis.

Bear in mind that a 60% chance of winning is, by no means, a slam
dunk.

John

John Conover writes:
>
> A note about the polls. The current polls show the Presidential race
> for the popular vote to be a dead heat between Bush and Kerry, (well,
> sort of, with a +/- 3 points margin of error,) at 48% of the popular
> vote, each.
>
> The +/- 3 points margin of error is a statistical estimate of the
> accuracy of the poll; its the estimated error created by sampling the
> population-which is used to generate the poll numbers.  It means that
> the "real" value of the aggregate population lies between a 52% for
> one candidate and 46% for the other, for one standard deviation of the
> time, or about 68% of the time, (and it would be more than a 52% win
> and less than a 46% loss for 16% of the time for one candidate, and
> 16% of the time for the other, too.) See:
>
>     http://www.johncon.com/ndustrix/utilities/tsstatest.txt
>
> for particulars.
>
> So, what does it mean for the election 25 days away?
>
> The statistical estimate is an uncertainty, and the uncertainty will
> increase by about sqrt (25) = 5% by election day, (relative to today,)
> meaning that the standard deviation on election day will be +/- 8%, or
> there is a 16% chance that Bush will win by more than 57% to Kerry's
> 41%, and likewise for Kerry. The chances of them being within +/- 1%
> on election day is one eighth of a standard deviation, or about 10%,
> (one eighth of a standard deviation is about 5%, but it can be for
> either candidate, or +/- 1%.)
>
> With probabilities that close, if either, or both, candidates feel
> losing the election will be an insurmountable loss, then its a
> Game-Theoretic "Mexican Standoff," which has an optimal solution to
> shoot first; meaning it will get ugly with last minute mud-slinging,
> (i.e., there is a 10% chance that moving the polls by as little as 1%
> in 25 days will win the election.)
>
>         John
>
> BTW, the reason I mention a method for handling the uncertainty of the
> poll data is that with 49% Bush, 49% for Kerry, and 2% for the third
> party, with 5% undecided, (which sums up to more than 100%,) not to
> mention that 2% +/- 3% means that there is a 25% chance, (2 / 3
> standard deviation,) that the third party could end up with a negative
> number of votes if the election were held today. All of which do not
> lend a lot of credibility to the data.
>
> John Conover writes:
> >
> > I swore after the 2K election that I would not do this again, but here
> > is the way you model the outcome of an election based on poll data; I
> > got beat up when I predicted-5 days before the election-that the 2K
> > Presidential Election would have an inconsistent popular and electoral
> > vote:
> >
> >     http://www.johncon.com/john/correspondence/001103172714.28588.html
> >
> > See:
> >
> >     http://www.johncon.com/john/correspondence/020508170137.5425.html#appendixIII
> >
> > for details of the methodology, (using the California Gubernatorial
> > recall as an example.)
> >
> > The recent polls say 46% of the voters, (in the popular vote,) would
> > vote for Bush, 43% for Kerry. There are 43 days left to the election,
> > and the polls move at about 1% per day, and are a zero-sum game, (in
> > the simple/lay sense-what Bush gets in a poll on one day, Kerry loses,
> > and vice verse.)
> >
> > What that means is that sqrt (43) = 6.5574385243% is the standard
> > deviation of the popular vote poll data on election day-43 days from
> > now-or there is a 14% chance that Bush's poll data will be above 46 +
> > 6.5574385243 = 52.6%, and a 14% chance of being below 46 -
> > 6.5574385243 = 39.4%. The numbers for Kerry on election day is a 14%
> > chance of being above 43 + 6.5574385243 = 49.6% and a 14% chance of
> > being below 43 - 6.5574385243 = 36.4.
> >
> > Since it is a zero-sum game, the winner of the popular vote will have
> > to have more than 43 + ((46 - 43) / 2 ) = 44.5%, or a 1.5 /
> > 6.5574385243 = 0.228747855 standard deviations.
> >
> > Referring to the CRC tables for the standard deviation, 0.228747855
> > corresponds to about a 59% chance for Bush, and a 41% chance for
> > Kerry, to win-call it a 60/40 chance, 43 days from now.
> >
> > So, going long on Bush, one would optimally bet a fraction (2 * 0.6) -
> > 1 = 20% of one's capital, (be it political, money, whatever,) on Bush
> > winning. And, the expected returns would be 2%, per election, iterated
> > over elections with similar statistical characteristics, from:
> >
> >     http://www.johncon.com/john/correspondence/020213233852.26478.html#equation1.20
> >
> > which is a 10% ROI in 43 days-not too bad, (that's equivalent to about
> > 2X per year.)
> >
> > Of course, there is uncertainty in the poll numbers, and a strategy of
> > watching the numbers to do a stop-limit, (e.g., betting on Kerry later
> > to go short on the Bush wager,) may be advisable-just like in options,
> > and possibly set up a no risk wager, (working backward on implied
> > volatility.)
> >
--

John Conover, john@email.johncon.com, http://www.johncon.com/

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