Decision Tree Probability - With Back Step For the below decision tree, I can see how the probabilities of each end state are calculated... simply multiply the previous decisions:

But for this one below, I'm totally stumped. It seems in my head the chance at resetting back to the first decision is completely negated because essentially the whole decision restarts like the first decision was never made.  But based on the end probabilities this gives the s3 a larger chance at being chosen?

What does the math behind this look like? How are those final probabilities calculated given the reset in the decision tree?
 A: The probability of arriving at S1 is 
$$\frac 12\cdot \frac 29 
+ \frac 12\cdot \left(\frac 49\cdot \frac 12\right)\cdot\frac 29
+\frac 12\cdot \left(\frac 49\cdot \frac 12\right)^2\cdot\frac 29
+\frac 12\cdot \left(\frac 49\cdot \frac 12\right)^3\cdot\frac 29
+ \cdots\\
= \frac 19\cdot \left[1 + \left(\frac 29\right) 
+ \left(\frac 29\right)^2 + \left(\frac 29\right)^3 + \cdots \right]\\
= \frac 19 \cdot \frac{1}{1-\frac 29} = \frac 17 = \frac{2}{14}.$$
A similar calculation (replacing the trailing $\displaystyle\frac 29$'s
by
$\displaystyle\frac 39$'s gives the probability of arriving at S2 as 
$\displaystyle \frac{3}{14}$.
At this point, we can jump to the conclusion that the probability of
arriving at S3 must be $\displaystyle\frac{9}{14}$ without sullying our hands with
more summations of geometric series, but more skeptical folks can work with
$$\frac 12 
+ \frac 12\cdot \left(\frac 49\cdot \frac 12\right)
+\frac 12\cdot \left(\frac 49\cdot \frac 12\right)^2
+\frac 12\cdot \left(\frac 49\cdot \frac 12\right)^3
+ \cdots\\$$
which looks a lot like the sum on the second line of this
answer except for those trailing $\displaystyle \frac 29$'s, and so
we get that the probability of arriving at S3 is $\displaystyle\frac 92$
times the probability of arriving at S2, which gives us
$\displaystyle \frac 92\cdot \frac{2}{14} = \frac{9}{14}$.
Look, Ma! No more summations of geometric series! 
