Problem: A cereal company gives 2 images in each cereal box it has. There are a total of 5 images. Once a buyer have 5 images she wins a prize. No box contains 2 images that are the same. What is the expected value of the number of packages you'll have to buy to win a prize?
Quesiton: I wrote down the matrix P for this markov chain.
P=
1/10*
[
1 6 3 0;
0 3 6 1;
0 0 6 4;
0 0 0 1
]
Going from 0 to 2 unique card is always done in the firs round. In the second round, going from 2 to 4 cards har the chance $P_{2,4}=3/10$.
I then calculated $S=(I-P_T)^{-1}$ as described on page 5 here http://sites.stat.psu.edu/~jiali/course/stat416/notes/meantime.pdf
But now that I have this matrix, should I sum all entries in matrix $S$ or just in one of the rows.
(Just as I finished typing, I realised the answer. Correct me if I'm wrong).
Since your matrix start on 2 then going 0->2 (in the first round) must count as Expected value = $1$.
Then going from 2->4 is seen by summing up the first row, In this case $10/9+20/21+95/42$ logic: 2->2 + 2->3 + 2->4 ... no, wait, why are we not summing up the whole matrix? Please help me with the logic here.
The expected time is $1+10/9+20/21+95/42=5,3254$ You need to buy about 5 cereal boxes.