John S.

# Probability 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.

By:

Tutor
5 (79)

Engineering manager professional, proficient in all levels of Math

John S.

1. In the question it states that if  k=1, so there can be k or more 4's before the 5 or 6 is rolled.
2. Because Homer rolls on the odd throws (he starts the competition), thats why he wins because the 5 or 6 is thrown on the odd throw.
3. It is asking for the probability of Homer winning for each of the cases of k
4. When k=0, it means there can be 0 or more 4's before a 5 or 6 is rolled. So the probability can be found.
Report

03/17/18

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.