How to come up with different ways to find the same solution? 
A coin has probability $p$ of heads. Adam flips it first, then Becca, then Adam, etc., and the winner is the first to flip heads. Compute the probability that Adam wins.

So I saw this problem this afternoon and I tried solving it on my own. I used the following reasoning to get to the solution:
If $n$ denotes the number of tosses done in total. Then the only way Adam can win is if the head appears at an "odd" trial ($2n-1$). For the head to occur at the $2n - 1$ trial the first $(2n-1) - 1$ tosses must be equal to tails, so I expressed this as
$$p\sum_{n=1}^\infty (1-p)^{2n - 2}$$
I then realized that this is a geometric series, which converges to the following value
$$\frac{p}{1 - (1-p)^2} = \frac{1}{2 - p}$$
From my perspective I thought this was the only way you can get to this result, but the paper from which I got the question from used a very different approach (recursion). So I guess my question is, how can you train yourself to think about problems like this in more than one way?
 A: Think about different ways the problem can be represented. The more math you know, the more tools from which you'll have to choose. Also don't try to judge ahead of time which methods will be useful, because you never know what you may discover with a given approach.
For this particular problem, what about drawing out a tree, like this? Let's see...Adam's probability of winning is $0.5 + (0.5)^{3} + (0.5)^{5} + \ ... = 0.5 + 0.125 + 0.03125 + \ ...$, it's not too tough to see that this is steadily going to get closer and closer to a limit of $\frac{2}{3}$.
Are you familiar with Markov chains? That might be a bit quicker. We define 4 states, which are applied to the horizontal and vertical a with the same labels: Adam's flip, Adam wins, Becca's flip, Becca wins. The initial state distribution matrix, then, looks like this:
$$\begin{pmatrix}
0 &0.5  &0.5  &0 \\ 
0 &1  &0  &0 \\ 
0.5 &0  &0  &0.5 \\ 
0 &0  &0  &1 
\end{pmatrix}$$
Translation:
First row says: there's no chance of going from Adam's flip to Adam's flip, a 50% chance of moving from Adam's flip to Adam's win, a 50% chance of moving from Adam's flip to Becca's flip, and no chance of moving from Adam's flip to Becca's win.
Second row says: There's a 100% chance of moving from Adam's win to Adam's win (absorbing state).
Third row says: there's a 50% chance of moving from Becca's flip to Adam's flip, no chance of moving from Becca's flip to Adam's win, no chance of moving from Becca's flip to Becca's flip, and a 50% chance of moving from Becca's flip to Becca's win.
Fourth row says: There's a 100% chance of moving from Becca's win to Becca's win (absorbing state).
To find out how this plays over the long run, we raise this to infinity (or, more realistically, an arbitrarily high power), to get the stability distribution.
Overall, I would recommend 3 things to help learn to vary your mathematical perspective:


*

*Read Martin Gardner's books, especially his 15 collections of Scientific American columns. Persi Diaconis put it best when he said, "Martin Gardner has turned thousands of children into mathematicians, and has turned thousands of mathematicians into children." Martin Gardner's 2 books, Aha! Gotcha! and Aha! Insight! are also great lessons in changing your mathematical perspective.

*Study James Tanton's Curriculum Inspirations material closely.

*Learn that you will never be stumped if you change your point of view.
A: It's not really a different approach.   Demonstrating recursion is a method by which the closed form of a geometric series can be obtained.
$$\begin{align}S & = \sum_{k=0}^\infty r^k  \\ &= 1+r\sum_{k-1=0}^\infty r^{k-1} \\[1ex] &= 1+ rS \tag{Recurrence} \\[2ex] \therefore \quad S &= \frac{1}{1-r}\end{align}$$
So if a geometric series is involved in evaluating the probability, there will be a recurrence in the set-up.   If you can spot the recursion, you can use it to directly evaluate the probability.   It may be easier to find one or the other depending on how you look at it.
You just need to be familiar with the proofs of the results you use and have practice spotting patterns.
Practice, practice, practice.  
Also practice some more.
And when you are not practicing, practice.
A: Let $A$ be Adam's win probability. 
He has a probability  $p$ of winning on the first flip, and a $(1-p)^2$ probability of surviving the second - at which point his probability of winning would be $A$, because it's the same game.
So, $A = p + (1-p)^2A$, which reduces to the answer in the paper. 
Even more tersely, the probability of Adam winning when it's not his flip is $1-A$.  
So, $A = p + (1-p)(1-A)$; that gives $A(2-p) = 1$ directly.
