In a 10 team league, the draft order is chosen by weighted lottery. The best team in the league is given ten ballots, the next best-nine ballots, etc. until the worst team, which recieves only one ballot. All 55 ballots are placed in a box and one is drawn out. This in the team that will draft 10th. All of this teams remaining ballots are removed, and another draw happens for the 9th spot.

I know the chance for each team to be drawn 1st and thereby recieve the 10th draft spot. I was also able (I think) to determine the probability for each team to get the 9th spot, regardless of who gets the 10th. I am having trouble getting any further. Everything I have tried comes up with total not equal to 100%. I know I can do it by simply pounding out the 10^10 calculations for each possible combination, but was looking for a better way.

Sir Elfman

How many ways can the tickets be chosen? We fix some permutation of the first 10 positive integers: i.e. the permutation <3,2,4,6,5,1,10,9,8,7> means that the third-best team got 10th rank, the 2nd-best team got 9th rank, etc. For each of these cells, we multiply by a number saying how many ballots could have been chosen for each team. Obviously, team #1 has 10 ballots, team #2 has 9 ballots, etc. This means that there are (10!) permutations times (10!) ways to pick the tickets, or (10!)^2.

Now, let's see how many ways team #1 can go in each spot. To be picked 10th, they go into the first spot of the permutation, leaving 9! ways to arrange the other 9 teams, and then they had 10 choices of ballot, whereas there are 9! ways for the other teams to take their ballots. So the number of ways in which they could be 10th is
10*(9!)^2, and the probability is 10*(9!)^2 / (10!)^2 = 10/10/10 = 1 / 10.

This analysis works no matter which cell you pick. So team #1 has a 1/10 chance to get any spot.

Now, for team #2, there are 9! factorial ways to arrange the other teams around our chosen spot. These teams are every team except for #2, so there are 10*8*7*6... ways to pick their ballots, or (10! / 9). Once we have chosen a spot for team #2, then there were 9 possible ballots they could have taken. This means that the total number of arrangements for them is 9*9!*(10! / 9) = 9!*10!, and the probability of getting the spot is 9!10! / (10!)^2 = 1/10 once again.

This works for any team. This seems very counter-intuitive to me. It's possible that, although I'm counting all the possibilites, I'm making the mistake that each of the possiblities occurs with the same probability.

Email me to let me know if this problem gets anywhere good: mbk@andrew.cmu.edu.