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    Probability of throwing balls into bins

    Ok. I think I've got this. For n balls distributed randomly in N bins, the probability of picking a bin at random and finding x balls in it is: C(n-x+N-2, N-2)/C(n+N-1, N-1) which is the same as (n-x+N-2)!(N-1)!n!/[(N-2)!(n-x)!(n+N-1)!] for 50 balls distributed randomly between 20 bins the...
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    Probability of throwing balls into bins

    I am part of the way to calculating the numerator but unfortunately here in the uk it's time for bed.
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    Probability of throwing balls into bins

    It's interesting to note that for 1 bin there is only one way. All the balls go in the one bin. For two cups there are k+1 ways (n=number of balls, k = number of cups) taking 7 balls for some examples: 70 61 51 41 31 21 11 n+1 ways for 3 cups it is the sum of the natural numbers from 1 to...
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    Probability of throwing balls into bins

    I just wrote a long reply which I lost (that's what comes with being a noob). I have a similar answer to you but slightly different: C((n+k-1),(k-1)) which is (n+k-1)!/(k-1)!(n-2)! This gives 10 for putting 3 balls into 3 cups which is what I mentioned above. Here are the 10 ways: 300 210 201...
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    Probability of throwing balls into bins

    SW VandeCarr: I like your way of doing this but am having some problems understanding it fully. You seem to suggest a normal distribution around a peak 2.5. I have a problem with this because if you imagine a 3 ball deviation from the mean then you get a problem. Of course it is possible to have...
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    Probability of throwing balls into bins

    I believe solving this problem is what you do to derive the Maxwell-Boltzman distribution of energies in molecules at equilibrium, although in that case you have a very large number of bins (molecules) and balls (energy). That might be a good starting point for researching for a combinatorial...
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