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July 21st, 2016, 06:46 PM  #1 
Newbie Joined: Jul 2016 From: In a bus Posts: 1 Thanks: 1  The Monty Hall problem with multiple idiots
So, you're all familiar with the Monty Hall problem. Let's say I have 3 test subjects, none of whom have any awareness of the other subjects. They each select a different door. Say A selects A, B selects B and C selects C. Monty opens door C, revealing a goat. Person C starts crying because they didn't think that was how the game worked. Ignore them for the rest of the problem. Now, A and B both know about the Monty Hall Problem, and so they both switch doors. Hence they each have a 2/3 chance of their new door being the one with the car behind. But this would mean the total probability was more than 1. So how can they both have a 2/3 chance? 
July 22nd, 2016, 01:28 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723 
Probabilities can be added only for disjoint events. These events are not, both include door C being opened.

July 22nd, 2016, 07:45 PM  #3  
Senior Member Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration  Quote:
 
July 23rd, 2016, 01:41 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723 
I posted the paradox on Probability forum.

July 29th, 2016, 01:05 PM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
Or were they in different rooms? In other words: silly  
July 29th, 2016, 05:51 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723 
The problem lies with Monte Hall (see my post in probability forum). With one contestant, he can pick a door to favor the contestant to have a 2/3 chance of winning. In this case he is simply picking a door at random, leaving the other contestants each with 1/2 probability of winning.


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hall, idiots, monty, multiple, problem 
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