What is the probability of this?

  1. Hi,

    Suppose that there are M persons, and K balls number from 1 to K. Each person pulls k balls at the same time, and return them back. What is the probability that all persons pull the same k balls?

    Thanks
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Hi S_David! :wink:

    tell us what you think, and then we'll comment! :smile:
     
  4. Well, if we suppose that the probability of each ball is p, then the probability of a specific k-tuple is:

    [tex]{K\choose k}p^k(1-p)^{K-k}[/tex]

    I guess the probability that all persons will pull the same k-tuple, will be:

    [tex]\left({K\choose k}p^k(1-p)^{K-k}\right)^M[/tex]

    Is that true?
     
  5. Are there K balls in total for everyone or K balls per person?
     
  6. haruspex

    haruspex 12,795
    Science Advisor
    Homework Helper
    Gold Member
    2014 Award

    No, that's quite wrong. How many equally likely k-tuples are there? So what is the chance of picking a particular one?
     
  7. We have K balls in total, where each person pick up k balls, and then return them back to the box. So each person has the same set of balls to pick from them.
     
  8. We have K choose k distinct k-tuple patterns. Right?
     
  9. The solution you gave is the probability that M persons choose the same number of balls regardless their label.

    But you already know they choose the same number of balls, soooooo...
     
    Last edited: May 25, 2012
  10. I am sot sure!! Another tip?
     
  11. In this case, this is wrong.

    I think what you mean and correct me if I'm wrong, is that you have K balls and everyone takes turns in picking the balls out, not that they all pick out the balls at once.

    If they all pick them out at once, then there is no way they can pick out the same balls and the question is not even valid.

    So lets assume that they take turns and they pick out the balls each turn for each person and the process is random in line with the assumptions of a binomial distribution. We assume that after the person has picked the balls, they put them back in and the whole thing starts again.

    Then under this situation, you will need to calculate the probability of getting the specific number of balls taking into account that the balls are in fact the same. This should be done from first principles to clarify and not just using a distribution in ad-hoc manner.

    So in terms of calculating the probability of getting a particular set of balls you need to calculate the following probability:

    Let your balls be a,b,c and let A be the first ball, B the second, and C the third. Let X = (A = a) XOR (B = a) XOR (C = a), Y = (A = b) XOR (B = b) XOR (C = b) and Z = (A = c) XOR (B = c) XOR (C = c). Then you need to calculate the probability P(X AND Y AND Z).

    Where the XOR is the symmetric difference which is given by (A XOR B) = (A\B) OR (B\A).

    You can use the fact that A,B,C are not the same since picking up a ball and not replacing it until the round has finished will change what happens to later balls in the same round.

    If you expand the probabilities then you should get a simpler expression which will give you the actual probability and you will have to use the Kolmogorov axioms at some point for calculating P(A OR B) and so on.

    You could also do it with the binomial in the way that you calculate getting k balls out of K and then factoring in that you only get one type of permutation of your balls out of all permutations, but still account for getting them in any order. This would probably be easier to calculate in contrast with the raw probability calculation above.
     
  12. Well, I guess that one was not a very good tip anyway :tongue:

    OK, imaging you have 5 balls 1,2,3,4,5 what is the probability you pick, let's say 2 and 4? Well, then you would go the pair 2,4 divided by all the possible pairs in 1,2,3,4,5.

    So that is the probability you choose one particular pair, so that is the probability anyone else chooses that one particular pair... sooo... what is your guess now?
     
  13. If we suppose that probability of ball i is pi, then the probability of choosing 2 and 4 is:

    [tex]p_2p_4(1-p_1)(1-p_3)(1-p_5)[/tex]

    Right?

    Then why we need to divide by the number of pairs?
     
  14. What? :yuck: that's a new problem! in the way you stated your problem every ball is equally likely to be chosen...

    OK, the solution for the this problem as it is stated is [itex] {{K}\choose{k}}^{-M}[/itex]
     
  15. Dude, just do a binomial distribution and adjust for the different permutations so that you only select 1 permutation since all permutations are equally likely.

    How many permutations can you have when you select 3 balls from 6? (Hint: Use combinatorics)
     
  16. What if they are not equally likely? Why p does not appear in the probability?

    Actually, this is not the real question. The real question is: if each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?
     
  17. Maybe you should ask the real question first off :)

    Use probability axioms for situations that don't fit existing models.
     
  18. I am sorry, but it is just so confusing to me. I am not so good in probability, and I am trying to figure this out, since I am working on an engineering model, that has the same issue.

    Thanks
     
  19. It does not appear because every ball is supposed to have the same probability and that simplifies the problem.

    In the new stated problem, again, every ball has the same probability! so you already know your new solution won't have any p either.... :wink:
     
  20. OK, I will try to figure it out. Thanks
     
  21. No problem, but anyway, the answer for your new problem, assuming equally likely every cardinality, is [itex]K^{-M}[/itex].
    I think you've just suffered enough :tongue:
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?