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Topic: Mathematics of Voting
Replies: 17   Last Post: Nov 24, 2000 4:13 AM

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Steven E. Landsburg

Posts: 154
Registered: 12/8/04
Mathematics of Voting
Posted: Nov 10, 2000 12:50 PM
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Events have overtaken me---an eternity ago, when Bush's lead
in Florida was 12,000 votes and everyone was talking about how
"achingly close" it was and how this really proves that
everyone's vote counts, I wrote a brief column about how it
actually proves exactly the opposite: Your vote
counts when the margin is 1, and 12,000 is a very poor
approximation to 1.

More precisely, I calculated as follows: Assume each voter
votes for Bush with probability p and Gore with probability
1-p, and each choice is made independently. There are 2N
voters. The probability of a tie is then (pretty easily)
seen to be about

(4p(1-p))^n / Sqrt(Pi n)

Plugging in the actual value of N (call it 3,000,000) and
assuming the final observed percentages to be a good
approximation to p, I get a probability of about 1 in
500,000,000. In other words, your vote doesn't count.

But now that Bush's lead is down in the neighborhood of 300, a
recalculation gives a probability of about 1 in 3000, which
starts to make voting seem worthwhile.

Surely this is the wrong model, in the sense that I should
really start with a prior distribution for p, update on the
basis of the observed outcome, and then integrate against the
updated prior---but I'd guess this oversimplified model gives
a pretty good approximation.

Anyway, since the column is no longer relevant, I'll publish
in the place most appropriate for an irrelevancy: Usenet!
Herewith the column:

Which is stupider---playing the lottery or
voting for president? Let's look at the

To make the best possible case for voting,
suppose the entire election hinges on your
state alone---in other words, suppose you
live in Florida. And suppose your state is
so evenly divided that the margin of victory
is just 12,000 votes out of 6 million cast.
Then what's the probabilty that in the next
election, with a similarly divided
electorate, your vote will actually decide
the outcome? The answer turns out to be
about 1 in 500 million. (Click here for the
calculation.) That's roughly 10 times worse
than your chance of winning the jackpot in
the New York State Millenium Millions game.

That won't stop Bob Schieffer of CBS News
(and presumably other anchors on other
networks that I wasn't watching) from telling
you that the closeness of the election proves
``every vote can make a difference''. This
is of a piece with the lottery propaganda
that says ``If you don't play, you can't
win''. True enough, but the right question
is: what are the odds? And the odds are

Moreover, if you'd hit the Millennium
Millions jackpot last week, you'd have won
$130 million. If you'd hit the jackpot in
the voting booth by casting the deciding
vote, you'd have won nothing more than the
opportunity to choose the next president of
the United States. That's not an
inconsiderable prize, but I'm willing to bet
that most people would rather have the
$130 million.

And look at the cost of playing. A Millenium
Millions ticket costs $2. Voting, by
contrast, costs most people at least 15
minutes (including travel time). If your
time is worth at least $8 an hour, then
voting is more expensive than buying a
lottery ticket. So compared to the lottery,
voting offers a smaller chance at a less
valuable prize for a higher price.

If you vote to affect the outcome, you're
far more foolish than your neighbor who buys
a lottery ticket to win the jackpot. On the
other hand, maybe your neighbor isn't so
foolish after all---maybe he's playing the
lottery just for the fun of it, clearly
understanding that he has no realistic chance
of winning. And maybe that's why you voted,
which is fine. I voted too. But I didn't
fool myself into thinking my vote might

Steven E. Landsburg


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