Game Show Theory: Should You Switch Choices?

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The game show theory, often illustrated by the Monty Hall problem, suggests that contestants should switch their choice after one non-winning option is revealed. Initially, the contestant has a 1/3 chance of picking the car and a 2/3 chance of picking a goat. When the host, who knows where the car is, opens a door to reveal a goat, switching increases the contestant's chances of winning to 2/3. This is because in two out of three scenarios, switching leads to a win, while staying with the original choice only wins in one scenario. The discussion highlights the importance of the host's knowledge in determining the optimal strategy.
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So I would imagine everyone has heard of the game show theory as it has been in various tv/movies including 21 and numb3rs.

But to refresh your mind.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I have heard some people say yes you do have an advantage because you start with a 1/3 chance and when the card is flipped you have double the chance of winning if you swap choices.

But I have heard other people argue the other way.

Just curious to hear what you all have to say about it.
 
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What's confusing about this is that most probability problems assume no "prior knowledge". However, the MC (how many of us remember Monty Hall?) does know which door the grand prize is behind. He intentionally opens only the door a does not show the prize. That gives the contestant the chance to use the MC's knowledge to increase his chances.

Since we can label the doors any way we wish, we can, without loss of generality, assume the prize is behind door A and that the other two doors are labeled B and C. Without the knowledge of which door the prize is behind, to the contestant, "a-priori", the prize is equally likely to be behind any door- choosing any door, you have a 1/3 chance of being right.

Now let's look at cases:
The contestant initially chooses door A. That is the door the prize is behind so the MC can open either remaining door. If the contestant does NOT change, he wins. If he does change, he loses.

The contestant initially chooses door B. The MC, knowing the prize is behind door A, opens door C. Now if the contestant changes to A, he wins, if he does not change he loses.

The contestant initially chooses door C. The MC, knowing the prize is behind door A, opens door B. Now, if the contestant changes to A, he wins, if he does not change, he loses.

Those three cases are equally likely. In two cases, by changing the contestant wins. In only one does he win by not changing. He increases his chance of winning by changing- in effect using the MC's "inside" knowledge to increase his chances.
 
What's confusing about this is that most probability problems assume no "prior knowledge". However, the MC (how many of us remember Monty Hall?) does know which door the grand prize is behind. He intentionally opens only the door a does not show the prize. That gives the contestant the chance to use the MC's knowledge to increase his chances.

Since we can label the doors any way we wish, we can, without loss of generality, assume the prize is behind door A and that the other two doors are labeled B and C. Without the knowledge of which door the prize is behind, to the contestant, "a-priori", the prize is equally likely to be behind any door- choosing any door, you have a 1/3 chance of being right.

Now let's look at cases:
The contestant initially chooses door A. That is the door the prize is behind so the MC can open either remaining door. If the contestant does NOT change, he wins. If he does change, he loses.

The contestant initially chooses door B. The MC, knowing the prize is behind door A, opens door C. Now if the contestant changes to A, he wins, if he does not change he loses.

The contestant initially chooses door C. The MC, knowing the prize is behind door A, opens door B. Now, if the contestant changes to A, he wins, if he does not change, he loses.

Those three cases are equally likely. In two cases, by changing the contestant wins. In only one does he win by not changing. He increases his chance of winning by changing- in effect using the MC's "inside" knowledge to increase his chances.

By the way, many years ago, long before Marilyn Savant made it famous, I saw this problem as an exercise in the first chapter of an introductory probability text!
 
So this works because the game show host knows where it is.
If the game show host did not know where it was and randomly picked a goat leaving the car and the second goat remaining, would it still be in your favor to change?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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