Tangled in the Rigging 2: Probabalistic Boogaloo

I can't believe I have to do this again.

It was two years ago, and the New Orleans Hornets had just won the NBA Draft Lottery. With the third-worst record in the league, the Hornets had a 13.7 percent chance to win the 2012 Anthony Davis sweepstakes. This is equivalent to a coin coming up heads three times in a row -- i.e., unusual but not super rare -- but of course everyone cried foul since the league owned the Hornets at the time and was working to sell them to a private owner.

The draft lottery being rigged is one of the oldest jokes known to man. That joke is so old, creationists have to insist it was created by the Devil to test our faith. That joke went stale so long ago, the mold growing on that joke was used to discover penicillin*. Enough.

* - If you like this construction, here are some more examples.

And even reasonable people -- people who love them some numbers! -- fall into this trap. Here's Grantland's Bill Barnwell on Twitter tonight:

Odds of the Cavaliers winning 2011, 2013, and 2014 lotteries: 13,467-to-1.

— Bill Barnwell (@billbarnwell) May 21, 2014

This is technically true:

P(Cavs win in 2011)*P(Cavs win in 2013)*P(Cavs win in 2014) = (2.8%)(15.6%)(1.7%) = .0000743,

or 13,467-to-1. But really, that's disingenous for a couple reasons. First off, the Cavs actually had two chances to win in the 2011 lottery: their own chance, based on their own abysmal performance (19.9%) and the Clippers' chance, acquired by trade (2.8%). So their actual probability of winning the first pick in 2011 was 22.7%, improving their odds to 1,661-to-1.

@Doctor_Bryan @Alex_Kienholz Yep. I should have said "with the picks they won with".

— Bill Barnwell (@billbarnwell) May 21, 2014

Okay, fine. But even then, we're still not talking about the odds of the Cavs winning three of four lotteries. We're talking about the odds of the Cavs winning those three lotteries in that order. If we really wanted to find the odds of the Cavs winning three of the last four lotteries, we'd need to calculate something like this:

P(2011)*P(2012)*P(2013)*P(~2014) + P(2011)*P(2012)*P(~2013)*P(2014) + P(2011)*P(~2012)*P(2013)*P(2014) + P(~2011)*P(2012)*P(2013)*P(2014),

where P(x) is the probability the Cavs won in year X, and P(~x) is the probability the Cavs didn't win in year X.

Also, this sequence of lottery selections isn't even the least improbable four-year stretch in the last decade. From 2005-2008, four straight lottery winners had less than a 10% chance of winning. The odds of those four teams winning is 200,194-to-1. But because the picks went to four different teams, no one thought it was that unusual.

That's because every lottery outcome is unlikely until it happens. Here's another example: the Rhode Island lottery has a four-digit numbers game, where you have to pick four single-digit numbers in order to win the grand prize. Sunday's winning numbers were 9-2-9-9. Now, the odds of drawing three nines in a four-number drawing could be expressed as 1,000-to-1, and that's correct, but are the three nines really so unlikely? No. They're no more or less unlikely than any other three-digit combination*; we're just wired to pick out those patterns.

* - We're assuming it's a fair lottery here, but it is Rhode Island, so who knows.

How Much Is a Win Worth to an NBA Team?

Last month, I used J.C. Bradbury's free agent valuation method to determine how many wins the Red Sox expected Mike Napoli and Shane Victorino to contribute to the team in 2013. That worked fine, but suppose we want to build a similar model for the NBA. Again, we'll use the basic system Bradbury outlines in "The Baseball Economist" (ch. 13). Here, Bradbury found a relationship between revenue, wins, and the size of the city a franchise plays in.

All three of those variables are readily available. For city size, we'll use the population of the metropolitan statistical area (MSA) each team plays its home games in, as reported in the 2010 U.S. Census*. Revenue is available through Forbes' Business of Basketball listings. This data is almost exactly one year old -- suggesting that it covers the 2010-2011 season, and not the recent lockout-shortened 2011-2012 season. This is better for our purposes; I don't want the compressed schedules and reduced number of games to interfere with my results.

* - And the Canadian equivalent for Toronto, with the hope that the two have very similar methodologies.

Tangled in the Rigging: Defending the NBA Draft Lottery

The NBA conference finals brings with it one of the best sideshows in sports: the NBA draft lottery, in which 14 grown men stand around awkwardly for half an hour to figure out how a bunch of ping pong balls bounced. We*, the viewing audience, are treated to a half-hour special containing some 15 minutes of talking heads speculating wildly, 2 minutes of commisioner David Stern reading franchise names, and 5 minutes of awkward interviews with team representatives. Fascinating.

* - Maybe "we" is the wrong pronoun; I mean, I didn't watch it.

But while the presentation of the lottery may not be especially compelling, the lottery itself sure is. The lottery teams (i.e., those that miss the playoffs) are ranked in inverse order of record, with the worst teams receiving the best chances of a high pick. So the team with the worst record has a 25% chance of winning the lottery, the second-worst team has a 19.9% chance of winning the lottery, and so on down to the 14th-worst team (the last team out of the playoffs) who has a 0.5% chance of winning the lottery. The whole list of probabilities for this year's draft is available here.

Some have argued (with varying degrees of seriousness) that the lottery system is rigged*, and point to the fact that the worst team in the league hasn't won a lottery since the Orlando Magic won and picked Dwight Howard in 2004. But I want to stress this again, because it's important: the team with the highest probability will still lose the lottery (i.e., not get the first overall pick) 75% of the time.