AMC10 2008 10A, Problem 22

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1. If it comes up tails, he takes half of the previous term and subtracts 1. What is the probability that the fourth term in Jacob's sequence is an integer?

Solution: We can see that as long as the last flip is heads, it will be an integer, so there is a $\dfrac{1}{2}$ chance of this happening. Doing a little casework, we see that the only possibility when the last flip is tails is when the third flip brings it to 0. There is a $\dfrac{1}{8}$ chance of this happening. Therefore, there is a $\dfrac{1}{2}+\dfrac{1}{8}=\fbox{\frac{5}{8}}$ possibility of ending with an integer.

Answer: \frac{5}{8}