Keep rolling while it is high
You roll a fair die and are paid its face value in dollars. If the roll was 4, 5 or 6, you roll again and are paid again, and so on. The game ends right after the first roll showing 1, 2 or 3, which is also paid.
Find the expected total payout.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Condition on the first roll. After a 4, 5 or 6, the rest of the game looks exactly like the whole game.
Alternatively: how many rolls are there on average, and what is the average value of a roll?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. After a high roll the game starts over, so the expected payout appears on both sides of a one-step equation.
Step 1: condition on the first roll. With probability the roll is 4, 5 or 6, worth 5 on average, and the game starts again; with probability it is 1, 2 or 3, worth 2 on average, and the game ends:
Step 2: solve. , so .
Check with Wald's identity. The number of rolls is geometric with mean 2, and each roll is worth 3.5 on average, so the total is worth . (The number of rolls depends on the values rolled, but only through the past, which is what Wald's identity needs.)
The trap. Using 3.5 for the rolls that continue the game. Given that the game continues, the roll was high and is worth 5 on average; given that it stops, the roll is worth 2.