Expected Value of a Spinner Game where you can Spin Again! Suppose I wanted to calculate the expected value of the spinner game below, where both options are equally likely: you can either spin and win \$2 or spin and win \$1 with the addition that you can now spin again.

The probability of gaining \$2 is $\frac{1}{2}$, the probability of gaining \$3 (from getting \$1 and spinning again to get \$2) is $\frac{1}{4}$. Continuing in this way, the expected value should be
$\sum_{k=1}^{\infty} \frac{1}{2^k}(k+1) = \sum_{k=1}^{\infty} \frac{k}{2^k} + \sum_{k=1}^{\infty} \frac{1}{2^k}$.
Now the first term evaluates to 2 (see How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?) and the second term evaluates to 1, so that makes the expected value 3.
However, I ran into considerably more difficulty when dealing with the following game, this time involving four options: spin again, earn $x_1$ dollars and spin again, earn $x_2$ dollars and spin again, and just earn $x_3$ dollars. What is the expected value?

Okay, my first thought was to get rid of the "only spin again" piece, and just view the problem as 3 regions of probability $\frac{1}{3}$, where 2 of the regions required additional consideration.
Next, I started viewing the problem in terms of number of spins the person makes, so for one spin, there was only one way that happens, with probability $\frac{1}{3}$. To get two spins, this happens with probability $\frac{2}{3}*\frac{1}{3}$. There are 2 scenarios with which this occurs: $(x_1, x_3), (x_2, x_3)$.
For three spins, this happens with probability $\frac{2}{3}*\frac{2}{3}*\frac{1}{3}$. There are 4 scenarios with which this occurs: $(x_1, x_1, x_3), (x_1, x_2, x_3), (x_2, x_1, x_3), (x_2, x_2, x_3)$.
So far, building my expected value to come up with an infinite series looks promising: $\frac{1}{3}*x_3 + \frac{2}{3}*\frac{1}{3}*[1(x_1+x_2) + 2(x_3)] + \frac{2}{3}*\frac{2}{3}*\frac{1}{3}*[4(x_1+x_2) + 4(x_3)] + \frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{1}{3}*[12(x_1+x_2) + 8(x_3)] + ...$.
Unfortunately, I am unable to see a pattern with the number of $x_1, x_2, x_3$ terms. Either it is a simple pattern I failed to recognize, or I completely messed up this entire analysis. I really appreciate the help!
P.S. I put the terms $x_1+x_2$ together because I noticed with the first game example that the expected value was just the sum of the two options. Obviously it isn't as simple, but my conjecture is that this grouping is involved somehow.
 A: For the 2nd game, let $E$ denote the expected value.
Then, assuming that I am interpreting the question correctly, there are $4$ possible outcomes of the 1st spin:

*

*$(1/4)$th of the time, you win $x_1$ and spin again. 
In this event, your expectation must be $x_1 + E$, because after the 1st spin, you are in exactly the same position as before the 1st spin, except that you are $x_1$ richer.


*$(1/4)$th of the time, you win $x_2$ and spin again.
Expectation must be $x_2 + E$.


*$(1/4)$th of the time, you win $0$ and spin again.
Expectation must be $0 + E$.


*$(1/4)$th of the time you win $x_3$.
Expectation must be $x_3$.
Adding all of these expectations up, you get that
$$E = (3/4)E + [(1/4) \times (x_1 + x_2 + x_3)] \implies E = (x_1 + x_2 + x_3).$$

Addendum
I just can't resist:
Consider the following variation of the 2nd problem: 
It is assumed that the player may voluntarily stop spinning and walk away, at any time.
The $4$ possible spins are:

*

*Option-1 
You win $x_1$ and spin again.

*Option-2 
You win $x_2$ and spin again.

*Option-3 
You lose $x_1$ and spin again.

*Option-4 
You lose $x_2$ and spin again.

So, if you try to use the same approach that I used at the start of my answer, you get the equation that
$$E = (4/4)E + (1/4)(0) \implies 0 \times E = 0.$$
Good luck trying to compute $E$ using the above equation.
In my opinion, the only way to attack this variation is to conclude that your mathematical expectation on any individual spin is
$$(1/4) \times [(x_1 + x_2) - (x_1 + x_2)] = 0.$$
Then, you conclude that any individual spin is worthless.  Therefore, the prospect of spinning an infinite number of times must also be worthless.  Therefore, the game must be a break-even game.
So, this variation represents a problem that can not be attacked using the method that I used in the first part of my answer.
