Re: Octonion Power Rankings
To make my previous point concrete, let's take your statement: "That's why KRACH has rated Minnesota over Quinnipiac." But the question is: which one is better, and by how much? KRACH currently has ratings of 162.619 for QPac and 153.051 for Minny. This means that the probability of a Quinnipiac win over Minnesota on neutral ice is 162.619/(153.051+162.619) or 51.5 percent. I'm not sure exactly how your Poisson model works, of course, but if works like a typical Poisson model with offensive factors of 1.536 and 1.916 for QPac and Minny, with defensive factors of .398 and .453, then on neutral ice we get a Poisson score of 1.536-.453= 1.083 for QPac's goals and 1.916-.398 = 1.518 for Minny, then we get a probability of a QPac victory of 37.7 percent (I have taken the liberty of breaking ties proportional to net strength, but that should be relatively robust.) OK. So that's a pretty big difference between two methods right? One has QPac with a 37.7 percent chance of winning head to head and the other method has 51.5 percent. But it's really hard to separate those two probabilties even if they played 10 times. using a likelihood ratio of 8 to signify strong evidence, 0 wins would be evidence that you had the better argument and 9 or 10 wins would suggest KRACH did, the remainder of the cases would not be very strong evidence one way or the other. Even using a weak evidence standard of a likelihood ratio of 4, 3-6 QPac wins would be inconclusive. And that's playing 10 times!
To make my previous point concrete, let's take your statement: "That's why KRACH has rated Minnesota over Quinnipiac." But the question is: which one is better, and by how much? KRACH currently has ratings of 162.619 for QPac and 153.051 for Minny. This means that the probability of a Quinnipiac win over Minnesota on neutral ice is 162.619/(153.051+162.619) or 51.5 percent. I'm not sure exactly how your Poisson model works, of course, but if works like a typical Poisson model with offensive factors of 1.536 and 1.916 for QPac and Minny, with defensive factors of .398 and .453, then on neutral ice we get a Poisson score of 1.536-.453= 1.083 for QPac's goals and 1.916-.398 = 1.518 for Minny, then we get a probability of a QPac victory of 37.7 percent (I have taken the liberty of breaking ties proportional to net strength, but that should be relatively robust.) OK. So that's a pretty big difference between two methods right? One has QPac with a 37.7 percent chance of winning head to head and the other method has 51.5 percent. But it's really hard to separate those two probabilties even if they played 10 times. using a likelihood ratio of 8 to signify strong evidence, 0 wins would be evidence that you had the better argument and 9 or 10 wins would suggest KRACH did, the remainder of the cases would not be very strong evidence one way or the other. Even using a weak evidence standard of a likelihood ratio of 4, 3-6 QPac wins would be inconclusive. And that's playing 10 times!