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  #1  
Old 12-30-2005, 02:55 PM
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Default How many expected unique cards in 52 of 6 decks?

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?

So how do i figure this out for 6 decks or for N decks? What about an
infinite deck?
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  #2  
Old 12-30-2005, 06:41 PM
BruceZ BruceZ is offline
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Default Re: How many expected unique cards in 52 of 6 decks? (edited)

[ QUOTE ]
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?

[/ QUOTE ]

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.


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

[/ QUOTE ]

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


[ QUOTE ]
What about an infinite deck?

[/ QUOTE ]

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 -> infinity. So we have established the general result that:

lim N -> infinity C(k*(N-1),N)/C(k*N,N) = [(k-1)/k]^k.
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  #3  
Old 12-31-2005, 03:28 AM
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Default Re: How many expected unique cards in 52 of 6 decks? (edited)

Wow that's exactly what I needed... great approach and thank you.
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