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July 21st, 2016, 06:46 PM   #1
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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?
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July 22nd, 2016, 01:28 PM   #2
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Probabilities can be added only for disjoint events. These events are not, both include door C being opened.
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July 22nd, 2016, 07:45 PM   #3
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Math Focus: tetration
Quote:
Originally Posted by CLC2016 View Post
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?
This is bull the change for one is 2/3 for the other 1/3.
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July 23rd, 2016, 01:41 PM   #4
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I posted the paradox on Probability forum.
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July 29th, 2016, 01:05 PM   #5
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Quote:
Originally Posted by CLC2016 View Post
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.
Poooorrrrrrrr C; all he could pick was C: nothing else left...

Or were they in different rooms?

In other words: silly
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July 29th, 2016, 05:51 PM   #6
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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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