Thursday: Watch Out, Vandergriff Honda

You may remember from sometime in March that Vandergriff Honda promised free cars to certain customers if the Rangers start the season 4-0 and if the Yankees start the season 0-4.

I previously commented on the exact probability of that occurence here at HWC using BP's PECOTA season projections. As you might expect, the probability was extremely low, at .0008 (.08%).

As I'm sure Vandergriff Honda has noticed, the Rangers have started the season 2-0 and the Yankees have started the season 0-2 -- the event is still very possible, and much more likely, although not likely in itself (for an explanation of the method used, see my previous post):

P(Rangers start 4-0, given that they start 2-0)=.196047
P(Yankees start 0-4, given that they start 0-2)=.151967

Thus,

P(Rangers start 4-0, given that they start 0-2 AND Yankees start 0-4, given that they start 0-2)=.196047*.151967=.02979

The probability that Vandergriff Honda will give away free vehicles is now .02, a 2% chance. I don't think anyone at the dealership is contemplating paying someone to throw a game . . . . yet.

UPDATE: at 3:35 pm, the Yankees were beating the Orioles 9-2 in the top of the ninth . . . . . I think it's safe to say that within the hour, this probability will change to 0. It was fun while it lasted, though.

Wednesday: The Not-So-Likely Way To Get a Honda for Free

You may have heard about the publicity stunt being pulled by Vandergriff Honda: anyone who buys a vehicle from them between March 9 and March 16 gets one free if the Yankees start the season 0-4 and the Rangers start the season 4-0. This, of course, sounds tremendously unlikely. But exactly how tremendously unlikely is it?

To calculate the exact probability of this occurence, we will use Bill James' Log5 method (which gives the probability that one team beats another, given their winning percentages). Baseball Prospectus' 2009 PECOTA predictions have been used for winning percentages.

The Rangers play their first four games against Cleveland (three games) and Detroit (one game). The Yankees play their first four games against Baltimore (three games) and Kansas City (one game). According to PECOTA, these six teams are predicted to win the following fraction of their games this year:

Texas Rangers: 73/162
New York Yankees: 95/162
Cleveland Indians: 85/162
Detroit Tigers: 80/162
Baltimore Orioles: 77/162
Kansas City Royals: 77/162

Using the Log5 Method (with the above numbers) yields the following single-game probabilities:

P(Texas beats Cleveland)=.426285
P(Texas beats Detroit)=.4598955
P(Baltimore beats New York)=.389829
P(Kansas City beats New York)=.389829

Since each game can be considered as an independent event, the desired probability follows from the following:

P(Texas starts 4-0 and New York starts 0-4)
=[(.426285)^3]*[(.4598955)]*[(.389829)^4]
=.0008

So, Vandergriff Honda, you can sleep soundly the first few weeks of April. There is only a .08% chance that you'll be giving away free vehicles.

UPDATE (4/9/09): The Rangers have started the season 2-0 and the Yankees have started 0-2. Go here for my latest post, which updates these probabilities based on this new information.