TITLE: Probability of covering all vertices of a square
QUESTION [5 upvotes]: Let $~A_1 = (0,0), ~~A_2 = (1,0),~~ A_3 = (1,1)~$ and $~A_4 = (0,1)~$ be the four vertices of a square. 
A particle starts from the point $~A_1~$ at time $~0~$ and moves either to $~A_2~$ or to $~A_4~$ with equal probability. 
Similarly, in each of the subsequent steps, it randomly chooses one of its adjacent vertices and moves there. Let $~T~$ be the minimum number of steps required to cover all four vertices. 
The probability $~P(T = 4)~$ is 
$(A) ~~~~0$
$(B) ~~~~\frac{1}{16}$ 
$(C) ~~~~\frac{1}{8}$
$(D) ~~~~\frac{1}{4}$
I am getting it as $~\frac{3}{4}~$ but answer is $~\frac{1}{8}~$ please help!

REPLY [5 votes]: To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.
It could have been $12143$ or $14123$.
Hence $\frac{2}{2^4}=\frac18$