Once the value, y, of ticket B is decided, ticket C will be worth 2y in 4/7 of cases, and y/2 in 3/7 of cases. So the gain of switching from B to C is 1/7( 4(2y) + 3y/2 - 7y ) = 5y/14. Since y > 0 the gain in each case is positive for any value of x. Therefore, if given the chance, we should swap A for B and then B for C.
The problem is that ticket C is none other than ticket A. We have a paradox because after switching twice with positive gain each time we end up with the original value, ie x.