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This question already has an answer here:

I tried to find the probability distribution for this problem and then calculate the expectation of $n$. I suppose the total states of our problem would be $52^n$ and our desired states would be $52$ combination of n with permutation $52!$ for distinct cards and our probability function will be $p(n)=\frac{n!}{(n-52)!(52^n)}$ with $n = 52,53, ....$ But the summation over $p(n)$ is not $1$. I don't know what is wrong with this solution. Can anybody help? Thank you so muchenter image description here

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marked as duplicate by joriki probability Jun 13 '16 at 13:12

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    $\begingroup$ "I suppose the total states of our problem would be 52^n" ?? The probability space cannot depend on n the value of the random variable defined on it. $\endgroup$ – Did Dec 24 '14 at 11:16
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    $\begingroup$ If you are only asked to calculate this expectation then I advice you not to use the distribution. $\endgroup$ – drhab Dec 24 '14 at 12:38
  • $\begingroup$ Why does the space not depend on n? @ Did $\endgroup$ – user203021 Dec 24 '14 at 13:12
  • $\begingroup$ How can we calculate expectation without distribution? @ drhab $\endgroup$ – user203021 Dec 24 '14 at 13:13
  • $\begingroup$ As @drhab wrote, it's almost never a good idea to go for the distribution if you're only interested in the expectation; the expectation is usually far easier to determine than the distribution. I marked this as a duplicate of a question asking for the expectation; if you're still interest in the distribution, see this question as well as the links in the comments there. $\endgroup$ – joriki Jun 13 '16 at 13:14
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Have a look at $p(53)$: $$p\left(53\right)=\frac{52}{52}\frac{\color{red}1}{\color{red}{52}}\frac{51}{52}\frac{50}{52}\cdots\frac{1}{52}+\frac{52}{52}\frac{51}{52}\frac{\color{red}2}{\color{red}{52}}\frac{50}{52}\cdots\frac{1}{52}+\cdots+\frac{52}{52}\frac{51}{52}\frac{50}{52}\cdots\frac{2}{52}\frac{\color{red}{51}}{\color{red}{52}}\frac{1}{52}$$

The colored factors correspond with the unique non-hit.

This is not the expression for $p(53)$ that you suggest.

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This is the Coupon Collector's Problem

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  • $\begingroup$ Thank you very much. That is correct. $\endgroup$ – user203021 Dec 24 '14 at 14:14

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