You accumulate 1 point when rolling a 1,2, or 3. You get paid out your points when rolling a 4 or 5. You get nothing if you roll a 6. What is your expected value?
I came up with the following solution. I get 1 point with probability (1/6), 2 points with probability (1/12), 3 points with probability (1/24), and so on.
So then the EV is: $\frac{1}{6}\sum_{n = 0}^\infty \frac{1}{2}^n \times (n+1)$
Two question: how do I actually evaluate that sum, and second how can I set up a recursive equation to solve the problem. It seemed like recursion would be easier but I couldn't set up the equation properly
You can do the sum using the differentiation trick $$ \frac{1}{6}\sum_{n=0}^\infty (n+1)x^{n} = \frac{1}{6}\frac{d}{dx}\sum_{n=0}^\infty x^{n+1} = \frac{1}{6} \frac{d}{dx} \frac{x}{1-x} = \frac{1}{6}\frac{1}{(1-x)^2}.$$ Plugging in $x=1/2$ gives an answer of $2/3.$
As for recursion, this is trickier. The probability the game ends on a six is $1/3$ by symmetry since it is the probability you roll a six before you roll a four or five. So we may work in an imaginary world where sixes never occur. Let $V$ be the value of the game. We have a $3/5$ chance of landing on a one, two or three, in which case we gain one point and continuation. Otherwise, if we land on $4$ or $5,$ we gain nothing. The recursive equation is that $$V = \frac{3}{5}(\frac{2}{3}V+1)$$ (where the $2/3$ on the inner $V$ is to cut the continuation value to take into account the possibility of landing on a six) so that $V = 1,$ which we then multiply by $2/3$ to take into account the sixes, and get the answer of $2/3.$