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 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: As you said, if the decision resets, it's as if it never happened.  Let $V$ be the event "The system returns to the initial state after the current iteration", then the probability of each decision $P(D_i)$ can be replaced by the probability $P(D_i|\neg V)$, and the loop can be removed.  This only affects decisions where looping is still a possibility, so the tree shown is equivalent to a tree with no back-step (but the same structure otherwise), where the upper branch has probability $\frac 5 {14}$, the lower one has probability $\frac 9 {14}$, and the upper and lower branches of the second stage have probability $\frac 2 5$ and $\frac 3 5$ respectively.
Breaking it down further, we can start by looking at the first step, and observing that our chances of coming back are $\frac 1 2 \frac 4 9=\frac 2 9$. Thus probability of not looping back is $1- \frac 2 9=\frac 7 9$.  Since the bottom branch has probability $\frac 1 2$ the probability of taking it given that we don't loop back is $\frac {\frac 1 2} {\frac 7 9}=\frac 1 2 \frac 9 7=\frac 9 {14}$ meaning the upper branch has probability $1-\frac 9{14}=\frac 5 {14}$. For the second stage we repeat the process, except the probability of looping is now $\frac 4 9$, so our conditional probabilities are $\frac {\frac 2 9} {\frac 5 9}=\frac 2 5$ and $\frac {\frac 3 9} {\frac 5 9}=\frac 3 5$.
One way of understanding why the loop affects the outcome is that, while both branches are equally likely to be chosen, if you choose the upper branch, your choice only has a $\frac 5 9$ probability of "sticking", but if you choose the lower branch, it sticks with probability $1$ , thus we should expect the upper branch to be our "final" (that is non-looped) choice only $\frac 5 9$ as often as the lower branch, which is exactly the case. 
