Given the bottom 5 players are up for elimination in a given round I'm wondering how I can find the likelihood that a given player would be safe from elimination given each player may vote upto 1 time for who could be eliminated.
The catch is that the closer a person is to the bottom of the up for elimination players rankings, they get a penalty vote for each of those players that are above them that is cast by no one but still count for their total.
I have listed the players A-E from best to worst and the non-penalty votes are in this equation
A+B+C+D+E+N=19
where if say A got 4 votes, B got 3 votes, C got 2 votes, D got 1 vote, and the other players abstained from voting, this would result in a 5 way tie and any of them could be voted out in a re-vote by the player ranked number one (not in the up for elimination players at all) in a tiebreaker vote.
Also if possible maybe also add the logical constraint that every up for elimination person does vote for another up for elimination player than themselves.
(What I tried)
I used the stars and bars technique to get the total sample space of the votes to be
A+B+C+D+E+N=19
(19+6-1 choose 19)/(1^5) = 24 choose 19= 42504
I then did a constraint setup using A as the player I wanted to be safe from elim and made the constraints so A either tied with everyone or other people got extra votes.
A<= B+1, A<= C+2, A<= D+3, A<= E+4.
A+xB=B+1, A+xC=C+2, A+xD=D+3, A+xE= E+4
B=A+xB-1, C=A+xC-2, D= A+xD-3, E=A+xE-4
(xB,xC,xD,xE are extra votes for each for those players)
A+A+xB-1+A+xC-2+A+xD-3+E+xE-4+N=19
5A+xB+xC+xD+xE+n-10=19
5A+xB+xC+xD+xE+n=29
Though after doing this I noticed using stars and bars here
(10+29-1 choose 29)/((5)(1^4)) = 38 choose 29 /5 = 32602328
I found this was significantly bigger than what we started with 42504
then adding the constraint there is at least 1 extra vote
with xB+xC+xD+xE>=1
(x is the total extra votes in excess of A being safe)
xB+xC+xD+xE=1+x
x=xB+xC+xD+xE-1
5A+xB+xC+xD+xE+n=29
5A+xB+xC+xD+xE+n-1=29-1
5A+x+n=28
and using stars and bars 1 more time
(28+7-1 choose 28) /((5)(1^2)) = 34 choose 28 /5 = 268980
This is still more than 42504 and am wondering what needs to be improved over the method to get the correct probability.