Geometric Probability with Lines A point is on a number line, but it is limited to being at 0, 1, or 2. At each step, the point moves 1 unit left or right. If the point is at 1, it moves to either 0 or 2 with equal probability. However, if the point is at 2, it must move to 1 with its next step.
For a point currently at 1, let $t_1$ be the expected number of steps the point will take before it reaches 0 for the first time. Similarly, for a point currently at 2, let $t_2$ be the expected number of steps for the point to first reach 0.
Determine the ordered pair $(t_1,t_2)$.
I am thinking that because this problem deals with expected value, I should find the average moves it will take to get to 0. But I am stuck their. How should I find how many steps it will take to get from 2 to 0, because the average move would be move to 1, then move back to 2, then move to 1, and move back to 2, etcetera etcetera. So how would you progress from 2 to 0?
 A: If you start at $1$, then: with probability $\frac{1}{2}$ the point will arrive at $0$ in $1$ step; also with probability $\frac{1}{2}$ the point will go from $1$ to $2$ and back to $1$ in the first two steps (so the expectation given you are in that case is $2 + t_1$). Hence $t_1$ must satisfy:
$$t_1 = \frac{1}{2}\cdot 1 + \frac{1}{2}(2 + t_1)$$
which you can solve by rearrangement to find $t_1 = 3$. A similar argument gives you that $t_2 = 1 + t_1 = 4$ (since if the point starts at $2$, with probability $1$ it moves to $1$ on the first step and then the expectation is $t_1$).
This method is equivalent to the suggested development of geometric series in another answer, but gives you the sum of those series directly. (As an aside, the UK newspaper the Guardian has a puzzle column on Saturdays that has more interesting maths problems than most newspapers: it gave repeated examples of this kind of problem a few months ago.)
A: Hint: Each time you are at $1$, you may terminate with probability $1/2$ by making one more step to 0, or else you will make 2 more steps by going to $2$ and then automatically return to $1$, also with probability $1/2$. So you should be able to formulate a series resembling a geometric series with powers of $1/2$ for your answer. Also, note that $t_2 = t_1 + 1$ by the rules.
