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By Jszab (Jszab) on Sunday, October 12, 2003 - 11:27 am: Edit |

Could You tell me what is probability of being alone in randomly aggregating group?

By Fairyofwind (Fairyofwind) on Sunday, October 12, 2003 - 11:41 am: Edit |

I assume that you mean that if there are n people (including you), what is the probability that if the n people randomly get into groups of arbitrary size (and each different group arrangement is equally likely), you will be alone.

The number of group arrangements is B_n, where B_k's are Bell numbers.

Now if you were alone, the number of group arrangements is B_{n-1} (Why?).

So the probability is (B_{n-1})/(B_n).

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