I will defer to Serge's comment that the host always opens a door with the goat. That was the original version I encountered. The most recent version I encountered Monty randomly opened a door revealing a goat. In terms of the problem (and I may be wrong) it doesn't matter whether Monty knows or not so long as it's a goat. Serge can correct me on that if I am way off, I guess I need to assess whether we are dealing with family wise or individual case error. (For those in the know I am intentionally using vague phrasing to avoid giving away to much).
-Josh
3 comments:
Josh is correct. The problem works just fine as long as we ignore the case of the host picking the prize.
The question has an interesting/funny history. Marilyn Vos Savant once posed the question in her newspaper column, and it sparked a major controversy. Several mathematics departments weighed in on different sides (basically calling those of the opposite viewpoint idiots) and the controversy raged for months. The difficulty was the vague wording which was being interpreted in two different ways.
So the host picks a door that he knows has the goat. so now you are down to fifty fifty? My hunch is stay with your first choice. But then if he didn't pick your door because he had to pick the goat and your door had a goat behind it, too...that might change things. Before he picks a door, you had a 1 in 3 chance. HMMMM! I'm not sure which perspective to focus on. Your chances don't improve by switching. I say stay with your first choice. But I don't think I can articulate why...statistics, btw, bite!
Don't your chances improve by switching? The first time you picked you had one in three of picking the car. If you switch you have one in two.
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