A trading card series has 200 different cards in it, which are sold in 5-card packages.
Each package has a random sampling of the cards (assume that any card of the 200 has an equal chance of being in a package).
On the average, how many packages will need to be bought to collect the complete series if...
A: all the cards in a package will always be different
B: a package can have repeats
Problem A is a more annoying version of problem B. It can be solved in the same iterative way.
Let s equal the number of sets so far, n equal the number of unique cards so far, and p(s,n) equal the probability.
p(1,5) = 1
p(s,n) = the sum from i=0 to 5 of f(i,n-i) * p(s-1,n-i)
where x=n-i
f(0,x) = (200-x) * (200-x-1) * (200-x-2) * (200-x-3) * (200-x-4) / denominator
f(1,x) = 5 * x * (200-x) * (200-x-1) * (200-x-2) * (200-x-3) / denominator
f(2,x) = 10 * x * (x-1) * (200-x) * (200-x-1) * (200-x-2) / denominator
f(3,x) = 10 * x * (x-1) * (x-2) * (200-x) * (200-x-1) / denominator
f(4,x) = 5 * x * (x-1) * (x-2) * (x-3) * (200-x) / denominator
f(5,x) = x * (x-1) * (x-2) * (x-3) * (x-4) / denominator
denominator = (200-x) * (200-x-1) * (200-x-2) * (200-x-3) * (200-x-4)
For the special case of n=200:
p(s,200) = the sum from i=1 to 5 of g(i,200-i) * p(s-1,200-i)
g(1,199) = 5 * (1/200)
g(2,198) = 10 * (2/200) * (1/199)
g(3,197) = 10 * (3/200) * (2/199) * (1/198)
g(4,196) = 5 * (4/200) * (3/199) * (2/198) * (1/197)
g(5,195) = (5/200) * (4/199) * (3/198) * (2/197) * (1/196)
The expected value is the sum from s=40 to infinity of s * p(s,200).
a method of solving problem A
