This is an interesting puzzle in that it shows a basic technique used in solving these kinds of probability puzzles, as well as in models for some derivatives pricing, without the need for conditional probability calculations. Hint: start with the last event where we know the expected winnings, and work backwards.
Question: It’s your birthday, and a generous uncle tells you that he’ll give you in dollars whatever number comes up when you roll a die – so you will win between $1 and $6. He makes it more interesting – if you don’t like what you roll the first time, you can roll again. And if you don’t like that, you can roll one more time – for a maximum of three rolls. Note that if you choose to throw again, you can’t go back to an earlier better roll. What’s your strategy to maximize your winnings?
Answer: The trick here is to realize that on the third roll (if you get that far) you’re really stuck with whatever comes up, and that on average on your third roll you’ll win $3.50. So on your second roll, you should stick with anything 4 or higher, anything less you go on to the third roll. So, on average, your winnings on the second roll are ($6 + $5 + $4 + $3.50 + $3.50 + $3.50)/6 which equals $4.25. So then on your first roll, you should stick with a 5 or a 6, but go on to your second roll with anything less. On average, your winnings overall if you stick with this strategy will be ($6 + $5 + $4.25 + $4.25 + $4.25 + $4.25) /6 which equals $4.67.