How many tables get at least one guest?

Question

A reception hall has 6 tables. Four guests arrive and each one independently sits at one of the 6 tables chosen uniformly at random. On average, how many tables end up with at least one guest?

Accepted Answer

Skip the bookkeeping of how the guests cluster and instead put one indicator on each table. Fix a table. It is empty only if all 4 guests choose a different table. Each guest avoids a given table with probability 	frac56, so P(	ext{a given table is empty}) = \left(	frac56ight)^4 = \frac{625}{1296}. So a given table is occupied with probability 1 - 	frac{625}{1296} = 	frac{671}{1296}. Let Xi = 1 if table i is occupied; E[Xi] = 	frac{671}{1296}. Summing over 6 tables, E[X] = 6 	imes \frac{671}{1296} = \frac{671}{216} \approx 3.106. Takeaway. The number of occupied bins is a sum of one …

How many tables get at least one guest?

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