When it comes to ties in record, the NFL determines draft position by strength of schedule (SOS). The team with the weaker strength of schedule picks first. What happens when there are two teams with the same SOS? There is a coin flip.
We found out this morning what the league does when there are 3 teams with the same SOS, and unless there is more information than is provided in this ESPN article, the statistician in me says the Kansas City Chiefs got a raw deal. Here's what happened:
The Falcons had the first call and won the toss. Under that scenario, the Falcons got the third pick and the Raiders the fourth pick. Had the Falcons lost the toss and the Raiders won, the Raiders would have drafted third, and then the Falcons would have flipped against the Chiefs for the No. 4 and No. 5 picks.
The Falcons had the first call. If they got it right, they go #3. If they get it wrong, they go against the Chiefs for #4. That means the Falcons have a 50% chance of having #3 and 25% chance each of having #4 and #5.
The Raiders are in even better shape than the Falcons. They had chances of 50% for #3, 50% for #4, and 0% for #5.
Where does this leave the Chiefs? It's easy from the math above that they have a 0% chance for #3, 25% chance for #4, and 75% chance for #5.
Here are the expected draft positions for the three teams under this scenario:
Falcons = (.5 * 3) + (.25 * 4) + (.25 * 5) = 3.75
Raiders = (.5 * 3) + (.5 * 4) + (0 * 5) = 3.5
Chiefs = (0 * 3) + (.25 * 4) + (.75 * 5) = 4.75
I am assuming that there were some other behind-the-scenes tie-breakers not mentioned in the article to determine the coin flip position, because if not the Chiefs did indeed get hosed.*
*There is another possible explanation. Mainstream media reporters are notoriously horrible at understanding statistics (they are just journalism majors after all), so the author probably doesn't comprehend that his article gives the impression that the Chiefs had terrible odds. This may be worth an e-mail to the league for clarification.
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2 comments:
Email the league. There is no way this is how it is set up. The Chiefs totally got screwed if this is right. Wouldn't they flip the coin all at the same time and the odd one out gets the #3 pick and then the other two flip for #4?
First of all, great post, Clint. NFL and probabilities. Two things I'm very interested in. The only thing that could have piqued my interest more is if you could have worked in an SMG reference.
NFL.com had more info. "All three teams finished the 2007 season with identical 4-12 records, but the Raiders were awarded the fourth overall selection based on their finish behind the Chiefs in the AFC West standings. Kansas City finished ahead of Oakland in the division standings based on a better record against common opponents."
So, it sounds like the league wanted to force KC to pick behind Oak. There are 3 possible outcomes. 1: Atl-Oak-KC; 2: Oak-Atl-KC; 3: Oak-KC-Atl.
At first I thought each should have an equal chance and the Falcons had an unfair advantage the way it was done. But then again, why should Oakland have a 2/3 chance of getting the #3 pick?
I think it was fair the way they did it.
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