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#2
12-30-2005, 06:41 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: How many expected unique cards in 52 of 6 decks? (edited)

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6 decks of 52 cards are shuffled together randomly. How many UNIQUE
cards (i.e. rank and suit) would you then expect in the first 52 dealt?

9 is the minimum, and 52 is the maximum. What is the expected number?

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The probability that a particular card, say the ace of spades, appears in the first 52 is 1 minus the probability that it doesn't appear which is 1 minus the probability that all 6 copies of this card are in the remaining 5 sets of 52 cards, or 1 - C(5*52,6)/C(6*52,6). The sum of these probabilities over all 52 cards gives the expected value of the number of unique cards, which is 52*[1 - C(52*5,6)/C(52*6,6)] =~ 34.75.

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So how do i figure this out for 6 decks or for N decks?

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N decks: 52*[1 - C(52*(N-1),N)/C(52*N,N) ]

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What about an infinite deck?

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I'll assume you mean an infinite number of 52 card decks. In this case, each card has a probability of 51/52 of not being chosen on each draw, and the probability of not being chosen the first 52 draws is (51/52)^52 since an infinite number of each card means there is no effect of removal. The probability of each card being chosen in the first 52 is 1 - (51/52)^52, so the expected value of the number of cards chosen is 52*[1 - (51/52)^52] =~ 33.06. You can confirm that this is the limit of the answer to the second part as N -&gt; infinity. So we have established the general result that:

lim N -&gt; infinity C(k*(N-1),N)/C(k*N,N) = [(k-1)/k]^k.