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So an interesting question was brought to me today and I'm not sure how to formulate the equation to answer it. A person draws $5$ cards from a deck, writes the cards down, puts the cards back in the deck and reshuffles. What is the average number of draws they have to perform before all $52$ cards are drawn?

What about before before $45$ out of the $52$ cards are drawn?

Finally, how many draws if red cards were twice as likely to be drawn as black cards?

If you could show the equation for each it would be much appreciated.

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  • $\begingroup$ Just a note: The probability of drawing a card that hasn't been drawn before falls of exponentially. $\endgroup$
    – Bobson Dugnutt
    Commented Feb 10, 2016 at 17:20
  • $\begingroup$ This problem seems somewhat difficult at first glance. Do you have any reason to believe there is a reasonably simple expression giving the answer? $\endgroup$
    – David
    Commented Feb 10, 2016 at 17:23
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    $\begingroup$ I think you can set this up as an absorbing Markov chain. See en.wikipedia.org/wiki/Absorbing_Markov_chain . For example, for your first question, let state $S_n$ be the state in which $n$ of the $52$ cards have been discovered. The initial state is $S_0$, and the sole absorbing state is $S_{52}$. The probability of moving from state $S_n\to S_{n+i}$ is $\dfrac{{52-n\choose i}{n \choose 5-i}}{52\choose5}$. $\endgroup$
    – Steve Kass
    Commented Feb 10, 2016 at 18:07

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This sounds like the coupon collector problem. https://en.wikipedia.org/wiki/Coupon_collector%27s_problem

While the 5 card draw does make it easier, for a single card hand it will take 236 draws on average to collect all 52 cards.

So as an upper bound for the 5 card version I would say 48 hands.

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