Coming off of my last Switcheroo puzzle, here's another similar one in which you have to figure out the maximum number of options for the riddle.
The Monty Hall problem goes as follows: you're on a game show, and you can pick one of three doors. One has a car, and two have goats. There's an equal probability of any prize being behind any door. After selecting one door, the host reveals a goat behind one of the other doors, and offers you the option to switch. Should you? The answer, again in spoilers for those who don't know:
is to switch. Most people assume there's a 50-50 chance that your door has the car, and thus you shouldn't switch. They don't realize that by switching, you increase your odds to 2/3: since there are three doors, two of which contain goats, there's a 66% chance that you did not select the car.
My question is as follows: What if, instead of there being three doors, there were more? There's still only one car, and the rest of the doors contain goats. Each time, the host reveals a door with a goat, and you have the option to switch. The process repeats itself until you're left with one door. How many doors can there be for the logic above,
that you should switch every time,
to still apply? To prevent the answer from being infinity, say that once you switch from a door, you're not allowed to pick it ever again, unless you've already picked all of the remaining doors.