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Tuesday, February 5, 2008
posted by Kyle Hampton | 4:50 PM | permalink
Ian Ayres, a professor at both Yale's law school and business school, has an interesting post on the Freakonomics blog about the Zogby poll that has Mitt leading California. Ayres says:
But what is the probability that more likely voters in the state actually support Romney? Given the 2.9 percent margin of error, it’s possible that Romney just got lucky and the pollsters happened to ask an unrepresentative group that disproportionately favored Mitt.

It turns out that it is really easy to use the raw information of the poll the leader percent, the follower percent, and the size of the poll) to calculate the probability of leading in the population. In winner-take-all elections (which are not the case for many of the primaries), this “probability of leading” is crucially what we should care about –- because if people don’t change their minds (and, if undecided, break evenly), this is the probability that the poll leader will win the election. But most people have a very hard time making the calculation in their head.

So take a shot: what do you think is the probability that Romney is leading McCain in the population of likely Republican California voters?

Turns out that Romney’s probability of leading is a whopping 92.7 percent. If you want to calculate your own leader probability, I’ve created a simple Excel spreadsheet where you can plug in the numbers and generate an answer for any poll you want.

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1 Comments:


There's math now? No one said anything about math, I'm going to support Paul....kidding, just kidding.

By Anonymous Anonymous, at February 5, 2008 at 5:16 PM  



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