From: John Conover <john@email.johncon.com>

Subject: Re: Bush/Kerry optimal wagers

Date: 8 Oct 2004 01:03:10 -0000

The new Reuters/Zogby poll, (they are about the most reputable,) was just released: http://www.zogby.com/news/ReadNews.dbm?ID=877 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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