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.
Feb 16, 2022·edited Feb 16, 2022Liked by dynomight
I don't think you can do better than the naive solution.
Two thirds of the time, Monty's hands are tied: that's what yields the traditional solution. When his hands aren't tied, he doesn't have an avenue to communicate the difference between "I have to pick this door" and "We conspired to pick this door".
I think a clever thing about this variation is that it re-introduces a 50/50 likelihood in a manner similar to the false intuition of the original problem. If you conspire to pick a specific door under the scenario when you have already chosen the car, then when Monty selects that door it really is 50/50 for keeping your choice or switching.
I've verified this via Monte Carlo simulation, but I'd be very happy to be proven wrong. I'm one of the people who's lost productivity to this problem. :)
Last night I read your observations about writing and commenting on the internet, and felt both nourished and entertained. So OK, Dynomight, I'm going to think about these doors. But if playing with the problem fucks up my week, I'm going to lay a curse on you: recurrent dreams of making paper chains out of mobius strips. Really really LONG paper chains.
I don't think this is doable. To my understanding the only thing that matters is what Monty does when you pick the car, and therefore an algorithm comes down to where he picks when you pick car.. There are other ways to get to 2/3, i.e. "if I pick the car, always pick the one on the right", but fundamentally you always provide as much false information as true information. I imagine you might have some probabilistic solution but I am skeptical it will work.
1/3 You pick the Car, he is free to tell you to hold, you win.
1/3 You pick a Goat, he has no choice which door to open but it corresponds to switch, you win.
1/3 You pick a Goat, he has no choice which door to open but it corresponds to hold, you lose.
I don't think you can do better than the naive solution.
Two thirds of the time, Monty's hands are tied: that's what yields the traditional solution. When his hands aren't tied, he doesn't have an avenue to communicate the difference between "I have to pick this door" and "We conspired to pick this door".
I think a clever thing about this variation is that it re-introduces a 50/50 likelihood in a manner similar to the false intuition of the original problem. If you conspire to pick a specific door under the scenario when you have already chosen the car, then when Monty selects that door it really is 50/50 for keeping your choice or switching.
I've verified this via Monte Carlo simulation, but I'd be very happy to be proven wrong. I'm one of the people who's lost productivity to this problem. :)
Last night I read your observations about writing and commenting on the internet, and felt both nourished and entertained. So OK, Dynomight, I'm going to think about these doors. But if playing with the problem fucks up my week, I'm going to lay a curse on you: recurrent dreams of making paper chains out of mobius strips. Really really LONG paper chains.
I don't think this is doable. To my understanding the only thing that matters is what Monty does when you pick the car, and therefore an algorithm comes down to where he picks when you pick car.. There are other ways to get to 2/3, i.e. "if I pick the car, always pick the one on the right", but fundamentally you always provide as much false information as true information. I imagine you might have some probabilistic solution but I am skeptical it will work.
1/3 You pick the Car, he is free to tell you to hold, you win.
1/3 You pick a Goat, he has no choice which door to open but it corresponds to switch, you win.
1/3 You pick a Goat, he has no choice which door to open but it corresponds to hold, you lose.