Author: Uri Blass
Date: 06:45:47 09/27/02
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On September 27, 2002 at 08:50:27, Gerrit Reubold wrote: >On September 27, 2002 at 08:22:06, Sune Fischer wrote: > >>On September 27, 2002 at 08:14:28, Gerrit Reubold wrote: >> >>>Hi Sune, >>> >>>On September 26, 2002 at 14:47:32, Sune Fischer wrote: >>> >>>I disagree, I am sure it doesn't matter whether the host opens a door (with a >>>goat) KNOWING where the car is or by SHEAR LUCK. The candidate should switch in >>>either case, thus improving his winning odds from 1/3 to 2/3. Of course, if the >>>host doesn't know and opens the door with the car (accidently), the game is over >>>:-) >> >>Ok, well at least we agree that if he KNOWS he should switch. >> >>So suppose the host doesn't know, and the host opens a door to a goat. >> >>What does this tell us? >>Well, since he didn't know, it means he either had a 50% probability of opening >>to a goat, or a 100% probability. >> >>He gets 100% probability if we are sitting on the car, and he gets 50% if the >>car is behind the one of the doors he had to choose from. >> >>The odds of picking a goat if you have 100% probability is higher than if you >>only have 50% (obviously), okay? > >I don't care what that probability was _before_ he opened the door. The door >with the goat is _now_ open. Suppose for the discussion that the candidate also guess door 1 and the host always open door 3 in case that door 1 has a goat. If I understand correctly you say that the candidate should switch and get probability of 2/3. Do you agree? Let assume case i means that the car is behind door i. In case 1 the car is behind door 1 so the candidate win only if he does not switch. In case 2 the car is behind door 2 so the candidate win only if he switchs. In case 3 the car is behind door 3 so the host expose the car so there is no game. Do you agree? The probability for every case is 1/3 before the game start. Do you agree? If we know that case 3 does not happen and this is what we know after the host opened door 3 and found no car then it means that there are only 2 cases and the probability for every case is (1/3)/(2/3)=1/2. Do you agree? It mean that switching is going to give you probability 1/2 Do you agree? I did not spend some minutes to try to understand sune's post but I believe that he also understood the point. Uri
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