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Hey.

If I remember correctly, it was an example of the "Monty Hall Problem" which is a probability based problem.

A host presents 3 curtains to a player. A prize is behind 1 of the three curtains, randomly assigned. The player is asked to choose one curtain. It remains unopened. the host, who knows where the prize is located, then reveals one of the remaining curtains, revealing NO prize. The Host then asks the player if they would like to switch with the remaining unopened curtain before revealing if they are a winner.

Most people would think that they have a 50 / 50 chance at this point, and it didn't really statistically matter whether or not they switched or not, so, they "go with their gut". statistically it has been proven that the majority of people will stick with their original choice at this point.

However, a seasoned mathematician would know that they should SWITCH their choice.

So, why you say?

Since you are only picking the curtain containing the prize 1 out of 3 times on your initial choice, that means that there is 2 out of 3 times that the prize will be behind one of the other curtains. Since the host MUST show you one of those curtains that does NOT contain the prize, that means that the remaining curtain will contain the prize 2 out of 3 times.

That means that 66.666666% of the time, when you switch, you win! Much, much better than 50/50!

The problem is much easier to rationalize when you deal with larger numbers. Say there are 100 curtains. You pick curtain #1, and then the host reveals all the curtains except curtain #64. He then asks if you want to switch your choice (#1) with #64. The clear answer here is to switch! He knows where the prize is, and the likelihood of it being the one that you chose was only 1 in 100. Using the info he just gave you, you can win the car 99 out of 100 times by switching!

Neat stuff!

-Brad M.

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Q: In the movie called 21 there was a scene where Ben Campbell chooses right board behind which is price using probabilities Can someone explain how he did it?
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