junior college 2 chevron_right H1 Maths

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.

I believe the way to solve for k = 0 is as follows. Since Homer is going first the probability he wins in the first throw would be the probability of a five or a six so 1/3. If he doesn't get it, the probability of him getting it on his second turn would be the probability of both him and Marge failing (2/3)(2/3) times the probability of him succeeding again so (1/3)(2/3)^2. This would repeat adding factors of (2/3)^2 each time until he eventually won. The total probability of him then winning would be the sum of all these. Looking at Marge's probability of winning the same way we can see her probability of winning is 2/3 of Homer's probability. Since either of them has to win the total off their probabilities would be 1. So Marge + Homer = 1, 2/3 Homer + Homer = 1, 5/3 Homer = 1 -> Homer's probability of winning = 3/5.

This is where I am stuck, I am unsure of how I would incorporate the other k values into the solution and how to go about answering the rest of the question.

Thank you for your help :)