conditional probability: climbing ladders up or down with 1/2 chance a robot can climb up the ladder or climb down. There are 3 bars of the ladder. Each time robot climb it, there is 1/2 chance to go up or down (except he is at the first bar of the stairs). 
Question: what's the chance for the robot to successfully reach the 3rd bar of the ladder?
(it's surely not just 1/2 * 1/2 * 1/2, cuz there is chance he can climb down during the process, so that's really bugging me... )
 A: Supposing reaching the $0^{th}$ floor means the robot stops, let $P(1)$ and $P(2)$ be the chance the robot reaches the third floor from starting at the first and second floor respectively.  Then
$$P(1) = \frac{1}{2} P(2)$$
$$P(2) = \frac{1}{2} P(1) + \frac{1}{2}$$
Use these to solve for $P(1)$
Edit:  I just noticed you said he gets $3$ moves.  Then yes, the chances would simply be $\left(\frac{1}{2}\right)^3$ since he needs to move up 3 times to reach level $3$ from level $0$.
A: It seems that in the model you have in mind, the robot starts at bar 1, goes to bar 2, flips an even coin to decide if it goes back to bar 1 (heads) or to bar 3 (tails), stops if it is at bar 3, and starts all over again if it is back at bar 1.
Thus the robot goes to bar 3 at the first tails in a sequence of heads and tails. The event that the $n$ first draws are heads has probability $\frac1{2^n}$ and $\frac1{2^n}\to0$ hence sooner or later a tails occurs, almost surely. Thus, the probability to reach bar 3 is $1$.
Edit: This also indicates that the probability the robot reaches bar 3 at its $(2n)$th move or before is the probability that at least one tails occurs during $n$ draws, that is, $1-\frac1{2^n}$.
A: The robot will reach the top of the ladder with probability $1$. 
