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View Diary: WYFMP What's your f'ing MATH problem (with poll) (27 comments)

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  •  Legitimate math question (1+ / 0-)
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    In a committee election to choose fill five seats, each voter chooses five and the top five vote getters win. So far, so good. Now one of the five winners resigns and the sixth highest vote getter claims that he should be the replacement. What is the best way to explain  that with each voter choosing only the top five candidates, the sixth highest vote getter would not be the same person if the voters had actually voted for an alternate, i.e. voted for the top six.
    •  This gets very tricky (0+ / 0-)

      I don't think you CAN prove thise; in fact, I don't think it's necessarily true.  

      Republicans believe government is the enemy. When they're in charge, they're right

      by plf515 on Fri Sep 22, 2006 at 06:09:41 AM PDT

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    •  Prove it by contradiction (1+ / 0-)
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      I think you can rigorously show that the sixth place winner would not necessarily be 6th if the top 6 were voted on instead of top 5.  Basically all you say is that it is mathematically possible that there was some other person who voters would have definitely given their votes to had the top 5 not been there (to put this in a realistic setting say there were 6 Democrats on the ballot and 1 Republican and the 5 Democrats overwhelmingly came in the top 5, the 1 Republican came in 6th and the, let's assume for the sake of the story, political neophyte Democrat came in 7th - one could definitely see that in this electorate it is entirely possible that if 6 candidates were picked on the ballot then the Democrat would probably get the huge bunch of extra votes that he lost out to the Democrats who actually got elected and which would presumably propel him/her over the lone Republican's total).

      That said this is a mathematical possibility in that there are configurations of voting patterns that ensure that the 6th place finisher in a pick-top-5 voting scheme would not necessarily finish 6th in a pick-top-6 scheme.  But the statistical likelihood of this kind of configuration of voters is probably much lower than the outcome that the 6th place finisher would finish 6th in a pick-top-6 scheme I think...

      Hope that helps.

      Give me liberty, or give me death!

      by salsa0000 on Fri Sep 22, 2006 at 06:22:54 AM PDT

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      •  Why would you care? (0+ / 0-)

        Sometimes you should just count the votes.  

        IOKIYAR! They believe markets and competition solve everything AND that the universe is centrally planned.

        by No One No Where on Fri Sep 22, 2006 at 07:42:47 AM PDT

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      •  Yeah (1+ / 0-)
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        Simple, reductio ad absurdum, proof by contradiction:

        Let there be 7 candidates, 3 voters.  Let the voters pick 1,2,3,4, and 2@5, 1@6


        and let 7 be the 6th choice of all the voters - everyone who likes 5 hates 6 and vice versa.

        This way, 6 would be the replacement candidate under the first proposal, but 7 would be the replacement if a new vote were held.


        -7.50 -6.56 | Why is it that those who can remember that those who forget history are bound to repeat it are bound to repeat it?

        by cmanaster on Sat Sep 30, 2006 at 02:46:01 PM PDT

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    •  Preliminaries (1+ / 0-)
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      Let's see, I think it might help to try an formalize the problem. I don't know that I have a solution, but there is a way to state this more rigorously:

      1.) First let's establish the number of candidates. I'll pick an arbitrary number and say 7. To avoid confusion I'll use a letter to represent each candidate. I kind of have a nagging feeling that the total number of candidates could affect the outcome, but we'll see if that's actually borne out. Now, let's put our candidates into a set, appropriately called "c":

      c = {a, b, c, d, e, f, g}

      2.) Now, we have a nonrandom collection of subsets of c. These are the "voters", with one subset representing each person's vote for five of the candidates. We don't know how many people are participating in the election, but let's suppose that it's greater than the number of candidates. I will arbitrarily say there are 15 voters. Here are their votes:

      {b, d, c, f, g}
      {c, e, a, g, d}
      {a, e, b, c, f}
      {f, a, c, d, b}
      {e, c, b, a, g}
      {g, a, d, e, b}
      {d, a, c, e, g}
      {a, b, f, c, e}
      {d, b, a, f, g}
      {a, b, d, e, f}
      {f, g, c, b, d}
      {b, a, d, f, e}
      {g, d, c, b, a}
      {a, c, d, b, e}
      {d, f, e, g, b}

      3.) Now, those are in no way random; people aren't very good at picking random numbers or sets of things. We like patterns too much. In this case, however, that's fine. If anything an election isn't random, so it shouldn't be a factor in figuring out the problem.

      4.) Here's the vote tally for all seven candidates:

      b = 13
      a = 12
      d = 11
      c = 10
      e = 10
      f = 9
      g = 9

      5.) A couple of observations. First off, the total number of votes cast in the election (not the total number of voters, that's a different thing) is 78. Second, If we consider the "average" votes possible for each candidate, assuming voting were random (which it's not, but bear with me) is 78 / 7, or about 11. If we were to remove one candidate from this election, the "average" votes possible for each candidate would go up to 13. Due to the size of the cadidate pool and the voter pool, removing one candidate would have a fairly major impact on the number of votes a candidate could receive. I don't know enough about stats to say whether or not that's a problem, but it seems to me that having a nice spread of "winners" and "losers" around our average of 11 is better than having almost everyone be a "loser" because our new average is 13, and only candidate "b" received that many votes. Intuitively I'd think replacement would only make sense if the candidate who's being replaced had a vote total to the sixth-place finisher.

      I'd have to think about this alot more to come up with something more concrete, but perhaps this will get the juices going for someone else.

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