A note about statistics

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.