The Monty Hall problem.

[Dotard]

More of my reasoning behind this:

You are basicly setting up two games. One is 1 out of 3. 33.33333% chance of 'winning'.

You make your choice, but instead of sticking to that choice the host removes one of the losing choices. Whether you picked a winning or losing door, one of the losing doors will be removed from the equation. So being more likely to have chosen a goat first time around is irrelevant. That would only be relevant if one of the two unpicked doors was removed without knowledge of it's contents (by the player).

Now you get a 2nd game. You are presented with two doors. One wins, one loses. Pick one. What are your chances or picking the winning one?

50/50