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Topic: Electoral college probablities
Replies: 9   Last Post: Nov 2, 2000 6:02 PM

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-Mammel,L.H.

Posts: 90
Registered: 12/12/04
Re: Electoral college probablities
Posted: Nov 2, 2000 6:02 PM
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In article <8tngpg$bsk$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
[...]
>So the probability that the winning candidate loses in the popular vote
>is approximately F(-m s/sqrt(SN-(ms)^2)), where F is the standard normal
>cdf.
>
>For example, with S=51 and N=10^6 we would have s = 5.726033806
>and m = 797.8847608, and F(-m s/sqrt(SN-(ms)^2)) = F(-.8323716341)
>= .2025996041.
>
>Asymptotically, m = sqrt(2 N/Pi) and s = sqrt(2 S/Pi), so
>-m s/sqrt(SN - (ms)^2) -> -2/sqrt(Pi^2-4) = -.8255161606.


Very slick, I like it. Just as a reference point for
some of us who may be slightly inclined towards the
concrete, here is a simulation of 20 elections with
N=101, S=11, with the state by state and national totals,
and electoral result, each marked + or - accordingly
as the given candidate won or lost:

48- 57+ 41- 45- 44- 54+ 51+ 47- 51+ 42- 51+ 531- 5-
49- 48- 45- 54+ 48- 43- 53+ 56+ 48- 51+ 49- 544- 4-
50- 49- 45- 48- 48- 58+ 47- 58+ 55+ 55+ 52+ 565+ 5-
56+ 56+ 58+ 49- 41- 56+ 57+ 52+ 49- 43- 50- 567+ 6+
47- 44- 44- 39- 55+ 48- 58+ 47- 55+ 52+ 46- 535- 4-
53+ 46- 46- 47- 64+ 53+ 47- 62+ 49- 53+ 48- 568+ 5-
43- 49- 49- 54+ 55+ 47- 55+ 53+ 54+ 50- 39- 548- 5-
50- 55+ 57+ 47- 48- 60+ 56+ 54+ 53+ 54+ 49- 583+ 7+
47- 53+ 50- 63+ 44- 51+ 48- 57+ 53+ 56+ 57+ 579+ 7+
54+ 56+ 54+ 48- 53+ 50- 57+ 61+ 59+ 50- 53+ 595+ 8+
50- 52+ 52+ 50- 48- 57+ 60+ 58+ 50- 45- 52+ 574+ 6+
50- 51+ 57+ 58+ 40- 53+ 42- 50- 51+ 50- 38- 540- 5-
51+ 51+ 48- 44- 51+ 51+ 38- 52+ 43- 45- 48- 522- 5-
57+ 51+ 48- 55+ 50- 50- 56+ 50- 54+ 42- 45- 558+ 5-
49- 45- 45- 48- 45- 49- 51+ 50- 47- 51+ 55+ 535- 3-
53+ 40- 54+ 53+ 42- 44- 47- 50- 43- 50- 48- 524- 3-
57+ 45- 56+ 49- 59+ 49- 49- 52+ 49- 57+ 48- 570+ 5-
45- 48- 53+ 53+ 54+ 48- 38- 50- 52+ 59+ 54+ 554- 6+
53+ 46- 53+ 51+ 58+ 50- 57+ 51+ 49- 58+ 53+ 579+ 8+
55+ 45- 54+ 50- 56+ 49- 45- 55+ 53+ 45- 49- 556+ 5-

... showing 6 anomalous results.

Three runs of 10000 elections gave anomalous results 1981, 1940,
and 1928 times. Your formula predicts 10000*F(-0.86185771) =
1943.83 . Almost too good! ... but it was fair and square.
( Although, one might fairly argue that if they hadn't looked
good, I would have tried some more. )

Lew Mammel, Jr.







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