# Re: Bush/Kerry optimal wagers

From: John Conover <john@email.johncon.com>
Subject: Re: Bush/Kerry optimal wagers
Date: 28 Oct 2004 04:42:45 -0000

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There are 11 States in question in the Bush vs. Kerry run for the
Presidency in 2004, (these are the average of three poles, rounded to
nearest percentage integer-NV, PA, and AZ are considered decided):

State     Bush over Kerry
FL        +4
OH        +4
WI        +4
IA        +4
MN        +2
MI         0
HI        +2
NM        +5
CO        +5
NH        -5
NJ         0

Considering the polls to be a simple zero sum game, (i.e., what Bush
loses, Kerry wins, and vice versa, and ignoring utility,) from the
median where the poll would "cross":

State     Bush    Kerry
FL        +2      -2
OH        +2      -2
WI        +2      -2
IA        +2      -2
MN        +1      -1
MI         0       0
HI        +1      -1
NM        +2.5    -2.5
CO        +2.5    -2.5
NH        -2.5    +2.5
NJ         0       0

There are 7 days to election day, November 2, and the polls move about
a percent a day, so the standard deviation at 7 days from now would be
about sqrt (7) = 2.65. Converting to Bush's chance of winning by
taking Bush's numbers, as a fraction of the standard deviation on
November 2, and looking up the normal probability:

State     Bush's Chance of Winning State
FL        0.76
OH        0.76
WI        0.76
IA        0.76
MN        0.65
MI        0.50
HI        0.65
NM        0.83
CO        0.83
NH        0.17
NJ        0.50
--        ----
11        7.17

Or the chances are that Bush will win 7 of the 11 States in
question. (Note that Bush's percentage of the popular vote, at about
60% has not changed since mid September and 7.17 / 11 = 65%, which is

FL        27              * 0.76   = 20.5
OH        20              * 0.76   = 15.2
WI        10              * 0.76   =  7.6
IA         7              * 0.76   =  5.3
MN        10              * 0.65   =  6.5
MI        17              * 0.50   =  8.5
HI         4              * 0.65   =  2.6
NM         5              * 0.83   =  4.2
CO         9              * 0.83   =  7.5
NH         4              * 0.17   =  0.7
NJ        15              * 0.50   =  7.5
--       ---                         ----
11       128                         86.1

Bush currently has 213 electoral votes, against Kerry's 171,
(excluding the State's in question,) and it takes 271 to win.

So, Bush should win with about 299, or about 55% of the electoral

John

BTW, rather than base a projection on only 1 or 2 states, (like OH and
FL,) I increased the number of states in question, averaged the polls
of those states, and chose those states that were marginal, and then
worked with all 11 of them. Its an attempt to get the "granularity"
up, so the projection is not based on only states with large electoral
votes. Just winning FL or OH is not enough in a worst-case-scenario,
(for example FL, OH, and MI, have 64 electoral votes-the remainder of
the 11 have 64 which would still give Bush the win, if he got them
all.)

As an estimation, with a 55% probability of winning, one would
optimally wager 2 * 0.55 - 1 = 0.1 = 10% of one's capital on Bush
winning. In the long run, making a lot of similar such wagers, the
expected returns would be about 0.5% per iteration. (Since
Presidential elections occur every 4 years, that's about an eighth of
a percent a year-a savings account would give better ROI, so its not
worth the effort; except for political zealots.)

--

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

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