Author: Rolf Tueschen
Date: 12:05:03 09/25/02
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On September 25, 2002 at 14:41:59, David Hanley wrote: >It's very very simple and well-understood. You switch doors. > >Let's suppose you choose door 1 > >If the prize is behind that one, and you switch, you lose >If the prize is behind #2 the host opens door #3, you switch, you win >if the prize is begine #3 the host opens door #2, you switch, you win > >You can make arguements about a manipulative host, but that is just pointless >thrashing. The problem illustrates a very important point about probability. > >dave Thanks for your opinion. But the point here is exactly if and why the conditioned probability should be applied. You are a few steps too far already. Of course if the questions have been answered in your favor then it's all simple. Very simple. I don't have objections to that question. But I see now how difficult the question is I asked. Therefore I'd like to remind you that I said, that the chances increased indeed _both_ for your first choice and also the reamaining door. But why should your first choice now be "influenced" or better "infected" by the "virus" or the past situation of the one third? My point is that the actual situation with 2 remaining doors is relevant. And since you can be sure [in our thought experiment] that the host is honest and does not influence you intentiously or unintentiously you have the completely new situation now. No longer the 1/3 but 1/2 now. Or: Could you prove why conditioned probability should be applied in the case of a unique situation??? (Also think of the roulette example with 10 times red and still no increase for black for the next trial.) Hope I could inspire you. Rolf Tueschen P.S. For those who think that this is all way off CC topics, I might remind you of the importance of the questions of methods in statistics, in special in the early stages when you must decide what you want to measure... Monty's Dilemma is a good excercise for questions about SSDF etc.
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