A new Reuters/C-SPAN/Zogby Poll has Mitt Romney ahead of John McCain by 37 percent to 34 percent in a poll of 1185 likely Republican voters in California (2.9 percent margin of error). 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.
The same poll found that Barack Obama led Hillary Clinton in California “by 45 percent to 41 percent, with a margin of error of 2.9 percentage points.” The same analysis suggests that, at the time of the poll, there is a 94.2 percent chance that more probable Democratic voters support Obama than Clinton.
Of course, these probabilities may end up being widely off — either because the poll was poorly done, or because people change their minds. But another advantage of calculating the probable leader statistic is that it builds a better bridge to the prediction markets. Just after the poll was announced, Intrade had Romney’s probability of winning in California as 94 percent (pretty close to the 92.7 percent leader probability). But Obama’s InTrade bond for California was only trading at 59.9 percent — substantially below his leader probability of 94.2 percent.
MSNBC reports that “Clinton held statistically insignificant 1-point leads on Obama in New Jersey and Missouri, well within the margin of error of 3.4 percentage points in both surveys.” But instead of saying that these races are statistical dead heats, it might be more useful to report that Clinton’s probability of leading is 63.6 percent in New Jersey (InTrade comparison: 60 percent) and 63.5 percent in Missouri (InTrade comparison: 67 percent).
The margin of error and the sample size tell the general public very little. How many people even know whether the margin of error represents one or two standard deviations? The probability of leading is much more intuitive, easy to calculate, and gives the public something much closer to the result they actually care about: the probability that the leading candidate will win the election.