There is a lot of brouhaha about pollsters (most notably Nate Silver) predicting an Obama win in the coming election. As of this post, Nate has the chance of Obama winning at 83%. Various news personalities are complaining that Nate must be in the bag for Obama to dare predict that Obama will win and that a Romney victory will prove that Nate Silver is a lying liar who lies. A Romney victory would prove nothing of the sort, though.
Why? Nate Silver has been eerily accurate at predicting electoral outcomes. He has a model that he follows and he is mostly public about his methods. His methods haven’t changed and they predict an over 80% chance of Obama winning. Does that mean Obama will win? No. Romney still has a 20% chance of winning. If you were to bet on the outcome, though, you’d be wise to put your money on Obama.
Look at it this way. Take a 10-sided die. What? You don’t have a 10-sided die? Ok, go back in time and play Dungeons & Dragons as a kid. Poof, you now have a 10-sided die. Pretend that an Obama win is represented by 1-8 and Romney is represented by 9-10. Roll the die. There’s an 80% chance you’d roll an Obama. Roll it a couple of times. Most of the time, you’ll roll a Romney in only a few rolls. If there could be two elections, there is a 36% chance of rolling a Romney one of those times: 2/10 + (2/10 * 8/10). A third election would give you a 49% chance of rolling at least one Romney: 2/10 + (2/10 * 8/10) + (2/10 * 64/100). Generally, you have a 1 – (8/10)^n chance of rolling a Romney in n rolls.
Of course, this is all just a useless exercise in math and probability because there is only one election and Obama currently has an 83% chance of winning according to the polls. But don’t go attacking Nate Silver or math if the die rolls a 9 or a 10.