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A085813
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Number of cards needed to be drawn (with replacement) from a deck of n cards to have a 95% or greater chance of seeing each card at least once.
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1
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1, 6, 11, 16, 21, 27, 33, 38, 44, 51, 57, 63, 70, 76, 83, 90, 96, 103, 110, 117, 124, 131, 138, 145, 152, 159, 167, 174, 181, 189, 196, 203, 211, 218, 226, 233, 241, 248, 256, 264, 271, 279, 287, 294, 302, 310, 318, 326, 333, 341, 349, 357, 365, 373, 381, 389
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| The probability that there are at most k different cards in t drawings is (k/m)^t * m_over_k, where m_over_k means the binomial-coefficient m!/(k!*(m-k)!). This contains also the cases with k-1 different cards, which we want to subtract. Inclusion and exclusion leads to the formula Sum[k=1..m; (-1)^(m-k) (k/m)^t * m_over_k].
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EXAMPLE
| a(2)=6 because you have to throw a coin 6 times to get both sides at least once with probability equal or higher than 0.95. (The probability of getting only one side in a series of 6 throws is (1/2)^6 * 2 = 1/32 = 0.03125 < 0.05.)
a(6)=27 because you have to roll a dice 27 times to see all 6 possible outcomes with a probability over 0.95. (If you roll a dice 27 times the probability to get all 6 sides at least once is 0,95658638... . If you roll the dice only 26 times, the probability is 0,94798274... .)
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MATHEMATICA
| f[1] = 1; f[n_] := f[n] = Block[{k = f[n - 1]}, While[ 2StirlingS2[k, n]*n!/n^k < 19/10, k++ ]; k]; Table[ f[n], {n, 1, 56}]
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CROSSREFS
| Cf. A073593 (number of drawings for a 50% probability to see each card, = median) and A060293 (expected value for the number of drawings until each card is drawn once).
Sequence in context: A145287 A140232 A184487 * A140527 A191158 A024730
Adjacent sequences: A085810 A085811 A085812 * A085814 A085815 A085816
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KEYWORD
| nonn
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AUTHOR
| Alfred Heiligenbrunner (alfred.heiligenbrunner(AT)aon.at), Jul 25 2003
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EXTENSIONS
| More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Sep 07 2003
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