Expected value of dice game You throw a die. Each time you get a 4,5, or 6, you get the value on the face of the die. When you get a 1, 2, or 3, you quit the game (but keep any winnings). What's the expected value of this game?
Here's how I'm approaching this. I tried to first write an equation for the EV, as so:
$EV = (1/2)\cdot (0) + (1/2) \cdot (5+EV)$ - however, this assumes that you don't get to keep your winnings; how do I account for the winnings?
Next, I tried approaching it like this, where I sum up all of the possibility; you can get a $\{1,2,3\}$ on your first roll, second roll, third role, etc. where the EV of a non-$\{1,2,3\}$ roll is $5$.
$(1/2)(0) + (1/2)(1/2)(5) + (1/2)(1/2)(1/2)(10) + (1/2)(1/2)(1/2)(1/2)\cdot 15 ....$ how do I find a closed for expression for this series (it looks both like an arithmetic and geometric series)
 A: Both approaches look correct! The $5+{}$in the first equation does account for the winnings you get to keep.
The series you found in the second case is
$$
\frac52 \sum_{k=1}^\infty \bigg( \frac12 \bigg)^k k.
$$
This is indeed a notch more complicated than a simple geometric series, but we can still handle it! Starting from
$$
\sum_{k=0}^\infty x^k = \frac1{1-x},
$$
valid for $|x|<1$, we can take the derivative of both sides to get
$$
\sum_{k=0}^\infty kx^{k-1} = \frac1{(1-x)^2},
$$
which implies
$$
c\sum_{k=1}^\infty kx^k = \frac{cx}{(1-x)^2}.
$$
This is exactly the series you want, with $c=\frac52$ and $x=\frac12$, and therefore your series equals
$$
\frac{(5/2)(1/2)}{(1-1/2)^2} = 5,
$$
which is the correct expected value.
A: Here is another approach.
Let's say your expected payoff is $E$. On your first throw, you roll a 4,5,or 6 with probability $1/2$ and an initial expected gain of
$$\frac{1}{6}(4+5+6) = \frac{15}{6}$$
and keep going, or you roll a 1,2,or 3 and stop.  If you get to keep going, then you are in exactly the same state you were in before the first throw, with an expected gain of $E$.  So
$$E = \frac{15}{6} + \frac{1}{2} E$$
Solve for $E$.
