I'm working on something just for fun in my fantasy sports league and am running into a little trouble with one aspect of my lottery system. Basically, I want the bottom half of the league (6 teams) to be entered into a lottery for the following year's draft selections. The way I have it structured is that the team that finished in last place has the highest probability of getting the first overall draft pick, second to last place has the second highest probability, and so on. After the first selection is awarded, we do the same thing for the second selection, then the third and so on through the sixth. This is all straightforward enough.
The issue that I'm running into is that I want to guarantee the the worst team gets one of the first three selections and that the second to worst team gets a top five selection. Obviously I could do this by just saying 'if worst team does not have selections 1 or 2, then automatically award them selection 3', but I was wondering if there was a way to work with the probabilities that would make it always work out that way. (If you're familiar with the NBA draft lottery, I'm more or less trying to mimic that with 6 teams instead of 14).
Thank you for any and all help you can offer.
There are $6! = 6\times 5\times 4\times 3\times 2\times 1 = 720$ different possible orders for six people without restriction. You don't want any possible order. You want the order where the worst person is in one of the first three spots and the second worst is in one of the first five spots.
Start with the first worst person. They can be in any of the first three positions. There are 3 ways to place him/her. Next, out of the first five positions, one position is already taken (it happens to be in one of the first three positions, but that is still among the first five). So, there are only four positions from the first five still available. That leaves four positions from the six. You can permute the remaining four individuals any which way. That gives a total number of "valid" permutations as:
$$3\times 4\times 4\times 3 \times 2 \times 1 = 288$$
So, there are 288 permutations you care about. Rank these full permutations in order of how likely you want them to occur. Now, when drawing, you are not drawing a single position, but all six positions at the same time. The permutations that favor the players that did the worst will have the highest probability of getting picked while the permutations that favor the players who did the best will have the lowest probability of getting picked.
Suppose your scheme is the worst team has a 30% chance of getting picked. The second worst has a 23% chance. The third worst has a 17% chance. Third best has 13% chance. Second best has a 10% chance, and best has 7% chance. So, you "grade" each permutation as to how likely you want it to be that it happens.
For a permutation, you look at the position of each team's pick. The first pick weighs a factor of the individual probabilities. The second pick weighs a factor of the individual probabilities out of 1 minus the probability of the first pick. The third pick weighs a factor of the individual probabilities out of 1 minus the probability of the first two picks. Example: 1 = worst team, 2 = 2nd worst, etc.
The permutation: 1,2,3,4,5,6 would have a weight of:
$$(0.3)\left(\dfrac{0.23}{1-0.3}\right)\left(\dfrac{0.17}{1-0.3-0.23}\right)\left(\dfrac{0.13}{1-0.3-0.23-0.17}\right)\left(\dfrac{0.1}{1-0.3-0.23-0.17-0.13}\right)\left(\dfrac{0.07}{1-0.3-0.23-0.17-0.13-0.1}\right) = \left(\dfrac{30}{100}\right)\left(\dfrac{23}{70}\right)\left(\dfrac{17}{47}\right)\left(\dfrac{13}{30}\right)\left(\dfrac{10}{17}\right)\left(\dfrac{7}{7}\right) = \dfrac{299}{32900}$$
Any permutation that violates your rules (team 1 is in position 4,5 or 6, or team 2 is in position 6) gets a weight of 0.
Once you weight each permutation, order them by weight. Remove any permutation with a weight of 0. The relative weights of whatever permutations are left will give you a probability space. You will want a permutation to have a relative probability to its weight. Now when you draw, you draw once and that completely determines the order of the six teams.