You participate in a game show where you must choose between 3 doors. Behind 2 doors is a goat, but behind 1 door is a car. You pick a door, say door no. 1, but before you open the door, the host opens another door, say door no. 3, which has the car behind it. "Oops," says the host.
You lose.
Yup. I hadn't done the full rationale, but it does seem to always break even. If chosen randomly, remaining doors minus no-prize doors, you're still left with an even chance versus your initial pick. In the three-door scenario, it's 1/3rd vs 1/3rd, and the last third is missing, so equal chances, i.e. 50/50.
And, that's a good point, even if you don't know how he's choosing, you always switch, because if he is choosing randomly it's a coin flip, and if he isn't you're increasing (sometimes very drastically depending on number of doors) your chances, so it's an overall gain.