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.*

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