Yes it is, go to http://www.willamette.edu/cla/math/articles/marilyn.htm for the full explaination
As the artice points out, If you had a million doors and you chose door 1 and the host opened all others except door 777,777 , would you swap?
I still like my explaination if you choose a goat and the other goat is eliniated
then swapping will get you the car, and your chances of choosing the goat...2/3
Neven MacEwan (B.E. E&E) Ph. 09 621 0001 Mob. 0274 749062
Paul M - ProSouth wrote:
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
soHere 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)
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
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