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, 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

Author

Hugo Pfoertner, Mar 07 2018

Keywords

• nonn
• tabl

Comments

This is the two-person version of the coupon collector's problem.

Examples

			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
  ...
		

Links

• IBM Research, Ponder This Challenge February 2018. Solution for 3 persons from Robert Lang.

Crossrefs

Cf. A073593, A090582, A135736 (first column in triangle), A300306 (diagonal in triangle).

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.