The Ghost of Jay Cutler, Ghost for short, got me turning some gears when he wrote a recent post comparing the stats of Ole Miss to Alabama in some sort of metric. I started thinking "What is the probability that Ole Miss will score a touchdown against the Tide?" (Answer after the jump.)None.
Well, that is not really true. Going back through Alabama's last 17 games (beginning of the 2011 season), the Tide has allowed 15 touchdowns. That is an average of .8824 touchdowns per game. I shudder to think what this ratio would be for Ole Miss's 2011 squad.
Anyway, there is a statistical tool called the Poisson probability distribution that helps us calculate the probability of the occurrence of a rare event in some sort of dimension (space or time typically). In other words, the Poisson distribution can help calculate the probability of a team scoring y=0,1,2,... touchdowns against Bama. (The formula is p(y) = (h^(y)/y!)e^(-h) where h is the average value of y.)
What I am saying is this: the probability of Ole Miss scoring 0 touchdowns against Alabama is
p(0) = ((.8824)^0/0!)e^(-.8824) = .4138.*
So the probability of Ole Miss scoring at least 1 touchdown is 1 - .4138 = .5862.
Below I have listed the probability of scoring exactly y touchdowns for a few values:
p(0) = .4138
p(1) = .3651
p(2) = .1611
p(3) = .0474
p(4) = .0105
* The fallacy here is that we are assuming that every offense has an equal chance of scoring and that there is no such thing as playing a good game or a bad game. Still, it is a pretty accurate measure, because we can think that an average offense playing the way they normally would against Alabama would score with these probabilities.
EDIT: I realized that I never really answered the question. You can make your guess with the probabilities listed above, but the estimated number of TD's for a Poisson distribution is h. So the expected number of TD's is .8... so since I would consider Ole Miss to be an above average offense, I'd say we will score one touchdown.