[BostonCard]

[quote author=Boston Card link=topic=13047.msg132754#msg132754 date=1444019651]
https://twitter.com/TedMillerRK/status/6...0459879428

It's hard to know what the quoted probability means.  It could be that they ran 10,000 simulations and looked at how many losses the Pac-12 champs had, whoever it was.  Alternatively, they could have looked at the maximum likelihood number of losses for each team (Stanford is projected to end at 9.9 - 3, which implies a 90% chance that we will make the Pac-12 championship) and weighted according to the probability that each team wins the championship (it's 40% for Stanford, by the way).  If it is the latter, they are doing it wrong.  For example, the fewer games we lose, the greater the likelihood that we wind up as Pac-12 champions (indeed, our probability of winning out is 8%).

In the extreme case, consider that Stanford and U$¢ run the table and meet in the Pac-12 championship.  Though this scenario guarantees that the Pac-12 champion will have only 1 loss, if the teams are viewed as more or less even, each team will be given a 50% chance of winning the championship, and a 50% chance of having two losses.  Thus, if you used a non-Bayesian approach, you would argue that the Pac-12 champion had a 50% chance of having two losses.

BC

I'm missing something.  Not for the first time.  In your scenario, where Stanford and U.Sc enter the championship game with one loss each, how can the Pac12 Champion have two losses?
[/quote]

That is exactly the point; in that scenario, both situations (Pac-12 champions and having only 1 loss is caused by the same thing, winning the game).  But if you are computing probabilities, and mistakenly assume that the two events are unrelated, you would say that both USC and Stanford have a 50% chance of having a second loss and a 50% chance of winning the Pac-12 championship.  Thus, by this mistaken logic, the Pac-12 champion has a 50% chance of having two losses.

BC

[BostonCard]

[quote author=Boston Card link=topic=13047.msg132754#msg132754 date=1444019651]
https://twitter.com/TedMillerRK/status/6...0459879428

It's hard to know what the quoted probability means.  It could be that they ran 10,000 simulations and looked at how many losses the Pac-12 champs had, whoever it was.  Alternatively, they could have looked at the maximum likelihood number of losses for each team (Stanford is projected to end at 9.9 - 3, which implies a 90% chance that we will make the Pac-12 championship) and weighted according to the probability that each team wins the championship (it's 40% for Stanford, by the way).  If it is the latter, they are doing it wrong.  For example, the fewer games we lose, the greater the likelihood that we wind up as Pac-12 champions (indeed, our probability of winning out is 8%).

In the extreme case, consider that Stanford and U$¢ run the table and meet in the Pac-12 championship.  Though this scenario guarantees that the Pac-12 champion will have only 1 loss, if the teams are viewed as more or less even, each team will be given a 50% chance of winning the championship, and a 50% chance of having two losses.  Thus, if you used a non-Bayesian approach, you would argue that the Pac-12 champion had a 50% chance of having two losses.

BC

I'm missing something.  Not for the first time.  In your scenario, where Stanford and U.Sc enter the championship game with one loss each, how can the Pac12 Champion have two losses?
[/quote]

That is exactly the point; in that scenario, both situations (Pac-12 champions and having only 1 loss is caused by the same thing, winning the game).  But if you are computing probabilities, and mistakenly assume that the two events are unrelated, you would say that both USC and Stanford have a 50% chance of having a second loss and a 50% chance of winning the Pac-12 championship.  Thus, by this mistaken logic, the Pac-12 champion has a 50% chance of having two losses.

BC