Calculating equivalent resistance 

In the given circuit, find the equivalent resistance between points A and B.

I solved this question in two different ways, out of which one gave the correct answer. The correct method is the most popular method: assuming the potential of given points and then calculating the others using the fact (or assumption) that in the absence of resistance, the potential difference is zero. This gives the correct answer: $R_{eq} = \frac{R}{3}$. I was confused regarding why the other method did not work.

Current follows the least resistance path when available. That is why you should not connect a conducting wire (R = 0) parallel to a resistor as it might lead to short circuit.

Using this, I can conclude that the current can flow in two possible paths: (1) from $V_A$ (leftmost) to $V_A$ (second from the right) to $V_B$ (rightmost) (2) from $V_A$ (leftmost) to $V_B$ (second from the left) to $V_B$ (rightmost).
Each of the paths has resistance R, so $R_{eq} = \frac{R}{2}$. But this is incorrect. Where is the mistake?

Update

Current follows the least resistance path when available.

The proof for this lies in the current divider rule. $$I_1 = I\frac{R_2}{R_1+R_2}$$ Putting $R_1 = 0$ we find that the entire current flows through $R_1$.
 A: (continued from the previous answer by  SG8)

Another way to think about this problem.
First:

Second:

Finally:

A: 
Current follows the least resistance path when available.

This is wrong. Recall the Ohm's Law
$$V=IR\Rightarrow I\propto \frac{1}{R}\ \  \ \ \ \ \text{if V constant.}$$
If there are two resistance one having resistance $1\ \Omega $ and the other $2\ \Omega$, connected in parallel then this doesn't mean that if the current will flow through $1\ \Omega$ resistance. Ohm's law says that there would be more current in the less resistant path.
A: If you re-arrange the diagram you will find the you have three resistors in parallel.  Another way to see this is to realise that (if we assume $V_A>V_B$) current flows from left to right through the first and third resistors, but from right to left through the middle resistor - so there are actually three paths from A to B, each with resistance $R$.
A: I am used to smoothing out badly shaped circuits by pulling the wires:

Then I get a better circuit by cutting the extra wires:

So there are three resistors in parallel, indicating that the current flows through three possible paths.
A: 
Using this, I can conclude that the current can flow in two possible paths: (1) from $V_A$ (leftmost) to $V_A$ (second from the right) to $V_B$ (rightmost) (2) from $V_A$ (leftmost) to $V_B$ (second from the left) to $V_B$ (rightmost).

The critical mistake that you've made is that there's one more path that you've overlooked. Namely, the path from $V_A$ (leftmost) through the wire to $V_A$ (second from right), "backwards" through the middle resistor to $V_B$ (second from left), through the wire to $V_B$ (rightmost).
Each of these three paths has resistance $R$, so $R_{eq} = \frac{R}{3}$.
This whole argument about the three equal paths is equivalent to simply noticing that the three resistors are actually in parallel with each other.
(There's also the misconception about the path of least resistance, which is covered in Young Kindaichi's answer.)
