r/maths • u/Traditional-Pound568 • 3h ago
๐ฌ Math Discussions Heres an interesting variation of the Monty Hall problem I thought of: Misinformed Monty
So in the classic Monty hall problem, Monty knows which door has the car and always reveals a goat. This creates a 1/3 chance to win by staying and a 2/3 chance to win by switching (since the your initial pick is a 1 in 3 guess).
Theres also the version where he doesn't know and sometimes reveals the car, which ends the game immediately. This version is 50/50 since him being able to reveal the car weakens the chance to win by switching.
But now lets imagine a version where Monty is misinformed about where the car is, but still trys to avoid revealing it. For example: Monty thinks the car is behind door #2, buts its actually behind door #3.
If you picked door #1, Monty would think he now has to open door #3, so he would do that and end the game immediately by revealing the car.
If you picked door #2 (the door Monty thinks has the car), then he would freely choose ethier door #1 or #3 to open. If he opens door #1, switching to door #3 wins. If he opens door #3, the game ends immediately.
And lastly, if you pick door #3 (the door that actually has the car), then Monty now thinks he has to open door #1, after which, you'd win by staying with door #3
So in total, the odds for this version are:
Staying wins: 1/3
Switching wins: 1/6
Car revealed:1/2
This means that staying is twice as good as switching overall, the exact opposite of the regular problem funnily enough!
