Actually, your winning chances would even increade to 2:3. (Which is more logical because 1:3 + 1:2 = 5:6 < 1)

I heard this proplem as "Ziegenproblem" and the "Zonks" were goats. The propability of the remeaining cup/envelope/door increases because it COULD have been a zonk/goat/nothing but it hasn't shown to be one. So if it's not the one you chose it's the remaining one and the propability that it's not the one you chose is 2:3.
It look a long time for me to figure that out, until I wrote a program that solves the propgram statistically and while looking at the code I understood it.