The Monty Hall problem has now been a pox on humanity for two generations, diverting perfectly good brains away from productive uses.
So it is with great trepidation that I’d like to announce a new and more pernicious variant.
As a reminder, here’s the standard problem:
- There are 3 doors. A car is randomly placed behind one, and goats behind the other 2.
- You pick one door.
- Monty looks behind the other 2 doors. He chooses one of them with a goat behind it, and opens it.
- You get two options:
- Option A: You get whatever is behind the door you picked.
- Option B: You get whatever is behind the other closed door.
The well-known solution is that you should always switch, which gets you the car 2/3 of the time.
In our variant, Monty is corrupt. The night before the game, you show up at his house with a suitcase of cash. He agrees that he will follow whatever algorithm you want when choosing the door to open. (Of course, he still can’t open a door with a car.)
What makes this challenging is:
- Monty doesn’t know where the car will be before the show starts.
- After the show starts, Monty can’t communicate with you other than by choosing what door to open.
The problem is: What is the best strategy for you and Monty to use? And if you use that strategy, how likely are you to get the car?
The naive solution would be to play as if there was no conspiracy: If Monty picks a random door to open and you always switch, then you get the car 2/3 of the time. The question is: Can you do better?
I’m pretty sure I have the answer, but
people always yell at me when I talk about this stuff I don’t yet have an elegant proof.
Do you have one? If so, please send it to me. When/if I get some nice solutions, I’ll post an update.
Update: I’ve gotten a few dozen proofs. Thanks everyone! I’m going to write an update with a summary of all the different proof strategies. If you have a particularly weird or elegant proof that you suspect other people haven’t thought of yet, please do send it, though!