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Three Curtains

You are a contestant on a game show. There are three curtains. Behind one of the curtains is a new car. You are asked to choose one of the curtains. Lets say that you choose curtain #1. The host of the show - who knows where the car is so as not to end the game prematurely - opens curtain #3 and of course there is no car behind it. The host now gives you a choice. You can stay with curtain #1 or you can change your choice to curtain #2. The question now is: would it be to your advantage to stay with curtain #1, or would it be to your advantage to change to curtain #2 or would there be no advantage either way?

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Originally Posted by rstrats
You are a contestant on a game show. There are three curtains. Behind one of the curtains is a new car. You are asked to choose one of the curtains. Lets say that you choose curtain #1. The host of the show - who knows where the car is so as not to end the game prematurely - opens curtain #3 and of course there is no car behind it. The host now gives you a choice. You can stay with curtain #1 or you can change your choice to curtain #2. The question now is: would it be to your advantage to stay with curtain #1, or would it be to your advantage to change to curtain #2 or would there be no advantage either way?

I cannot remember the exact calculation for the probabilities to prove this and that changing your mind is to your advantage. I will leave this for someone else to explain and supply the mathematics behind it.

David

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Anyone agree with David?

4. Originally Posted by rstrats
Anyone agree with David?
I believe David is correct. Imagine doing a computer simulation. If you played the game 1000 times, theoretical probability would predict about 333 winners for your first choice. Staying with your first choice does not change that outcome. You would still have about 333 winners. Therefore, that means the "other" choice will be a winner app. 666 times. I.e., you get the 333 winners from both of the other two curtains.

Does that make sense?

Steve

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Originally Posted by Ps 27:1
I believe David is correct. Imagine doing a computer simulation. If you played the game 1000 times, theoretical probability would predict about 333 winners for your first choice. Staying with your first choice does not change that outcome. You would still have about 333 winners. Therefore, that means the "other" choice will be a winner app. 666 times. I.e., you get the 333 winners from both of the other two curtains.

Does that make sense?

Steve
Hello Steve

My answer was correct, I could not remember the mathematics behind it.

I have found a website in which the explanation is given. The conclusion is this;

http://www.mathesonbayley.com/articl...?origin=lyrics

Stick with your original choice, and you have a ⅓ chance of being correct
Change your mind, and you have a ⅔ chance of being correct
All the best
David

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