[Frequitude]

Yup, what Enoch said. His table already considers all possible combinations of draft lottery outcomes and the probabilities of them occurring to come up with a distribution for each standings placing heading into the lottery.

For example, the chance of the last place team picking second is equal to:

0.2 x 0 (20% chance they pick first times 0% chance of them picking second since they already got first)
+
[0.2/(1-0.135)] x 0.135 (last's new chance of being picked next time if second last got the first pick times the probability of second last getting the first pick)
+
[0.2/(1-0.115)] x 0.115 (last's new chance of being picked next times if third last got the first pick times the probability of third last getting the first pick)
+
...
...
...
+
[0.2/(1-0.01)] x 0.01 (last's new chance of being picked next time if fourteenth last got the first pick times the probability of fourteenth last getting the first pick)
=17.49% (as shown in his table)

The math gets even crazier for the 3rd pick because you have 3 degrees of freedom.

Enoch's table (which I think came out to the same result as tankathon.com) has all of that math built in for every possible combination. It's very impressive.


I then took that table and overlaid it with sportsclubstats projections to come up with a probability distribution for what pick we'll get as of today. That math was infinitely easier!