Sweet Tooth and the Monte Hall problem

12 08 2014

I just read Ian McEwen’s ‘Sweet Tooth‘ (v gd), and it had a section about the Monty Hall problem which made me realise I’d gone back to not understanding it, goddammit!

So I had to go back to the good vid narrated with the ridiculous accent (is this what I sound like to Canadians?), and search for the type of astute comment (somewhat buried by angry people being rude to each other) that made me get it last year. Here is one (thanks, Joe Francis):

Look at it this way – if there were 100 doors instead of three, and you had to pick one at random, and then the host opens 98 of the other 99 doors to reveal goats behind each one, would you then swap to the last remaining closed door?  If you don’t, that means you’re assuming that you’ve picked right door first time, a one in a hundred shot.  By the logic of those who hold that the odds remain 50/50 after Monty Hall has opened one of the THREE doors, then the odds remain 50/50 after he’s opened NINETY-EIGHT doors, and it still makes no difference whether you swap or not, which is just ridiculous.  In the case of 100 doors, of course you’d swap, so why wouldn’t you swap when it’s just three doors?


Actions

Information