by **Tarsier** » 02 Mar 2018, 16:28

So I saw a post on Facebook about tricky math problems, and the illustration was for the Monty Hall problem. On my way to drop the kids off at school, I thought about it some. I posted that problem in this forum and we had some debate about it. Everyone's pretty logical here, though, and even those who struggled with it a bit saw their way into the logic eventually. However, I also posted on a different website/forum where I've seen a less logical basis for a lot of posts, and the arguments were more fierce. Some absolutely refused to accept the answer, even after the extreme scenarios of 1,000,000 doors were presented. Even with those odds against picking the ONE door with a car on the first try, some posters were insisting that, if Monty Hall eliminated 999,998 of the doors, the odds were still 50/50 and no door had any advantage with probability.

It occurred to me this morning that those same posters would (hopefully) have trouble with that conclusion, and realize the benefit of a switch, if we tried going in the opposite extreme:

Suppose you are on Let's Make a Deal, and Monty shows you ONE door. He tells you there is a car behind ONE of those doors. Which door would you pick?

Um ... I'd pick the only door possible - the ONE door.

That's right. And what would the probability be that you picked the door with the car.

100%, dude. It's the only door.

Okay. Now, as a fun twist, Monty Hall adds ANOTHER door. He reminds you that the car is only behind one of the doors. Any other doors hide a goat. He invites you to keep your original pick or switch. Are your odds better one way or the other? The car remains where it was at the start of the game.

I would stay the same - with the door I picked when there was only one door.

But there are two doors, now. The odds are 50/50, right?

No. I picked my door when I had 100% odds of getting it right. I'll stick with that one.

Just as, from this extreme, the odds don't change that you should keep the door you picked, from the other extreme (1,000,000 doors), the odds don't change that you probably picked wrong. Nor does it change with only three. In other words, your original pick was biased by a different probability that influences how correct you probably were with that selection (assuming that the rules are fair and the host knows what is behind the doors).

Thoughts? Is this a better explanation for those who struggle wrapping their mind around it?