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, $1-\frac1{2^n}$.