Wolves Draft History: Stats people please help?

When you are trying to figure out something like this, yes, you multiply the probability of them not doing something – but you’d go the instance of them not moving up for a single year, so if you had .924, .916, and .905 probabilities for picking 1st, 2nd, and 3rd respectively for a year; then you’d actually take 1-[(1-.924)(1-.916)(1-.905)]=.745 and not (.924*.916*.905)=.766

Kind of funny you ask, because I keep a sheet for the Timberwolves lottery history just to depress myself… here’s the cut and paste of it:

The Timberwolves lottery history to see if “they’re due” for the #1 pick.

http://www.draftexpress.com/nba-draft-lottery-history/

Originally the weighted lottery had 11 teams, but used a format where the worst team had 11 chances, and the best team had 1. It then progressed to the more current format in 1994 where they are heavily weighted, but still had 11 teams; then it progressed to 13 in 1996; and finally 14 in 2004.

Here’s the Timberwolves the overall odds that they would land the #1 pick in each season:

1990: 7 out of 66 chances for the #1 pick, or 10.61%
1991: 5 out of 66 chances, or 7.58%
1992: 11 out of 66 chances, or 16.67%
1993: 10 out of 66 chances, or 15.15%
1994: 164 out of 1000 chances, or 16.4%
1995: 182 out of 1000 chances, or 18.2%
1996: 77 out of 1000 chances, or 7.7%
1999: 89 out of 1000 chances, or 8.9%
2005: 5 out of 1000 chances, or 0.5%
2006: 53 out of 1000 chances, or 5.3%
2007: 53 out of 1000 chances, or 5.3%
2008: 138 out of 1000 chances, or 13.8%
2009: 76 out of 1000 chances, or 7.6%

Now for the calculation: Probability that they should have won by this point is one minus the probability they wouldn’t have won by this point. The probability they wouldn’t have won by this point is .8939*.9242*.8333*.8485*.836*.818*.923*.911*.995*.947*.947*.862*.924 which equals .23872. So the probability that they should have won the lottery by this point is 76.128%

How about their chances for a top two pick, since they’ve never won that before either.

Since DX doesn’t have the odds for the #2 pick on their database for the early years, I’m going to make my best guess (it’s conservative) at what the odds would be in those years for the Timberwolves to get a top-2 pick.

1990: 21.02% for a top-2 pick
1991: 15.29%
1992: 31.82%
1993: 29.29%
1994: 32.81%
1995: 36.00%
1996: 17.54%
1999: 18.67%
2005: 1.09%
2006: 11.32%
2007: 11.3%
2008: 28.04%
2009: 16.03%

Calculation: .7898*.8471*.6818*.7071*.6719*.64*.8246*.8133*.9891*.8868*.887*.7196*.8397, which equals.04373. So the probability that the Timberwolves should have got a top-2 pick by this point is 95.627%

but often, seemingly unlikely events are more likely than one initially suspects.

For example, I’m pretty sure that under current and previous lottery regimes, the worst team was more likely to get the third pick than the 1st pick and possibly most likely to get the fourth pick.

I’m convinced that the NBA does not understand the difference between “having the best chance to get the 1st pick” and “having the 1st pick be the pick your team is most likely to get.”