Distributing pebbles The rules to this "game" are simple, but after checking 120 starting positions, i can still not find a single pattern that consistantly holds. I am grateful for the smallest of suggestions.

Rules:
You have two bowls with pebbles in each of them.
Pick up the pebbles from one bowl and distribute them equally between both bowls. If the number of pebbles is odd, place the remaining pebble in the other bowl.
If the number of distributed pebbles was even, repeat the rule by picking up the pebbles from the same bowl. If the number of pebbles was odd, repeat the rule by picking up the pebbles from the other bowl.
Continue applying the previous rules until you have to pick up exactly 1 pebble from a bowl, at which point you win.

There are some starting positions, however, for which you will never be able to win. If the number of pebbles in the bowls are 5 and 3 respectively, you will cycle on forever.

Question:
Depending on the number of pebbles in each bowl at the starting position, can you easily predict if that position will be winnable/unwinnable?
Edit: Here is some python code i wrote to generate answers for given starting values: http://codepad.org/IC4pp2vH
Picking up $2^n$ pebbles will guarantee a win.
Edit:
As shown by didgogns, starting with n pebbles in both bowls always results in a win.

By the result of Simon, $T^k(n,n)=(1,2n−1)$ for some $k∈N$ if and only if $(1,2n−1)$ is in the orbit of $(n,n)$.
$(1,2n−1)$ is in the orbit of $(n,n)$ if and only if $(n,n)$ is in the orbit of $(1,2n−1)$.
Indeed, $$T^2(1,2n-1)=T(2n,0)=(n,n)$$ and it is obvious that there exist $a$ such that $T^a(1,2n-1)=(1,2n-1)$. $T^{a-2}(1,2n-1)=(n,n)$ because T is bijective and you always win the game at $(n,n)$.
