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Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer hereHenry's answer here, or below
$$P(\mbox{ every box has at least one bean after }T \mbox{ seconds}) = P^{-T}\ P\ !\ \left\lbrace {T\atop P} \right\rbrace.$$

In general, I think you should take Alekk's advice and use the Poisson approximation.

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here, or below
$$P(\mbox{ every box has at least one bean after }T \mbox{ seconds}) = P^{-T}\ P\ !\ \left\lbrace {T\atop P} \right\rbrace.$$

In general, I think you should take Alekk's advice and use the Poisson approximation.

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here, or below
$$P(\mbox{ every box has at least one bean after }T \mbox{ seconds}) = P^{-T}\ P\ !\ \left\lbrace {T\atop P} \right\rbrace.$$

In general, I think you should take Alekk's advice and use the Poisson approximation.

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user6096
user6096

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here., or below
$$P(\mbox{ every box has at least one bean after }T \mbox{ seconds}) = P^{-T}\ P\ !\ \left\lbrace {T\atop P} \right\rbrace.$$

In general, I think you should take Alekk's advice and use the Poisson approximation.

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here.

In general, I think you should take Alekk's advice and use the Poisson approximation.

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here, or below
$$P(\mbox{ every box has at least one bean after }T \mbox{ seconds}) = P^{-T}\ P\ !\ \left\lbrace {T\atop P} \right\rbrace.$$

In general, I think you should take Alekk's advice and use the Poisson approximation.

Source Link
user6096
user6096

Your problem falls under the general category of "coupon collector's problems". There is a large literature on such problems, mainly concerned with formulas for the mean and variance of the random variable $T$.

The exact distribution of the time $T$ to have at least one bean ($N=1$) in each box is known in terms of Stirling numbers of the second kind: see Henry's answer here.

In general, I think you should take Alekk's advice and use the Poisson approximation.