Author: Rolf Tueschen
Date: 15:42:21 09/25/02
Go up one level in this thread
On September 25, 2002 at 15:27:18, Joachim Rang wrote: >On September 25, 2002 at 15:05:03, Rolf Tueschen wrote: > >>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.) >> > >You can't look at the situation after the host opened a door isolated. In this >case history _does_ matter, because your first choice influences the decision of >the host. The host can't decide prior to your decision which door he will open, >because you may choose this door. So your first choice limits the options for >the host (because he had to open one of the _remaining_ doors _and_ the wrong >one). Not correct. There are two doors. Both closed. For the case here with n=1 you can't operate with probabilities from longer trials. Take a look into the table at the end of my Monty-page. There you can see that already for n=9 you get zero as a result for the confidence interval for no switches. If you lower n then you might get negative signs. ;) What does this mean? For case n=1 you have no solution. BTW this is well known. After some lessons in stats. Rolf Tueschen > >This is the difference to the roulette example where one trial is _independent_ >from others. Here the decision of the host, which door to open isn't independent >from your first choice and therefor probabilities changes. > >>Hope I could inspire you. > >so I do! > >> >>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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