Because MATH, that's why.

[theweaselking]

You roll two fair 6-sided dice.

One of them is a 6.

What are the odds that the other is also a six?

[Poll #1770498]

Comments

[theweaselking.livejournal.com]

That's the right answer.

And, then and here, the question is *not immediately clear*, not ambiguous. That's the thing about word problems, paradoxes, Monty Hall, and all the unintuitive results: Figuring out what the question is actually asking, and not jumping to the wrong conclusion, is an important part.

[ciphergoth.livejournal.com]

Hmm, I still think there is a duty on the asker to make the question totally, utterly clear, absolutely sealed against any but the most wilful misinterpretation. EG with Monty Hall, it's very important to make clear that he *always* opens a door, that he *always* shows you a goat. Otherwise you could end up at the "Monty Fall" problem - he falls on one of the doors by accident and it opens, revealing a goat - in which instance of course it makes no difference whether you stick or switch.

[theweaselking.livejournal.com]

I agree that it needs to be unambiguous, but I don't feel that restating Bob And Sue as "What percentage of Bob's wins come in the second turn" is the duty of the questioner, even though that's an equivalent question AND it's a much simpler statement of the problem.

The problem as stated is unambiguous.

Re Monty Fall: No, in *that instance*, Monty's revelation still gives you additional information, and still results in there being a 2/3 chance of it being not behind your picked door. It's simply not usable as a basis for a universal strategy.

If Monty Fall opens the door *before* you pick, you're right, it's 50/50.

And yes, it's important to explain that the behaviour in the puzzle is universal: Monty will ALWAYS open a door and he will ALWAYS reveal a goat by doing so. Sue will ALWAYS roll first. Monty will never accidentally reveal the car, and he will never decide not to open a door, and if you've chosen correctly, which door he opens is random.

[ciphergoth.livejournal.com]

Let me state Monty Fall as clearly as possible.

In this game, you pick a door. Then Monty taunts you about your choice for a bit to see if you change your mind, and finally opens the door you settle on at the end. Monty doesn't know which door the car is behind.

While he's taunting you, he slips on something and knocks open a door that as it happens, you didn't pick, revealing a goat. He hurriedly closes the door again and carries on as if nothing has happened, and asks you if you want to stick with your choice or change to another.

Does it make a difference what you do?

[theweaselking.livejournal.com]

Interesting.

I still want to say "I'm taking either my choice or the best of all possible doors that I did not pick" - which is to say, "1/3" or "2/3".

The fact that Monty *could* have slipped and showed the car doesn't change that I'm still getting "my door or all other doors" - except another part of me wants to say "yes, but all we've determined is that one of the doors is worthless. So you're getting either your door or another equally likely door."

Except THAT leads into "but isn't that the crux of the Monty Hall problem in the first place?"

So:
If I have 1000 doors, I have a 1/1000 chance of being correct and a 999/1000 chance of being wrong. Ignorant Robot Monty opens non-picked doors at random, without knowing where the car is - and in any case where he reveals the car, we throw that test out and start over. Given enough tests, he has opened all doors except my door and one door, and revealed no car... but all that shows is that the car was originally behind either my door or the one unopened door. Which does appear to compress to 50/50 nicely, because the lack of Robot Monty's knowledge means he COULD have revealed the car, and when he did we threw out those tests. So all we're left with is the subset of tests where he DIDN'T reveal the car: Where the car is either behind my door or the last door, with no indication of which.

As compared to the original Monty Hall, where he *cannot* reveal the car, and thus we're examining the set of all tests without ever discarding any of the tests.

So yes. Monty Fall appears to compress to 50/50. And *that* is a very neat nonintuitive result!

[ciphergoth.livejournal.com]

This is why Monty Fall is great fun to spring on people crowing about how people don't get Monty Hall :-) especially when many descriptions of Monty Hall don't make it crystal clear that people shouldn't be treating it as Fall instead. That's why you really have to go to town to seal the doors on what the sequence of events is.

[theweaselking.livejournal.com]

Yeah. Monty Fall is the same as the two beagle puppies problem: You're generating a set of possible universes, then reducing your set, then measuring.

Monty *could* have revealed the car, and if you run the test a million times he'll reveal the car in ~333K of them - but the fact that he didn't reveal it in this instance means we're not IN one of those ~333K worlds. That's kind of awesome.