My suggestion why the Monthy Hall reasoning is incorrect
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My suggestion why the Monthy Hall reasoning is incorrect

[From: ] [author: ] [Date: 13-09-28] [Hit: ]
When you intially choose your choice is actually 1/2. Why? Because you know that Monthy is always going to open a door with a goat after your choice, no matter what, and you know it! It doesnt matter if it is before you choose or after you choose,......
http://en.wikipedia.org/wiki/Monty_Hall_problem

They always say that initially when you choose the door, you have a 1/3 chance of getting the one with the car. That's the entire problem. When you intially choose your choice is actually 1/2. Why? Because you know that Monthy is always going to open a door with a goat after your choice, no matter what, and you know it! It doesn't matter if it is before you choose or after you choose, since one door will be gone no matter what. Why is the chance 1/3 then? That's an illusion. No matter which door you choose, your chances are 1/2 of getting the one with the car.
How can you counter that logic?

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Just try it with a friend. The friend sets up the doors randomly and plays the part of Monthy. Do it 100 times and see if you are closer to picking the car 50 times, or 33 times.

The friend can pick a random door for the car by rolling a die. Roll 1 or 2 ==> car behind door 1; roll 3 or 4 ==> car behind door 2; roll 5 or 6 ==> car behind door 3. Your friend rolls the die so that you can't see it, of course.

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Youre wrong. You are ignoring the case when you choose the car on the first guess... Monty will open up either of TWO doors. There are more possible combinations than you are accounting for.
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