Canada4Gold
Registered User
- Dec 22, 2010
- 43,051
- 9,237
Warning! Discussion of probabilities!
I think there isn't an exact percentage of the last place team drafting fourth, (the more easily discussed converse of drafting top three) because the odds change depending on the other teams that win the top places. Suppose that there are, in effect, three draws for the three top picks. The odds change, after pick one and two are selected and teams are no longer included in the following draw. As far as I can see, the odds are substantially different depending on the placement of the lucky team. So, for example, it makes a much bigger difference if the second last team, with its 13.5% chance, wins, as opposed to the 14th worst team with its 1% chance going in.
A quick back of the envelope calculation suggests the odds vary between 51% and about 43.5% for the last place team to draft 4th. The probabilities somewhat favour a number closer to the bottom than the top of that range.
I am semi- numerate at best and would like to see a comment on this by someone more knowledgeable tham I am.
No, there is an exact percentage, you just have to calculate the odds of every single team winning those initial picks and factor your odds afterwards. Enough work you'd need a computer program to do it and not want to do it by hand. Some guy on reddit did it
For example if the #2 team wins the first pick(they had 13.5% odds) then the last place team is left with a 0.20/(1-0.135) = 23.12% chance to win the 2nd pick.
The odds of both happening are 0.135 x 0.2312 = 3.12%
The you do it for the 3rd-14th place teams winning the 1st pick, and add them all together. The 3rd pick would have enough more combinations as you're doing every possible combination of the 1st 2 picks, which for each teams odds is 156 different possibilities, so to just fill out the 3rd pick part of the follow chart you'd need to do it 2184 times. Ergo you need a computer program to do the math for you.