In his first appearance as a pro, Ronald Jones ran for 9 yards on 8 carries. Reports from camp suggest he's struggling with pass protection and receiving. Now obviously it's far too early to pass judgment on any of the rookies, but it did get me thinking of a particularly famous problem in statistics. If you're paying attention to the numbers in this game then it's important that we avoid drawing conclusions they don't support, so let me get hypothetical and explain!

There are a bevy of backs that share a tier in the first round this year. The most recent DLF ADP has Guice, Penny, Michel, Jones, Chubb, Freeman and Johnson all ranked between 1.02 and 1.09, and I think we all know that some of these guys will bust. There has been a lot of research into the "hit" rate of 1st round draft picks: Rotoviz suggests 30%-47%, for example, depending on if you exclude guys with one year of success, like Trent Richardson. So here's an example of the question I want to look at:

If I drafted Chubb, and now Jones is a bust, does it raise the odds that Chubb is not a bust?

This is assuming I know nothing new about Chubb (imagine only Jones has played so far), actually do have these RBs in a similar tier (that is, a similar probability of success), didn't draft both of them, or any other complicating factor. On the surface it sure seems like I should be happy Jones is a bust, doesn't it? Only 30% of them are successful on average, and that's one down! I'll cut to the punch line, and then explain: Jones being a bust does not in any way raise the odds that Chubb is not a bust. Counterintuitive, isn't it?

-----

In statistics and probability, this is called the Monty Hall Problem. It is named for an old game show, and the version we're considering goes like this: a contestant is shown three doors, then told that behind two are goats and behind one is a new car. It is hopefully clear that their odds of getting a new car are 1/3 (33%), which just happens to be pretty similar to the 30% hit rate on first round draft picks. Now for the twist - the host opens one of the doors the contestant didn't pick, revealing a goat. The host then points at the two remaining doors and asks if you want to stick with your original guess, or change to the other door.

Should you change?

Let's get the punch line out of the way again: shockingly, the answer is unequivocally yes, you should change. If you stick with your original pick you only have a 33% chance of a car, but if you switch you have a 66% chance. What? How is it possible I get even better odds than 50% if I switch?! Two doors left, shouldn't it be 50/50 on my second choice? You can prove this yourself by setting up a version of the game and trying it over and over if you like, but I'll try to explain.

It comes down to the information you had

*when you made the choice.*Initially all doors looked the same, thus an even 33% chance for each. Crucially, we know that the host was never going to open our door,

*so we get no new information about it.*Its probability of being a car is fixed at 33%. The other two doors each had their own 33% chance of being the car - in other words, there was a 66% chance the car wasn't behind our door. When the host reveals the goat we didn't pick, recalling three sentences ago when we agreed we got no new information about the door we chose, there has to still be a 66% the car is not behind it. Therefore, the one remaining door now has a 66% chance of being the car.

There are dozens of explanations of this online, though this one is nice because it includes a simulator.

----

So bringing it back to football, learning that Jones is a bust gives us no new information about Chubb being a bust, even if we accept a fairly consistent bust rate amongst RBs in this tier. But here's a brain twister for you, now that you know about the Monte Hall problem: if you had Jones, Chubb and at least one other of these RBs in the same tier, and you own Chubb, and Jones is a bust, should you now trade Chubb for the other guy?

Have fun with that!