How to calculate how much profit a loan of a certain risk level (based on dice rolls) is likely to make?

Let R be the risk of the loan and L be the length of the loan. What is the probability that you receive exactly 0 payments? You would have to roll less than R on your first roll which has probability $\frac{R-1}{6} $

What is the probability that you receive exactly 1 payment? You would have to roll once at least R and then once below R. The probability of that is $\left(\frac{7-R}{6}\right)\left(\frac{R-1}{6}\right)$

In general, the probability that you get exactly n payments, for n strictly less than L, is $\left(\frac{7-R}{6}\right)^{n-1}\left(\frac{R-1}{6}\right)$

L is a special case. In order to get exactly L payments you need to never roll less that R, L times in a row. The probability of that is $\left ( \frac{7-R}{6} \right )^{L}$

Let X = the number of payments you recieve.

By the definition of expected value:

E(X) = 0*P(X = 0) + 1*P(X = 1) + ... + L*P(X = L)

= $\sum_{i = 0}^{L-1} i * \left ( \frac{7-R}{6} \right )^{i-1} \left ( \frac{R-1}{6}\right ) + L * \left ( \frac{7-R}{6} \right )^{L}$ 

If we let P = the amount of each payment, and Y = the total amount you receive in payments, then Y = X*P

By the Linearity of Expectation, E(Y) = E(X*P) = P*E(X) 

= $P * \left (\sum_{i = 0}^{L-1} i * \left ( \frac{7-R}{6} \right )^{i-1} \left ( \frac{R-1}{6}\right ) + L * \left ( \frac{7-R}{6} \right )^{L}\right)$ 

Let I = the amount of the inital loan, which is a constant you always have to pay. Therefore your total expected profit is

$P * \left (\sum_{i = 0}^{L-1} i * \left ( \frac{7-R}{6} \right )^{i-1} \left ( \frac{R-1}{6}\right ) + L * \left ( \frac{7-R}{6} \right )^{L}\right) - I$

You can play around with the 4 variables here:

https://www.desmos.com/calculator/etinwj2epc