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Math Help - chairs

  1. #1
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    chairs

    Six individuals, including  A and  B , take seats around a circular table in a completely random fashion. Suppose the seats are numbered  1, \ldots, 6 . Let  X = \text{A's seat number} and  Y = \text{B's seat number} . If  A sends a written message around the table to  B in the direction in which they are closest, how many individuals (including  A and  B ) would you expect to handle the message?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Six individuals, including  A and  B , take seats around a circular table in a completely random fashion. Suppose the seats are numbered  1, \ldots, 6 . Let  X = \text{A's seat number} and  Y = \text{B's seat number} . If  A sends a written message around the table to  B in the direction in which they are closest, how many individuals (including  A and  B ) would you expect to handle the message?
    Using a purely intuitive argument, the number of people who can handle the message is either less than or equal to 4. The worst case scenario would have A and B directly facing each other, which would leave 2 persons on either side between them.
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  3. #3
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    but you have to use expected values (e.g probability of sitting in a seat is  1/6 ).
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  4. #4
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    We can assume that X = 1.
    Then E(X) = 1 and
    E(Y) = (1/5)(2+3+4+5+6) = 20/5 = 4.

    You can then define a random variable Z = 4-abs(abs(X-Y)-3). This is the number of people who handled the letter.
    E(Z) = 4-abs(abs(E(X)-E(Y))-3) = 4-abs(abs(1-4)-3) = 4.


    I could be wrong though, since I don't know if you can just assume that wherever A is sitting will be where you start numbering the chairs. The method I used to find Z is kind of complicated to explain. A general solution is beyond my grasp of Prob/Stats.
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  5. #5
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    I disagree with the reasoning in the replies to this question.
    Also I see no need to confuse things by numbering the chairs.
    Just seat person A at table. Now the table is ordered.
    There are (5!) ways to seat the other people.
    If X is the number of people handling the note, the X=2,3, or 4.
    Where are two ways to seat B next to A and then (4!) ways to seat the others. So P(X = 2) = \frac{{2\left( {4!} \right)}}{{5!}} = \frac{2}{5}. Or more simply, B can sit in two chairs of the five so that X=2. By the same simple logic we get P(X = 3) = \frac{2}{5}\,\& \,P(X = 4) = \frac{1}{5}\,.
    E(X) = \frac{2}{5}\left( 2 \right) + \frac{2}{5}\left( 3 \right) + \frac{1}{5}\left( 4 \right) = 2.8.
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  6. #6
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    Thanks Plato, I knew my method was somehow flawed. Seems I let the constraints posted in the problem constrain my logic.
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  7. #7
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    Yeah it seems in problems where people are sitting in a circle, you always set one person down first to set an order (e.g. circular permutations).
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