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Here are two brainteaser:

You roll a 6-face even dice with following rule:

  1. If you roll 1,2,3, you get 1 dollar and the game continues; If you roll 4,5, you loss everything (go back to zero) and continue; If you roll 6, the game stops and you are paid the accumulated dollars you get.
  2. If you roll 1,2,3, you get 1 dollar and the game continues; If you roll 6, you loss everything and the game stops (get 0); If you roll 4,5, the game stop and you are paid the accumulated dollars you get.

I can only think that suppose $E(X)$ are the expected value when you have $X$ dollar. Then the iteration is (for 1 as an example): $$E(X) = \dfrac{1}{2}E(X+1) + \dfrac{1}{3}E(0) + \dfrac{1}{6}X.$$ And we need another equation to solve $E(0).$

Question 2 are discussed here: Dice Stopping Game (Law of Total Expectation). I don't understand the explanation and I think it is not a general solution.

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  • $\begingroup$ If you work out a solution by treating $E(0)$ as a constant $c$, you can then substitute in $E(0)=c$ at the end, and that should give the extra equation you need. $\endgroup$
    – Zoe Allen
    Commented Apr 15, 2023 at 16:54
  • $\begingroup$ Note that the outcomes in which you hit a $4$ or $5$ before hitting a $6$ have no impact on the expected value. Thus you can just compute the expected length of a sequence of tosses given that you get a $6$ before you get either a $4$ or a $5$. $\endgroup$
    – lulu
    Commented Apr 15, 2023 at 17:40

1 Answer 1

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Both games: you roll some number of 1,2,3s before rolling a 4,5,6. The expected number of rolls is 1 ( the sum of 1/2, 1/4, 1/8 ...). On average you'll have a single dollar when 4,5,6 comes up.

Game 1: if you roll 4,5 you get a fresh start - as if you just started playing with no accumulated winnings. If you roll 6 you can expect to keep 1. E = 1

Game 2: you roll 6 1/3 of the time and win 0. You will roll 4,5 2/3 of the time an expect to keep 1. E = 2/3

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