Suppose you repeatedly roll a fair six-sided die until you roll a 1 (and then you stop). Every time you roll a 2, you lose a points, and every time you roll a 6, you win b points. You do not win or lose any points if you roll a 3, 4, or a 5. What is the expected number of points (as a function of a and b) you will have when you stop?

In this questions what we are been asked is for the expected value E, to solve the problem we need to recognize the random variables and their respective probability.

x1 = roll a 1 -> P1 = 1/6
x2 = roll a 2 -> P2 = 1/6
x3 = roll a 6 -> P3 = 1/6
x4 = roll a 1, 4 or 5 -> P4 = 1/2

E = P1*x1 + P2*x2 + P3*x2 + P4*x4

x1 is equal to zero (no points and the game stop), x2 is equal to E-a (lose a points, affecting the expected value), x3 is equal to E+b (win b points, affecting the expected value) and x4 is equal E (no points are lost and the game continues, therefore the expected value is not affected)

E = 1/6*0 + 1/6*(E-a) + 1/6*(E+b) + 1/2*(E)

1/6*E = 1/6*b - 1/6a

Solving for E

E = b-a

Conclusion: If you get a 3, 4 or 5 the game is unaffected (in terms of points), you need to pay special attention to x1, x2 and x3 (all with the same probability) that are the results that are going to affect your game.