You ask for expert and i am here. Sorry for the delay friends, but this question is a case of binomial probability distribution. What we are suppose to do is to find probabilities for general cases and then add them for all the cases. Instead of finding all the cases first and then finding there probabilities.

Consider right side displacement as +1 nad left side as -1

After five moves x will be at +5,+3,+1,-1,-3,-5 depending upon how many right or left moves he makes.

 

In case he makes i right moves and 5-i left moves his position will be at 2i-1 with a probability of C(5,i)(2/3)ⁱ(1/3)^5-i  [C(5,i) ways to make i right and 5-i left moves]

I can write Px(2i-5)=C(5,i)(2/3)ⁱ(1/3)^5-i

Simillarly Py(2j-5)=C(5,j)(1/2)ⁱ(1/2)^5-i

 

I want y to be on right of x

Therefore j>i

So required probability is

_{[i=0 ]}^[5 (Px(2i-5)*_{[j=i+1 ]}^{[5 ]} Py(2j-5))

 

Add this to get final answer