Expected value in a dice roll game

Question

""Roll a die, and you get paid what the dice shows. But if you want, you can request a second chance & roll the die again; get paid what the second roll shows instead of the first. What is the expected value""
This question has been asked many times. But my question is whats wrong with my reasoning.
So a normal approach goes like this: Expected value on the first dice roll is $3.5$. Now if you get $4,5,6>3.5$, don't re-roll. Else, re-roll.
So this gives, $Exp.=(\frac36)(5)+(\frac36)(3.5)=4.25$.

Now my reasoning is just a bit different. Let the expected value of the whole game be $x$. Now if I play the game infinite times, I would expect to win $x$ out of it. So if on the first roll, I don't get at least $x$, then I would re-roll. (This was the difference).
Using this logic gives the equation (assuming $x$ to be a real number not an integer but a fraction): $$x=\frac{\lfloor x\rfloor}{6}(3.5)+\frac{6-\lfloor x\rfloor}{6}(\frac{\lfloor x \rfloor+1+6}{2})$$ This equation holds true for $x=\frac{25}{6}\approx4.1667$.
So what is wrong with my approach?

Edit: Looking back at the problem it seems that the question is missing one piece of information, and that is when does the player decide to re-roll. So assuming that threshold to be $k$, the expected value becomes: $$\frac{\lfloor k\rfloor}{6}(3.5)+\frac{6-\lfloor k\rfloor}{6}(\frac{\lfloor k \rfloor+1+6}{2})$$ which attains maximum at $\lfloor k\rfloor=3$. So this means that the question basically assumes that the player would take a step which basically maximises the total expected value.