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A Question About Dice

I was reading " Expected time to roll all 1 through 6 on a die " and it got me thinking...

There are various ways to generalize the coupon-collector's problem (for example, by requiring certain tickets to be collected a certain number of times minimally... I think you can abuse exponential generating functions for that one).

The generalization I had in mind is this: The coupons have an arbitrary distribution: e.g. some are rarer than others.

Thus, given a discrete distribution $p_{coupon=x}(x)$, is there a solution that is more elegant than doing a massive calculation over the entire tree of outcomes with factorially-many branches? Also does the solution generalize nicely to continuous distributions, and if so, what does it mean to have the "coupon collection" problem on a continuous distribution?

A simple case might be, for example, how many rolls it takes to see all outcomes (2-12) of rolling a pair of dice. Of course there's a nice solution, but if I were to give an arbitrary distribution...

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ninjagecko
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    The coupon collector's problem with unequal probabilities is discussed here: http://math.stackexchange.com/questions/25568/a-question-about-dice/25576#25576 –  Jun 08 '11 at 20:54
  • To be more explicit: The answer to your question for a discrete distribution is given in Byron's linked answer. – Mike Spivey Jun 08 '11 at 21:06
  • A recent related question here http://math.stackexchange.com/questions/42313/throwing-all-numbers-from-2-to-12-with-two-dice/42345#42345 – leonbloy Jun 08 '11 at 21:11
  • @Byron: ah interesting! I managed to find online the textbook example 5.17 on p322 in Introduction to Probability Models 10th edition by Sheldon Ross, as cited in your answer, and it does indeed give a general solution to the coupon collector's problem with unequal probabilities. I'd be happy to accept any answer which stated it with reference, along with whether it generalized to continuous distributions (and what that would mean). – ninjagecko Jun 08 '11 at 21:21
  • Yeah, it's an interesting problem. Many variations of the original problem have been investigated. I would use Google and start looking around. Have fun! –  Jun 08 '11 at 21:28

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