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That's no fun.

(I helped save b3ta! gone, Wed 27 Aug 2008, 11:21, archived)
statistics and probability generally isn't.

(I helped save b3ta! Moon Girl Technologies horrendous beanbag, Wed 27 Aug 2008, 11:22, archived)
It should be.

(I helped save b3ta! gone, Wed 27 Aug 2008, 11:23, archived)
What percent of it isn't?

(I helped save b3ta! WormuIus, Wed 27 Aug 2008, 11:24, archived)
It can be!
www.versificator.co.uk/misc/monty-hall-problem.html
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:24, archived)
I've never got my head around this.

(I helped save b3ta! WormuIus, Wed 27 Aug 2008, 11:26, archived)
I find this one really easy.
The chance that you'll win if you switch is the chance that you already picked a goat: 2/3.
It's the forgetful host variant that stumps me.
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:28, archived)
...
*head asplodes*
(I helped save b3ta! WormuIus, Wed 27 Aug 2008, 11:30, archived)
they're just not counting the results when the host chooses the car,
so he may as well have just known anyway.
(I helped save b3ta! Moon Girl Technologies horrendous beanbag, Wed 27 Aug 2008, 11:31, archived)
No.
It goes back to 1/3 discounting those results.
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:35, archived)
I'm getting 1/2.
I think it's because when the host fucks up, it means you definitely didn't choose the car the first time, so it's throwing away half the times that you didn't choose the car, leaving you more likely to have picked the car the first time, which reduces the benefit of switching.
(I helped save b3ta! Moon Girl Technologies horrendous beanbag, Wed 27 Aug 2008, 11:39, archived)
Wikipedia says 1/3.
Probably statistical clustering. Try it with more doors - if it were the same as the host-knows version, you'd win 9/10 times for ten doors.

Edit: no, sorry... it always ends up a 1/2 chance if you exclude those cases, yes, with any number of doors (assuming the host gets down to 2 doors without opening one on the car.) The point is that you don't gain anything by switching.
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:42, archived)
it's 1/3 overall, including the host fuck ups,
but it's 1/2 for each of switching and not switching.
(I helped save b3ta! Moon Girl Technologies horrendous beanbag, Wed 27 Aug 2008, 11:43, archived)
Of course it is.
Sorry.
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:44, archived)
I can follow that explanation when I see it written down (I can even come up with it if I think hard enough)
but I can't Get My Head Round It and make it seem intuitive and obvious to me like the normal version now does.
(I helped save b3ta! Newington, Wed 27 Aug 2008, 11:47, archived)
the probability of choosing the car the first time is 1/3.
the probability of choosing a goat the first time is 2/3.
If you chose a goat the first time, the probability of the host chooing the car is 1/2. So the probability of you choosing a goat and the result being counted is only 1/3.

So now you're in the situation where, if you chose the car the first time you will definitely win if you stick. But if you chose a goat the first time, you will definitely win if you swap. But the probability of each of those situations coming up is the same. i.e. 1/3, or 1/2 of the valid results.
(I helped save b3ta! Moon Girl Technologies horrendous beanbag, Wed 27 Aug 2008, 11:54, archived)