In Reply to: Re: I guess I will have to quit posted by tunenut on October 26, 2006 at 07:19:05:
I do not understand what your next set is supposed to represent, especially the "loseR3 loseR3" that you say I did not account for. Please explain what these 3 lines are meant to represent.My table is correct.
I don't see "loseR3 loseR3" in any of the tables. In the reduced door revealed table the first line is loseR2 loseR3 and this shows the possible revelations when door #1 is the winning door. The possible revelations when door #2 is the winner is winR2 losR3 and the possible revelations when door #3 is the winner is loseR2 winR3.
We don't know where the winning door is but from the table we can see there are 6 possible revelations of which only two of them can possibly indicate a winning door. However since there is only one winning door only one of the winRX states actually represents the winning door.
What I will tell you is that if you wish to write a short computer program and actually run a simulation you will not find the 1/6 chance the winning door is revealed, you will find this chance is 1 out of 3- because there are 2 separate doors to which this 1 out of 6 chance can apply.
No! Clearly if all three doors, including door 1, could be randomly selected the chances of the winning door being selected is 1/3. But since door 1 can't be selected the odds you present are obvious incorrect.
A program should run to deliver the required results - however properly specifying the correct required result is the battle. Writting the program should be trival.
Give me rhythm or give me death!
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Follow Ups
- Re: I guess I will have to quit - Don T 08:25:01 10/26/06 (7)
- here is your logical fallacy - tunenut 10:07:27 10/26/06 (6)
- It's still wrong. - Don T 11:41:47 10/26/06 (5)
- Please answer my three questions (NT) - tunenut 11:44:56 10/26/06 (4)
- Re: Please answer my three questions (NT) - Don T 12:04:12 10/26/06 (3)
- Re: Please answer my three questions (NT) - tunenut 12:45:51 10/26/06 (2)