Expected value for games where you can replay? Lets just say there's a game where you roll one fair die.  If you roll a 1 or 2, you pay 1.  If you roll a 3 or 4, you win 2.  If you roll a 5 or 6 you roll again until you get a 1, 2, 3, or 4.
How much are you expected to win?  I can't figure out how to think about this.
Thanks.
 A: Note that your expected gain given that you first rolled a $5$ or $6$ is the same as your expected gain initially… you just get to start over.
Using linearity of expectation, then, you can write your expected gain as
$$
E[G]=\sum_{i=1}^{6}E[G\;\vert\;X_1=i]\cdot P[X_1=1]=-\frac{1}{3}+\frac{2}{3}+\frac{1}{3}E[G]=\frac{1}{3}+\frac{1}{3}E[G],
$$
and then solve to yield $E[G]=1/2$.  
A: Does it really matter if you get a 5 or a 6? It doesn't really,  because you just end up back to where you started. At the end of this whole game, you can only lose 1 or win 2.
So there is a $\frac12$ chance you have a net profit of $-1$, and a $\frac12$ chance you have a net profit of $2$. 
$\frac12*-1+\frac12*2=\frac12$
In case you still think that the reroll affects the expected win rate, let's go through that process.
There is a 2/3 chances you get 1-4, and so we multiply the above fraction with 2/3. The other 1/3 chance means we do this all over again from the beginning. You end up with an infinite expression that looks like:
$$x=\frac23*\frac12+\frac13(\frac23*\frac12+\frac13(\frac23*\frac12+\frac13(...)))\\x=\frac23*\frac12+\frac13(x)\\6x=2*1+2x\\4x=2\\x=1/2$$
A: You pay one if either you roll $1$ or $2$, or you roll $5$ or $6$ for $k$ times and then you roll $1$ or $2$ again. Therefore the probability of paying one is:
$$
\Pr(W=-1)=\frac 13+\frac 13\left(\frac 13+(\frac{1}3)^2+...\right)=\frac 13\times \frac{1}{1-\frac 13}=\frac 12
$$
Similarly probability of winning two is equal to $\frac 12$ and you can find your expected value which is $\frac 12$.
The idea is that the probability of not winning anything is to roll $5$ or $6$ for infinite amount of times whose probability is zero and because of symmetry of the problem, the probability of each other cases are $\frac 12$.
