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Rumour Spreading Probabilities

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A hostel is occupied by 40 tourists and a hostel-keeper. The hostel-keeper spreads a rumour to one of the tourists, who then tells it to another one, and so on.

    1) Find the probability that the rumour is spreaded 15 times without returning to the hostel-keeper.

    2) Find the probability that the rumour is spreaded 15 times without nobody hearing it twice

    3. The attempt at a solution

    I'm a little stuck with this, because I don't know what kind of problem it is. For 1), am I suposed to think that the rumour can be heard more than once by the rest but not by the hostel keeper?

    If so, then the |[tex]\Omega[/tex]| = 4015, because each person can hear it twice, three times, etc. But it wouldn't make sense that a rumour is heard 15 times by the same person, because that means that the tourists told the rumour to itself 15 times.

    Then, otherwise, |[tex]\Omega[/tex]|= (40)15, that is 40!/25! But then nobody would hear it twice, because 40!/26! = 40*39*38*37*36*35*34*33*32*31*30*29*28*27*26, which is to say that the first person to hear the rumour can then tell it to 40 other people and the one to hear it can tell it to 38 other people (everybody else but itself and the person who told it to him/her, which forbids it to be shared twice with the same person).

    So I don't even know how many events can happen. Point 2) only brings more confusion. Any thoughts?

    Thanks.
     
  2. jcsd
  3. Mar 12, 2010 #2
    Anybody?
     
  4. Mar 12, 2010 #3

    Dick

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    It looks to me like you assume the person that has been told the rumor randomly selects another person to tell the rumor. In the first case just find the probability that that person is not the hostel keeper 15 times in a row. In the second case at each stage the person selected has to be not the hostel keeper and not one of the people who has previously heard the rumor. It's just multiplying probabilities, isn't it?
     
  5. Mar 13, 2010 #4
    I don't know, I haven't even been able to define every possible elemental event. That's what I can't do. And if I can't define every possible elemental event, I can't know the probability of any of these two events.

    I mean, how many cases are there? Is it possible that the rumour can be told to the same person 15 times?

    Supose the spreading of the rumour is symbolized as {1,2,3,...,15} where 1 is the tourist 1, 2 is the tourist 2 and so forth; and the position in the 15-uple is the order in which the rumour was heard: that is, tourist 1 heard it first and then told it to tourist 2, who then told it to tourist 3 and so on until tourist 15. This can be done with 40 tourists, 1,...,40. So another one could be {1,40,35,25,...,13}. The question is, could it be possible that the rumour spread is something like {1,1,1,1,...,1}? Or {1,2,1,1,1,...,1}? Because if that was possible, the number or total events is 4015. Otherwise, I have no idea.

    And if nobody can re-hear the rumour, then question b) is pointless, isn't it?
     
  6. Mar 13, 2010 #5

    Dick

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    You are trying to count events in too much detail. One person won't hear the rumour 15 times. Two people can pass it back and forth 15 times. Ok, the hostel-keeper tells the rumour to one person. What's the probability that person picks the hostel-keeper to tell it back to?
     
  7. Mar 13, 2010 #6
    1/40 (since it's not telling it to itself).
     
  8. Mar 13, 2010 #7

    Dick

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    Ok, so probability 39/40 it does not get back to the hostel keeper at this stage. What's the probability for the next stage?
     
  9. Mar 14, 2010 #8
    39/40 it doesn't go back to the hostel-keeper. OK, I get it. It's (39/40)15, right?
     
  10. Mar 14, 2010 #9
    And for the second point? Is it (39)15/(40)15
     
  11. Mar 14, 2010 #10

    Dick

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    Maybe. I'm not sure what the subscript on the 39 means.
     
  12. Mar 14, 2010 #11
    Sorry, (39)15 = 39!/(39-15)! = 39*38*37*36*35*34*33*32*31*30*29*28*27*26*25.
     
  13. Mar 14, 2010 #12

    Dick

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    Ok, I didn't check that all the numbers where there. But that looks right. Does it seem right to you?
     
  14. Mar 17, 2010 #13
    It does. Thanks.
     
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