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Possible Duplicate:
Expected time to roll all 1 through 6 on a die
A question about Poker (and probability in general)

Given a N-sided unbiased die. What is the expected number of throws of die so that each number is rolled at least once? Please anyone give a hint or solution to solve it.

Community
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Bharat Kul Ratan
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There's a Wikipedia article about this problem: http://en.wikipedia.org/wiki/Coupon_collector%27s_problem

Suppose $k$ of the $N$ numbers have appeared so far. The probability that you get a new number on the next trial is then $(N-k)/N$. Let $X$ be the number of trials until that new number appears. Then $X$ is geoemtrically distributed on $\{1,2,3,\ldots\}$ and has expected value $N/(N-k)$ (e.g., if the probability of success on each trial is $1/15$ then the expected number of trials needed to get one success is $15$).

You're asking about the expected value of the sum of random variables like this one. The first has expected value $1$ (on the first trial, you necessarily get a new number). The second has expected value $N/(N-1)$. The third has expected value $N/(N-2)$. And so on. So the expected number of trials needed is

$$ 1 + \frac{N}{N-1} + \frac{N}{N-2} + \frac{N}{N-3} + \cdots + \frac N3 + \frac N2 + \frac N1. $$

Michael Hardy
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  • thanks, At the moment I didn't have enough reputation to vote up your answer. :( – Bharat Kul Ratan Sep 03 '11 at 17:30
  • @Michael: I guess you didn't see my duplicate notice in time (which shows once again that new comments should cause notifications just like new answers do). I think it would be preferable to add this as an answer to the original question (if you feel it improves on the existing answers there) so we can keep everything in one place and close this one. – joriki Sep 03 '11 at 20:06
  • @joriki, yes, but what is the original question? See Expected time to roll all 1 through 6 on a die (asked and answered on March 24). – Did Sep 03 '11 at 20:26
  • @Didier: Sure, we can use that one instead -- as long as we don't keep producing more duplicates. I can't change my vote, but I presume if you and others vote to close as a duplicate of that one, that will be the one linked to in the end. – joriki Sep 03 '11 at 20:28