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I have N balls and M boxes. The balls are thrown at random onto the boxes. What is the probability that some box contains at least 3 balls?

Based on the Birthday problem, I know how to find the solution to the problem if we are finding the probability that some box contains at least 2 balls.

I am struggling to find the solution to this problem. Any help is much appreciated.

Here is what I have tried: If we have M boxes and N balls, then the probability that some box contains at least two balls is approximately equal to $$ 1-e ^{ (-n^2 / 2m)} $$

Let us suppose that we have 4 balls and 6 boxes, then the probability that some box has at least 2 balls is approximately equal to $$ 1 - e ^ {(-16/12)} .$$ This is approximately equal to 0.7364.

Now I would like to find the probability that some box will contain at least 3 balls, when 4 balls are thrown at random onto 6 boxes.

Thanks Sekhar

Henry
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user263229
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  • What have you actually tried? It's no use just telling that you're struggling; you need to show where you are struggling. (We don't want to reinvent the ball.) – Graham Kemp Aug 20 '15 at 05:21
  • I apologize for not being specific. This is my first post in this website. Here is what I tried: – user263229 Aug 20 '15 at 05:29
  • You can find some exact expressions here: http://math.stackexchange.com/questions/25876/probability-of-3-people-in-a-room-of-30-having-the-same-birthday –  Aug 21 '15 at 15:06
  • Thanks Byron. This really helps. I was struggling to understand how to solve this problem with Poisson distribution or Binomial dist. For three pairs I was dividing the Nc3 with M. But now I realized that I should use square (M) for three pairs. Thanks again. – user263229 Aug 21 '15 at 15:15
  • If we have 4 balls and 6 buckets, applying Poisson Distribution to find the probability of having at least 2 balls in at least 1 bucket ...I am getting a different answer. Using Poisson dist., the $$ Lambda = (4c2)/6 = 1 $$. Hence the Probability of having at least 2 balls in at least one bucket is $$ 1 - e^(-1) = 0.6321206 $$, but the correct answer is approximately 0.7222. Is it because the number of balls being too small and probability of a ball hitting a bucket being too large (a value of 1/6)? – user263229 Aug 21 '15 at 18:46

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There are $6^4=1296$ ways of distributing four balls among six boxes.

There are $6\times 5 \times 4 \times 3 = 360$ of having each ball in a different box, so the probability of at least two balls in the same box is $1-\frac{360}{1296} \approx 0.7222$ which is not far from your approximate calculation.

In addition, there are:

  • ${4 \choose 3}\times 6 \times 5=120 $ ways of having have three balls in one box and on in another.

  • ${4 \choose 4}\times 6=6$ ways of having four balls in one box.

So the probability of at least three balls in the same box is $\frac{126}{1296} \approx 0.0972$

Henry
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