This is not correct - at least based on your explanation. > We have 1 > car and 49 goats. I choose in the first place door A. The game master > opens 48 other doors: > > .00000000000000000000000000000000.0000000000000000 > > Where is the car? The proability for door A is still 0.02, but for door No > 34 it is 0.98. Anybody who still thinks it is 50:50?
Please explain why, in this scenario, the probabilities for doors 1 and 34 are different? Regards, Paul Matthews ProSouth Technology Solutions Phone: +64 3 470 1321 DDI: +64 3 470 1324 0800 PROSOUTH (0800 776 768) 249 Cumberland Street, Dunedin, New Zealand Visit us at www.prosouth.co.nz ----- Original Message ----- From: "Dieter K�hler" <[EMAIL PROTECTED]> To: "NZ Borland Developers Group - Delphi List" <[EMAIL PROTECTED]> Sent: Thursday, August 05, 2004 4:55 PM Subject: Re: [DUG] Welcome back problem > Here is a clearification of my previous argumentation. > > What is important here the distinction between > a simple probability P(X), i.e. the probability for X, > a conditional probabilitiy P(X|Y), i.e. the probability for X, given Y. > > We have one car (signified by 1) and two goats (signified by 0). Each > permutation has the same probability. The different permutations are: > > 100 > 010 > 001 > > Let us suppose I choose in the first place door A (the first column) and > get the information that behind door B is a goat: > > .0. > > which decides between 010 and 001 only, but contains no additional > information about 100. As a result I have now more information about door > C than about door A. > > The trick is that the probabilities are not independent (there is no 0.33 > chance individually for each door): > > 100 --> P(A) = 0.33 > 010 --> P(B) = 0.33 > 001 --> P(C) = 0.33 > > is converted to > > 100 --> P(A) = 0.33 > 010 --> P(B) = 0.0 > 001 --> P(C|non P(B)) = 0.66 > > The condition does not effect P(A), because there was nothing to choose. > > Or in other words: What remains constant is P(A) as well as the sum P(B) + > P(C), but we know now that P(B) is 0 (the only new information provided) so > it follows that P(C) is 0.66. > > > For those who are still not convinced try this modified example. We have 1 > car and 49 goats. I choose in the first place door A. The game master > opens 48 other doors: > > .00000000000000000000000000000000.0000000000000000 > > Where is the car? The proability for door A is still 0.02, but for door No > 34 it is 0.98. Anybody who still thinks it is 50:50? > > Dieter > > > _______________________________________________ > Delphi mailing list > [EMAIL PROTECTED] > http://ns3.123.co.nz/mailman/listinfo/delphi > > _______________________________________________ Delphi mailing list [EMAIL PROTECTED] http://ns3.123.co.nz/mailman/listinfo/delphi
