0

A question from a probability book I saw:

You are given a 100-sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game? (There is no limit on the number of rolls.)

A few StackExchange posts talk about this problem, here's one: Expected value of game involving 100-sided die all use the same optimal strategy: reroll until you have a die value greater than V.

But I don't understand why that's optimal. For example, the answer they give is stop rolling when you reach a die value of 87 or more. But what if you always get rolls below that, and thus, never actually payout?

Three questions:

  • How can one prove this is the optimal strategy?
  • What does it mean to say this game or this die is memoryless?
  • Is this still the optimal strategy if you have a finite number of dice rolls?
APerson
  • 107
  • Let me point out that you will eventually roll at least an $87$ with probability $1$ (that is, almost surely). – Brian Tung May 18 '23 at 16:37
  • True. Since this logic doesn't hold with a finite number of die rolls, does this strategy no longer work? Is there some "memory" now? – APerson May 18 '23 at 17:17
  • Yes. If you're on your last roll, obviously you'll take whatever you have (you have no choice). If you're on your second-to-last roll, you'll take anything in the upper half, approximately. And so on. To get exact figures, we have to do some minor calculations. – Brian Tung May 18 '23 at 18:05
  • Thanks! Yeah that does make sense, I've seen that approach before. – APerson May 18 '23 at 18:20

0 Answers0