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October 24th, 2018, 12:39 PM  #1 
Newbie Joined: Oct 2018 From: United States Posts: 4 Thanks: 0  Legal Ranking Game
Reallife legal scenario: There is a list of 10 individuals from which one will be chosen to decide your case. There are two parties who each rank their top 6 choices in order and strike their bottom 4. How often is there a tie for the top position? (For example, if the parties rank "A" 1st and 2nd, respectively, and rank "B" 2nd and 1st, respectively, "A" and "B" will be tied for the top pick.) Thanks! 
October 24th, 2018, 03:16 PM  #2 
Member Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3 
Your problem is still underdefined. Who wins e.g. if candidate A is ranked 3 and 5 while candidate B is ranked 2 and 8? You need to specify criteria to decide all ranking combinations.

October 24th, 2018, 04:03 PM  #3 
Newbie Joined: Oct 2018 From: United States Posts: 4 Thanks: 0 
They average the two ranks and lowest rank is appointed.

October 26th, 2018, 11:23 AM  #4 
Newbie Joined: Oct 2018 From: United States Posts: 4 Thanks: 0 
No suggestions so far. If it helps simplify things, what if there were no strikes and both sides ranked 1 through 10? How often is there a tie for the top selection then? Very interested to hear answers for either question! Thanks!

October 26th, 2018, 12:37 PM  #5 
Member Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3 
It's not such an easy problem The sum average obstructs the common binomial approach and the strikes demand different weighting of permutations. On top of that, not just 2 but even 3 or 4 candidates may simultaneously cross the finish line. I am not sure your update challenge is easier but I'll take a shot at it. 
October 27th, 2018, 07:06 AM  #6  
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 497  Quote:
First of all, what do you mean by "how often"? If you are talking about how many ties in n instances of such a procedure, the answer may be any integer from zero through n. It is impossible to say which will happen with certainty. You probably are asking either for an expected value. In the absence of any information, so that the lists are random, an expected value can be computed if you specify the size of the pool that the lists can draw on. But there is absolutely no practical point to such a procedure if the two lists are random. You might as well pick the judge at random and thereby avoid the possibility of ties. Finally, if the lists are not random but influenced by additional information available to the list makers, you need to specify what that additional information is and how it will alter the probability that each party will give each judge a specific rank as well as the size of the pool. As a practical matter, if the list makers are well informed, the probability of each list being headed by the same candidates but in opposite orders is effectively zero. If the virtually impossible happens, just flip a coun because each party will get a judge that each party considers good.  
October 27th, 2018, 08:11 PM  #7 
Newbie Joined: Oct 2018 From: United States Posts: 4 Thanks: 0 
Thanks for the reply. My thought was that there is a finite number of possible ways to rank the 10 for each side, and therefore a finite (though very large) number of combinations. So I figured someone could come up with a way to formulate every one of those scenarios taking place and how many times the top result was a tie. I think the answer is that there are 151,200 different permutations for one side to rank 6 arbitrators and strike 4  using n!/n!r!. And therefore about 22.8 billion permuations when looking to both sides' rankings (151,200 x 151,200). I am thinking there must be a specific number out of the 22.8 billion that have a tie at the top. Each side is ranking the same list of 10 people in my question, to be clear. And I don't think it's important to the question, but in reality it is very common for the two sides to rank them in opposite or nearly opposite manners (plaintiff ranks judges who award damages not frequently at the top of their list, and defendant does the opposite). 

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