Double Monty - Action At A Distance?

[Simon 6]

This may mystify some.

Imagine two Monty Hall games taking place simultaneously on different sides of the world. Each game is completely independent and follows the same Monty Hall rules.
Imagine you're a contestant in one of them.

There are three doors - two of them empty, one has a prize.
The host knows what's behind each door.
You pick a door.
The host eliminates an empty door from the remaining two.
You are asked whether you'd like to stick or swap.
You're aware of the familiar solution - that there is a 2/3 chance that the door you originally chose is empty and that the other has the prize.

Before you make your final decision, a phone call is made. You are about to learn whether the contestant on the other side of the world originally chose the same as you or the opposite.

Depending on what you learn, does this knowledge in any way affect the odds of your game

 

[D H]

You need to explain what you mean by "You are about to learn whether the contestant on the other side of the world originally chose the same as you or the opposite." The question as stated is not very clear.

 

[Focus]

Each game is completely independent

You answered your own question here, this means that $$Pr(X_1=x|X_2=y)=Pr(X_1=x)$$. So it does not matter what door the other contestant chose, nor what door contained the price on the other game. The Monty Hall game is easier to analyze in terms of probability measures.

 

[Simon 6]

You need to explain what you mean by "You are about to learn whether the contestant on the other side of the world originally chose the same as you or the opposite." The question as stated is not very clear.

I mean you that are informed of one of the following:

a) "Whatever you originally chose, the other contestant picked the same*."

OR

b) "Whatever you originally chose, the other contestant picked the opposite**"


* Same = you both picked an empty door or a prize door but you don't know which.
** Opposite = neither of you picked the same

 

[D H]

With that clarification, the probabilities are now

case a) Probability that I chose the prize door originally is 1/5, so holding yields the prize door one time out of five which switching yields the prize door four times out of five.

case b) Probability that I chose the prize door originally is 1/2, so both strategies yield equal probabilities.

Since case (a) arises five times out of nine and (b) four times out of nine, the overall probability of winning with a strategy of "always hold" is 5/9*1/5 + 4/9*1/2 = 1/3 and the overall probability of winning with a strategy of "always switch" is 5/9*4/5 + 4/9*1/2 = 2/3, so this new game has the exact same overall probabilities as the original game.