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Please help me to judge whether a 'random' draw was fixed !

  1. Oct 11, 2012 #1
    40 teams are taking place in a knock-out competition in which there is no seeding. They all have rankings determined by previous performance. The pairings are decided by a completely random draw, e.g. all the team names are put in a hat and drawn by a neutral party. What are the chances of the 14 top-ranked teams not meeting any team ranked higher than 15th?

    Thanks,

    Mazza
     
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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    Welcome to PF;
    Sounds like a homework problem - to get the most out of these forums you need to show us your attempts at the problem or at least how you are thinking about it. That way we can target our assistance to your needs.
     
  4. Oct 11, 2012 #3
    I am not a statistician, just a regular person needing some help, but since you ask the question, I'm happy to show you my feeble efforts :

    The chances of team 1 meeting a team lower than 14th are 13/26, team 2 13/25, team 3 13/24 and so on, but this only applies if the teams are drawn in this order. I know my reasoning is flawed, but I don't know how to correct it.

    It is not actually a homework problem. My husband and I think that the draw in a recent bridge competition was flawed, but we want to get our facts straight before going back to the body concerned.

    Help will be hugely apreciated.

    Yours,

    Mazza
     
  5. Oct 11, 2012 #4

    chiro

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    Hey mazza47 and welcome to the forums.

    Just to clarify, are the pairings down in a 1-2,1-2 fashion? (So you draw out 20 pairs one at a time in succession)?

    From there on, do you basically just reduce by two each time? (So loser doesn't play anymore and winner goes to next level)? Are you just looking at the first level? If not what levels?
     
  6. Oct 11, 2012 #5
    Yes, the first two teams drawn out play each other, the next two play each other, etc, until eventually there are only two teams left in the hat and so they play each other. Thus there are 20 matches and the 20 winners go through to the next round. I am only interested in this round. I hope that answers your question.

    Mazza
     
  7. Oct 11, 2012 #6

    haruspex

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    If I've understood the question, it's the same as the prob that that the top 14 play each other and the bottom 26 play each other.
    First need to figure out how many ways 2N objects can be arranged as pairs. For ordered pairs, there are 2NCN ways of choosing the first of each pair, then N! Ways of assigning the second of each, giving 2N!/N!. But these are unordered pairs so need to divide by 2N. Call the resulting function q(N).
    The numbers of ways a particular 2R can be arranged as pairs is therefore q(R), and the prob is q(R)q(N-R)/q(N).
    However, there's a difficulty with applying this sort of logic after the event to a surprising occurrence. All sorts of equally surprising things could have happened, so the chances that something this surprising happened is actually much higher. So you really need to list all the outcomes that would have also made you suspect a fix, and add up the odds of all of them (insofar as they're disjoint).
     
  8. Oct 11, 2012 #7
    Thank you - that's all a bit advanced for me but I appreciate your taking the trouble to post.

    Mazza
     
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