LOL. You needed no computer for that. 1/3 to hold and 2/3 to stay was already given, so to actually apply this gain of position information to the second "choice" you'd have to add some code.
Let's just stick with your 2/3 result since we already know the carny rigs the game so as to make that result inevitable. The initial condition in terms of both winners and positions can be symbolized thus since you can really only choose Cup #1 and nothing is always revealed under Cup #3:
? X 0 : where "?" = maybe, dunno | "X" = dunno + no access | "0" = known loser
Obviously only 2 of the 3 can be winners and only 1 of the 3 if you could never switch to Cup #2. You've already stated as much w/o needing your app. However, the question remains as to whether knowing the position of one loser cup increases your odds when switching to better than 2 in 3?
Keep in mind that you can never switch to #3, only to #2. In other words, you really only have the supposed "choice" of #1 to begin with (no choice), and then only a really dumb choice between sticking with Cup #1 or switching to #2. At best, you are really only provided a choice between two, never 3, and only one informed choice.
Consider setting up your app to calculate the odds given one really had no choice to begin with and will definitely switch now, knowing that Cup #3 was never really an option.
100% (no choice) + 50% (1 in 2) yields ??? (more than 2/3?)