Fun Probability problem

[Mathew]

Imagine you have three doors....



Behind one of these three doors is a car and the other two are goats. Your goal is to win the car. After you have chosen your door, a goat on one of the other doors will be shown. Would it be in your interest to switch doors at this stage?

What is your chances of winning ?

[armourall]

If you stick with your original choice your chances remain 1 in 3.

If you reselect... 1 in 2.

Do I get the car?

[Mathew]

If you stick with your original choice your chances remain 1 in 3.

If you reselect... 1 in 2.

Do I get the car?

Nope...lol

[Mathew]

The fact is that your chances of chosing the goat is 2 in 3 and the car 1 in 3.

If you choose either goat, the other goat will be eliminated and if you then choose to switch - you will get the car. There is 2 in 3 chance of this occuring

If you initially pick the car, switching will come out incorrect. There is a 1 in 3 chance of this occuring.

Therefore switching doubles your chances from 1 in 3 into 2 in 3.

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Still confused? This is far more apparent of a case of 100 doors. Your chance of getting the first one right is 1 in 100. If once you had selected your door, 98 were then opened and eliminated, it would be far more likely that the other door left would be the correct door. The correct strategy would then be to switch....

[Toolish]

The fact is that your chances of chosing the goat is 2 in 3 and the car 1 in 3.

If you choose either goat, the other goat will be eliminated and if you then choose to switch - you will get the car. There is 2 in 3 chance of this occuring

If you initially pick the car, switching will come out incorrect. There is a 1 in 3 chance of this occuring.

Therefore switching doubles your chances from 1 in 3 into 2 in 3.

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Still confused? This is far more apparent of a case of 100 doors. Your chance of getting the first one right is 1 in 100. If once you had selected your door, 98 were then opened and eliminated, it would be far more likely that the other door left would be the correct door. The correct strategy would then be to switch....

Wrong as far as i see it.

They are mutually exclusive events...the outcome of the first goat being revealed does not change the second result...you have a 50% chance of being correct either way once the other doors are revealed.

With your 100 door example, once the other 98 are revealed you are still left with a 50% chance of either door being right...the fact that incorrect doors have been removed from the equation means nothing...in choosing between the final 2 doors switch or stay means nothing, the door you initially chose becomes irrelevant.

[Matt]

Wrong as far as i see it.

They are mutually exclusive events...the outcome of the first goat being revealed does not change the second result...you have a 50% chance of being correct either way once the other doors are revealed.

With your 100 door example, once the other 98 are revealed you are still left with a 50% chance of either door being right...the fact that incorrect doors have been removed from the equation means nothing...in choosing between the final 2 doors switch or stay means nothing, the door you initially chose becomes irrelevant.

The chances that you picked the correct door out of 100 is 1/100. They then eliminate 98 incorrect ones, leaving 2 doors. At this point, there is a great chance - 99/100 - that the car is behind the 'other' door. Your original guess still has a 1/100, or 1%, chance of being correct.

Same obviously works for 3 doors. Your 'guess' has a 1/3 chance of being correct, meaning there's a 2/3 chance it's incorrect. When they eliminate an incorrect door, your chances with the original door remain at 1/3. However, the 2/3 that before were assigned to the other two doors now are assigned to the only remaining door. So you have a 2/3 chance if you switch, and the same, original 1/3 chance if you stay.