It becomes clearer that switching is the winning move if you imagine that there are 100 doors, and Monty shows you 98 goats. You had a 1/100 chance of picking the car. Monty has left only one door closed, so 99/100, the door that Monty leaves closed must be the car. (This is basically Unstoppable Gravy Express’s argument, but with more doors.)

I wonder: does this “always switch” strategy work with *Deal or No Deal,* assuming that a high-value briefcase is still in play at the end of the game? I feel like it’s basically the same problem.

I mean, you could answer “Lady & the Tiger” problems with “I put a slab of meat right outside the door and listen for growling” but that kind of misses the point.

]]>If one of the rules of the game is “Monty is required to offer you a switch,” then you should definitely switch. But if Monty is allowed any discretion about when he makes a switch offer, it becomes a lot less clear (and in fact, you could argue that switching is a flat-out bad plan, on the “cider in your ear” principle).

]]>The “run 1000 times” button grinds out 67-33 pretty consistently.

]]>YOUR CHOICE………………………Goat / Goat / Car

MONTY REVEALS…………………Goat / Goat / Goat

THEREFORE OTHER DOOR HAS: Car / Car / Goat

So in 2 of 3 cases, switching gets you the car.

]]>If you originally chose correctly, the unchosen door is 100% for sure a goat.

If you originally chose wrong, the unchosen door is 100% for sure a car.

Since there are 2 wrong doors to choose and only 1 correct door, the unchosen door has the car 2/3 of the time, and you should switch.

]]>Or, pragmatically: simulation proves the case.

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