# Re: Bush/Kerry optimal wagers

From: John Conover <john@email.johncon.com>
Subject: Re: Bush/Kerry optimal wagers
Date: 8 Oct 2004 00:09:55 -0000

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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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