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A300305
Expected rounded number of draws until two persons simultaneously drawing cards with replacement from two decks of cards with m and n cards, respectively, both obtain complete collections, written as triangle T(m,n), 1 <= n <= m.
1
1, 3, 4, 6, 6, 7, 8, 8, 9, 10, 11, 11, 12, 12, 14, 15, 15, 15, 15, 16, 18, 18, 18, 18, 18, 19, 20, 22, 22, 22, 22, 22, 22, 23, 24, 26, 25, 25, 25, 26, 26, 26, 27, 29, 31, 29, 29, 29, 29, 29, 30, 31, 32, 33, 35, 33, 33, 33, 33, 33, 34, 34, 35, 36, 38, 40, 37, 37, 37, 37, 37, 37, 38, 38, 39, 41, 42, 45
OFFSET
1,2
COMMENTS
This is the two-person version of the coupon collector's problem.
LINKS
IBM Research, Ponder This Challenge February 2018. Solution for 3 persons from Robert Lang.
FORMULA
T(m,n) = round(1 - Sum_{j=0..m} Sum_{k=0..n} ( (-1)^(m-j+n-k) * binomial(m,j) * binomial(n,k) * j * k / (m*n-j*k) )) excluding term with j=m and k=n in summation.
EXAMPLE
T(1,1)=1, T(2,1)=3, T(2,2)=round(11/3)=4, T(3,1)=round(11/2)=6, T(3,2)=round(57/10)=6, T(3,3)=round(1909/280)=7.
The triangle starts:
1
3 4
6 6 7
8 8 9 10
11 11 12 12 14
15 15 15 15 16 18
18 18 18 18 19 20 22
22 22 22 22 22 23 24 26
25 25 25 26 26 26 27 29 31
...
CROSSREFS
Cf. A073593, A090582, A135736 (first column in triangle), A300306 (diagonal in triangle).
Sequence in context: A031131 A105321 A217032 * A160095 A135319 A329193
KEYWORD
nonn,tabl
AUTHOR
Hugo Pfoertner, Mar 07 2018
STATUS
approved