Probability? 1/3 or 1/2?

Knowing that one of the boxes, B and C, must be empty, does NOT increase the probability that the coin is in the box A. Because, we know that one of box B and C(or both) must be empty BEFORE you choose box A. So after you choose box A, the probability is 1/3.

Suppose another man, who know which box is empty, open one empty box, just say B. It doesn't increase the probability that the coin is in box A. But it does increase the probability that THE COIN IS IN BOX C. So the probability that the coin is in box A is still 1/3, and the probability that the coin is in box C is 2/3.

If you are still not convinced, suppose there are 100 boxes, labeled 1,2,3, to 100. One of them contains the gold coin. You pick box 1. Another man open 98 empty box. Do you think the probability that box 1 contain the gold is increase to 1/2 ? No. The probability is still 1/100. We know that there are 99 empty boxes before you pick box 1, so it doesn't change the probability.

I hope this will help.

Ok, the probability that you just choose the right box at random is 1/3. What you've done by talking about the other boxes possibly being empty is actualy taking away one of the 3 choices. Because you then know that 1 box is empty, the probability to choose the right box out of the other 2 is 1/2. So they're really two different situations. In the first one, choosing completely at random, the probability is 1/3 as there are 3 possibilities, but in the second situation you know that one box is empty, so you only have to choose from 2, making that probability 1/2. Hope that helps!

You got it all wrong!

To start off, there are three boxes! The probability of picking one box with gold coin is 1/3 (33.33%)and the probability of picking one box that is empty is 2/3(66.67%).

This is the probability for picking one box out of the 3 boxes.

Once you picked one box and if you get to pick another (assuming you picked an empty) The probability is based on two boxes. Then it would be 1/2 or 50%.

The problem gives you a known fact: either box B or box C is empty. Therefore, there are only two boxes to choose from in each situation:

If box B is empty, the coin could only be in A or C.

If box C is empty, the coin could only be in A or B.

Since you are to make one choice out of two, the probability is 1 to 2 or simply put, you have half a chance the coin would be in one box as it might be in the other.

Well, it's like taking a multiple choice test and increasing your odds by eliminating choices. If you know for sure that C is wrong, the only answers left are A and B. Therefore, guessing is only a 0.50 probability. However, if you don't know for sure that the coin is not in either box B or C, that is, you can't eliminate one, then the probability is still 1/3 that the coin is in box A... try not to think too hard about it.

Without choosing any other box first, the probability of choosing the correct box of gold is 1/3.

If you already choose another box that doesn't contain the box, then yes the probability increases to 1/2. But you're not starting from the beginning. If you were to do it this way then you are trying to pick a box that doesn't contain the gold 1/3, and then you are trying to pick a box that does contain the gold 1/2... So if you do it this way, then the probability of picking the wrong box and then picking the right box is 1/6

Garrr!!! Give me yer Gold!!!

The probability that B is empty is 2/3

If B is empty the prob that A has the coin is 1/2

So the probability that B is empty and A has a coin is 2/3 * 1/2 = 1/3

similar reasoning for C being empty.

That is not the monty hall problem

with MH. ...

he Picks A

someone who knows right box lifts an empty one , say B

He can now switch from A to C, should he?

Yes

What if instead of 3 boxes you have 100 (and still only 1 coin)

is prob 1/100 or 1/2 ?

Because you eliminated one of the boxes.

This chain of logic supposes that you are allowed to choose one of the boxes BEFORE you make your choice, tho. And it's faulty, also.

Suppose you DO pick one of the empty boxes.

Only AFTER the empty box is selected does the probability of A containing the coin go to 1/2 - NOT BEFORE.

This goes back to an old game show ad answer by Marilyn Vos Savant. In statistics we went over this. basically because you now now that one of the other boxes has nothing in it you are left with two. So now there are only 2 boxes you don't know about, and therefor 1/2. It can get more complicated but that is the simplified answer.