# expected value of a rolling dice game

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.

• 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. Apr 15 at 16:54
• 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 $5$.
– lulu
Apr 15 at 17:40