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Math Help - Probability of assigning people to things

  1. #1
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    Probability of assigning people to things

    If I have 5 groups called a,b,c,d,e. And I randomly assign these five groups to 20 people. What is the odds that there will be 4 people in each group?

    Or more specificially how many times do I have perform this cacluation until there are exactly 4 people in each group?


    What then is the odds if I have 20 different dates. So 20 people and 20 dates how many times will I have to perform that calculation to get exactly 4 people in each group on each day.
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  2. #2
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    Quote Originally Posted by RhysGM View Post
    If I have 5 groups called a,b,c,d,e. And I randomly assign these five groups to 20 people. What is the odds that there will be 4 people in each group?
    I do not fully understand what you want to do.
    But there are 5^{20} ways to assign twenty people to five different cells. Note that some cells could be empty.
    There are \frac{20!}{(4!)^5} ways to assign twenty people to five different cells such that each cell has exactly four people.
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  3. #3
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    Quote Originally Posted by Plato View Post
    I do not fully understand what you want to do.
    But there are 5^{20} ways to assign twenty people to five different cells. Note that some cells could be empty.
    There are \frac{20!}{(4!)^5} ways to assign twenty people to five different cells such that each cell has exactly four people.
    Spot on.


    I put 5^{20} into excel and it gives me 95,367,431,640,625.

    If I put \frac{20!}{(4!)^5} into excel I get 305,540,235,000.

    So if I have 9.54*10^{13} total combinations but only want 3.06*10^{11}

    then does \frac{9.54*10^{13}}{3.06*10^{11}} give me the odds of getting the combination I want. Which gives me 312.

    So I have a 312 to 1 chance of getting the combination I want for each day.

    But this has to match for 20 separate days. So I'm assuming 312^{20} which gives 7.7*10^{49} too large for me to comprehend.

    Basically I have a programme running a loop checking all these combinations. So it has to do 7.7*10^{49} loops to find a valid combination?

    1,000 loops took 72 seconds. So will my programme take 1.76*10^{41} years to find 1 single valid combination???
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  4. #4
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    After reading your remarks in another thread about school children, I wonder if you need probability.
    There are \frac{20!}{(4!)^5(5!)} ways to assign twenty people to five groups of exactly four each.
    The difference here is that the groups are not labeled.
    Above you were labeling the groups as a,b,c,d,e. That is what we call an ordered partition.
    But something like ‘play-groups’ form an unordered partition because the groups are driven by content alone.
    The name of the group does not matter, only who is in the group matters.
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