Probability Question

Question

Question: Homer and Marge play a game by taking turns rolling a standard six-sided die,
starting with Homer, until there occurs a sequence of k or more 4s immediately
followed by a 5 or 6 (e.g. 45, 245 or 5446, if k = 1). The last person to roll wins the
game. (E.g., if k = 2, then each of the sequences 446, 64445 and 465443445 results in
Homer winning.) Find the probability that Homer wins, for each k = 0, 1, 2, 3 and 99 

What I've done so far: 

for k=0
P[homer] = (1/3) + (1/3)(2/3)^2 + (1/3)(2/3)^4 + ....
P[marge] = (1/3)(2/3)^1 + (1/3)(2/3)^3 + (1/3)(2/3)^5 + ....

P[marge] = 2/3 * P[homer]
P[marge] + P[homer] = 1
5/3 P[homer] = 1
P[homer] = 3/5

 

I am not sure how to do the calculations for the other cases. Any help would be appreciated.              

Answer 1

I'd like to help you with this problem;  however, the wording of the problem is confusing.  Here are my questions:

1. You wrote "(e.g. 45, 245 or 5446, if k = 1)".   Isn't k=2 when the sequence 5446 occurs?
2. You wrote "E.g., if k = 2, then each of the sequences 446, 64445 and 465443445 results in
   Homer winning."  It's not clear why Homer would win.  Did you mean that the person (could be either Homer or Marge) who rolls the last 5 or 6  wins?
3. Is the question asking "What is the probability at the start of the game that Homer will win when K = 0, 1, 2, 3 and 99?
4. When k=0, the rules of the game imply that no one wins.  That is, you need at least one 4 to be rolled before anyone has the opportunity to win.  With k=0, it means that a 4 has not occurred.  Therefore, the probability of Homer winning (or Marge, for that matter) is zero.  Am I missing something?