The Monty Hall problem is famously unintuitive. This post starts with an extreme version where the solution is blindingly obvious. We then go through a series of small changes. It will be clear that these don’t affect the solution. At the end, we arrive at the classic Monty Hall problem.
For reference, the classic formulation goes:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Intuitively, many people guess it doesn’t matter if you switch. But it does. You get the car 2/3 of the time if you switch, and 1/3 of the time if you don’t. Why?
Game 1 (Dyno Might© Monty Hall)
Here’s our first game.
 There are 10 doors. A car is randomly placed behind one, and goats behind the other 9.
 You pick one door.
 You get two options:
 Option A: You get whatever is behind the door you picked.
 Option B: You get whatever is behind all of the other 9 doors.
There’s nothing mysterious here. You should choose option B. There’s only a 10% chance you picked the right door, so there’s a 90% chance the car is behind one of the others.
Game 2
Now, we slightly update the game (new part in bold).
 There are 10 doors. A car is randomly placed behind one, and goats behind the other 9.
 You pick one door.
 Monty says “Hey! I promise you that there is a goat behind at least 8 of the other 9 doors!”
 You get two options:
 Option A: You get whatever is behind the door you picked.
 Option B: You get whatever is behind all of the other 9 doors.
Monty’s statement changes nothing. You don’t need to rely on his trustworthy looks. You already knew there were at least 8 goats! Option B still gets you the car 90% of the time.
Game 3
Let’s update the game again (new part in bold).
 There are 10 doors. A car is randomly placed behind one, and goats behind the other 9.
 You pick one door.
 Monty looks behind the other 9 doors. He chooses 8 with goats behind them, and opens them.
 You get two options:
 Option A: You get whatever is behind the door you picked.
 Option B: You get whatever is behind all of the other 9 doors.
The key insight is this: When Monty shows you that 8 of the 9 other doors contain goats, you haven’t learned anything relevant to your decision. You already knew there were at least 8 goats behind the other doors! So this is just like game 2. Option B still gets you the car 90% of the time.
Want more intuition? Suppose you picked door 3. Imagne Monty walking past the doors, opening doors 1, 2, 4, 5, 6, skipping 7, then opening 8, 9, and 10. Doesn’t door 7 seem special?
Game 4
Let’s make another change. Finally, we arrive at a game very similar to Monty Hall.
 There are 10 doors. A car is randomly placed behind one, and goats behind the other 9.
 You pick one door.
 Monty looks behind the other 9 doors. He chooses 8 of them with goats behind them, and opens them.
 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 only difference with Game 3 is that option B doesn’t get you the 8 visible goats. Since you don’t care about goats, this makes no difference. This is still just like the game 3. You get the car 90% of the time by switching.
Game 5 (Classic Monty Hall)
Here is the last game. We just change the number of doors from 10 to 3.
 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 1 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.
Of course, you still want to choose option B. The chance of success is now 2/3 instead of 9/10. This game is exactly Monty Hall, so we’re done.
Side Notes

It’s important that Monty looked behind the doors before choosing which to open. This is where people’s intuition usually fails. If he had chosen a door at random — in a way that he risked possibly exposing a car, then the situation would be different. (In that case, there’s no advantage or harm in switching.) But he doesn’t choose the door at random. He deliberately chooses to show you goats. Since this is always possible, it tells you nothing. I think this is the crux of what makes this problem unintuitive. Many people intuitively think it doen’t matter if you switch. And that would be correct if the door had been opened at random!

It might be helpful to draw a diagram of the relationship of the different games, starting with classic Monty Hall and ending with the extreme version.
Game 5 (Classic Monty Hall)
↓
↓ (Use 10 doors instead of 3)
↓
Game 4
↓
↓ (If you switch, get the contents of all other doors, not just the other closed door.)
↓
Game 3
↓
↓ (Monty promises 8 goats behind the other doors instead of showing you.)
↓
Game 2
↓
↓ (Monty doesn’t bother promsising.)
↓
Game 1 (Dyno Might© Monty Hall)

There are some other attempts at variants of the Monty Hall problem, also intended to be more intuitive. These involve switching the doors for “boxers”.

Monty Hall was actually named “Monte” at birth! Given that Monte Carlo simulations are often used for exploring the Monty Hall problem, that’s either a tragedy for puns or a miracle for confused students.