Hypothesis: forward lottery is equvalent to reverse lottery
Proof:
First, I'm assuming that everyone here understands that if you have a set of n unique balls {1,2,3,4,... n-2, n-1, n}, each ball has an equal probability of being selected at any position (P_i = 1/(n-i+1)), 1<= i <= n, then there are n! possible outcomes, and the probability for each possible outcome is 1/(n!). In addition, the probability of an outcome, is the SAME as the probability of its reverse (ie, if we have 3 balls, and the order the balls are selected is {2,1,3}, the probability of this outcome = 1/3! = 1/(3*2*1) = 1/6. The probability of its reverse, {3,1,2} is also 1/3!, which is 1/6). This is not a proof for this claim, but I'm pretty sure you could look it up in any stats textbook (probably one of the first random variable properties that was ever proved), and while its simple to understand this, its probably a bit harder to prove it, and I'm not going there because this is obvious.
Okay, so we've established the fact that with n unique balls, that its equally likely that any order of n balls being selected is as likely as it's reverse being selected.
Suppose we have 48 balls, each numbered uniquely, ie {1, 2, 3, ... , 47, 48}. Any outcome of the 48 balls is equally likely as it's reverse. So in a forward draft, if we get a certain outcome, the outcome's reverse is as equally likely to occur in a reverse draft.
We can easily assign each team to its proper amount of balls, ie ball #1 = BUF ball #1, ball #2 = BUF ball #2, ball #2 = BUF ball #2, ball #4 = CBJ ball #1, etc
So now we have to show that counting a team's ball first in the forward draft is equivalent to only counting the last ball in the reverse draft. This is pretty obvious-- lets just look at a smaller example, 5 balls with 3 teams (the draft is only for 3 picks)... say the forward order that we picked out the balls is {DET, NYR ball #2, COL, NYR ball #1, NYR ball #3}.. we proved that its probability would be equivalent to a reverse draft order of {NYR #3, NYR #1, COL, NYR #2, DET}. In the forward Draft, the first pick would be Detroit, second pick NYR (NYR #2), third pick COL, then we ignore the last 2 balls. In the reverse Draft, we'd ignore the first NYR ball (NYR #3), ignore the second NYR ball (NYR #1), then COL only has one ball left, so it gets the third pick, then NYR's last ball (NYR #2), so they get second pick, and Detroit has the last ball so they get first pick. You could extend this into some induction proof but I think this is good enough for the purposes of this board to prove that taking a forward order of 48 balls and flipping the order and doing the reverse order elimination draft is the same
Therefore Forward Draft is equivalent to Backwards Draft.
holy crap I can't believe I did that
Theres probably a way to prove it by anaylzing the statistics, but its a long ass way, and frankly, no statistician would do it unless they wanted to give their brain a work out