[bhc3]

My crack at it...

1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it.

But there's a 2/3 chance it's behind one of the other doors.

So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose).

You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B.

So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance

Make the switch every time.

[javajosh]

Dammit you beat me to it! Although maybe I can contribute because I was going to state it thusly...

Imagine a related Monty Hall problem, where you select a door, and then Monty immediately asks (without revealing anything), "Do you want to keep that door, or would you like to pick the other two doors?" Clearly you'd pick two doors instead of one.

When monty opens a door and gives you the "choice to switch" he is making noise designed to make picking two doors look like picking one door.

[beernutz]

Thank you. When you put it that way, it makes so much more sense to me. Opening the door does not change the odds. Why is that so counter intuitive?

[dperny]

I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.

[BerislavLopac]

I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.

[chrismcb]

One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.

[dllthomas]

Note that this depends on whether Monty's choice of door was informed by the location of the prize. If he picks randomly and just, this particular time, happens not to have revealed the car then it doesn't matter if you switch or not. Of course, might as well switch anyway in that case just in case you misunderstood...

[ecdavis]

I don't think that's correct.

The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.

[michaelcampbell]

> Note that this depends on whether Monty's choice of door was informed by the location of the prize

That's part of the setup. If it's not, it's not being told correctly. It's usually in the form of "...and Monty opens a door that is always a goat..."

[ntucker]

Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.

[ecdavis]

Fantastic explanation.

I think this is what the article was trying to get across with the example of the doors but I felt that just confused the matter by talking about "shifting" the probability.

[taeric]

Alternatively, there was a 1/3 chance you picked the winning door, which is the only scenario where you lose if you swap.

[pbreit]

What makes the answer most obvious is to play the game with 100 doors.