Voltage drop using resistor divider for ADC sensing
The cheapest feasible option is R1 only, R2 is not needed. 68k is about the right value to drop the 50k voltage span to 1.4v. However, potentiometers are notorious for poor tolerance, so the next value up would guarantee less than 1.4v on the pot.

To choose R1 for any given value of Rpot, supply voltage Vs, and voltage required on the top of the pot Vpot, solve the following equation.

$$V_{pot} = V_s \frac{R_{pot}}{R1 + R_{pot}}$$

I'm sure you can use straightforward algebra to manipulate the equation round so that R1 is by itself on the left hand side.

This equation expresses the fact that the current flowing through the two resistors is Vs/(R1+Rpot), so the voltage developed across Rpot is just Rpot times that current. I've separated out the resistor ratio however, as that's the way we'd normally use this potential divider expression, to multiply the input voltage by a function of the resistor values, to get the reduced output voltage.

Rather than flog through the algebra, you might notice that 1.4v is more or less half of 3.3v, so R1 will be a little bit bigger than Rpot. In fact I guessed at 68k (a little bit bigger than 50k) before I did the sums to calculate it exactly.