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In Reply to: I'm sorry but you are incorrect posted by tunenut on October 24, 2006 at 19:15:15:
In the case where a losing door is revealled the odds are1.) 1/3 if you stick with the original door.
2.) 2/3 if you switch doors. This is because revealling a losing door is not new information.In the case where one of the two non-selected doors is randomly revealled the odds are
1.) 1/3 if you stick with the original door.
2.) 1/2 if you switch doors.
3.) 1/6 of the time the winning door will randomly be revealed.
3a.) Proof. The winning door will be the selected door 1/3 of the time. So there is only a 2/3 chance that the winning door is behind the two non-selected doors. In the random selection the odds are 1/2 that the winning door is revealed so the odds of randomly selecting the winning door is 1/2 * 1/3 = 1/6.Thank you! I know this is correct.
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Follow Ups
- No I'm not! - Don T 17:42:23 10/25/06 (16)
- you are close but not correct - tunenut 19:29:36 10/25/06 (15)
- That's not the right way to count! - Don T 21:17:10 10/25/06 (14)
- I guess I will have to quit - tunenut 21:27:20 10/25/06 (13)
- Re: I guess I will have to quit - Don T 00:20:28 10/26/06 (12)
- Re: I guess I will have to quit - tunenut 07:19:05 10/26/06 (11)
- Re: I guess I will have to quit - Don T 11:56:28 10/26/06 (2)
- Re: I guess I will have to quit - tunenut 18:20:39 10/26/06 (1)
- Re: I guess I will have to quit - Don T 23:12:00 10/26/06 (0)
- 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)