I finally figured out the logic behind the Monty Hall problem and why it works the other day. Reflecting on other explanations I've seen, I hit upon the commonly provided one where instead of 3 doors there's 100, and after you choose one 98 goats get revealed.
This is a variant of that.
Let's take a set of 100 randomly-generated numbers, and assume that the car is behind the last number in the set. The rules are as follows:
- The player chooses a door
- The host opens the next door on the list
- If the player's current door gets opened before the last number, he loses immediately. Otherwise, he has a choice of whether to keep his current door or switch it.
- Repeat steps 2 and 3 until either the contestant's door is opened or the final door is opened.
Under these conditions, and assuming a truly random distribution of numbers being chosen, does the same principle still hold? Is it better to shift around every move or to wait for more favorable odds (e.g. perhaps sitting out the first 50 numbers or so)?
In short: Do the rules of the Monty Hall problem change when recursively opening the doors and allowing switching instead of jumping to revealing N-2 doors and allowing one switch?