#1
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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? |
#2
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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. |
#3
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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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