Ok, so stick with this thought, you almost have it. Recalculate your chance and the answer is ?

I can tell you the answer is not 100% / 2

They probably have some over the counter stuff that can help you out with that issue of yours.

All the best.

FYP

You can magnetise a pin with a hammer , think about that for a while.

I am not 100% sure, I think the answer might become 0.5/2

P=<1/2

=????/2

What if we take Box 3 and 4 from the moon and bring them to China and then we cause them to sink to the bottom of the Pacific Ocean off the coast of China? Now what are the answers:


Question 1: What are the chances there is a prize in Box 1?

Question 2: What are the chances there is a prize in Box 2?

Question 3: What are the chances there is a prize in Box 3?

Question 4: What are the chances there is a prize in Box 4?








Now what if we give Box 1 and 2 to a child and tell they child he/she can open one and if it has the prize he/she can keep it. What are the answers to the following questions:


Question 5: What are the chances there is a prize in Box 1?

Question 6: What are the chances there is a prize in Box 2?

Question 7: What are the chances there is a prize in Box 3?

Question 8: What are the chances there is a prize in Box 4?

Question 9: What are the chances the child ends up getting a prize?

FYI, 2/4 and 1/2 are the same thing.

2/4 = 0.5
1/2 = 0.5

I agree with this. 4 boxes that have a prize in 2 of them. We know for a certainty that there are exactly 2 prizes in the 4 boxes. Get rid of 2 of the boxes randomly so now we have 2 boxes. There could be 0, 1 or 2 prizes in the 2 boxes; there is no way for us to know which it is with certainty.

However, the probability of any of the four boxes having a prize in the first scenario is 1/2. And the probability of any of the two boxes having a prize in the second scenario is 1/2. Getting rid of 2 boxes doesn't change the probability that each of the other 2 boxes contains a prize.



PKDK:

I could propose a bet where we have 4 boxes with a prize in 2 of them and then we get rid of 2 of the boxes randomly and bet about how often the remaining 2 boxes will have a prize. I will predict a relatively small range of a number of occurrences within which the number of times the 2 boxes contain a prize. According to you this is impossible since there is no way to know how often the 2 boxes contain a prize, but I'll bet you that I can do it anyway.

It would be similar to the bet I proposed previously that you would not accept.

Are you interested? If so, I will write up the proposal.

Willing to back up your writings with your money at any point?

By strict definition they are not the same thing although you can get the same values.

You are so close to ''seeing'' the ostensible. Consider what you first said makes the second part logically false. By stating 1/2 you are stating 100% there is a prize in one of the two boxes which you already agree in your first statement this is not true.

added - I read that wrong originally so have edited, if you split the boxes you change the chance because your original information becomes invalid.

Ostensible is not a noun. It is an adjective. You keep using it as a noun. It needs to be modifying another word. It is meaningless the way you keep using it.

Wrong. You do not add the percentages for each box individually to determine the chance that there is a prize in any of the prizes.

Each box has a 1/2 (or 50%) chance of having a prize. This does not mean that there is a 1/1 (or 100%) chance that there is a prize in one of the 2 boxes. That is wrong. The chance that exactly one box has a prize and the other doesn't is 50% (or 1/2). The chance that one or both boxes have a prize is 75% (or 3/4).

Each box has a 50% chance of containing a prize and a 50% chance of not containing a prize, so:

Chance both boxes have a prize: 0.5 * 0.5 = 0.25 = 25% = 1/4
Chance neither box has a prize: 0.5 * 0.5 = 0.25 = 25% = 1/4
Chance one box has a prize and the other doesn't: 1 - 0.25 - 0.25 = 0.5 = 50% = 1/2


Chance one or both boxes have a prize: 1 - (0.5 * 0.5) = 0.75 = 75% = 3/4

Why does where the boxes are physically located affect what is inside the boxes?

Why does moving a box change its contents?


If this is true, Federal Express and other delivery companies need to be made aware of this problem. When I ship a package I want it to arrive at its destination with the same contents that I originally put into it.

With all due respect stop looking at the numbers as if they are some sort of magic. You can't logically agree with a statement then say the logic fails because the numbers work, If the logic fails , the numbers do not work and any calculation would be wrong. Your calculation is wrong , not the logic is what I am saying to you.
Consider you agreed that you do not know with a certainty any of the two boxes have a prize. So how can you state a first value when the value could be 0, 1, or 2?

Consider two Schrodinger's cat boxes,

What is the chance of picking a box with an alive cat in?

What is the chance there is a cat in either box considering the scenario we already have of four boxes?

1. The statement that there could be 0,1 or 2 prizes in the two boxes and we don’t know which is correct.

2. The statement that each box has a 50% chance to contain a prize is also correct.

3. The following statements are also correct:

There is a 25% chance that neither box has a prize.
There is a 25% chance that both boxes have a prize
There is a 50% chance that one box has a prize and the other does not.

There is a 75% chance that one box or both boxes contain a prize.
There is a 25% chance that neither box contain a prize.



All of the statements above are correct at the same time, including the statement in number 1.



I guess I would suggest that you review some elementary school textbooks on fractions (1/2 does in fact equal 2/4) and then when ready try find a county college, or something similar, that has a probability class that you can take.

Regrettably, I don't think you are even remotely close to getting it. I'll try once more in language that even a ten years old would understand.

Looking at remaining box 1, even you must agree that as it is the same box as it was when there were four in a row it still has the same chance that it always had of containing a prize. The fact that we don't know which box it is of the original four or whether it contains a prize or not is totally irrelevant when we are making our best guess as to whether or not it contains a prize. (Best guess is just yet another way of describing the probability or chance of an event.)

There were two prizes in the 4 boxes. As we don't know which particular boxes have prizes, we have to guess the contents each time we consider a different box. As each box has exactly the same chance as any other to include a prize, then our best guess is obviously the total number of prizes divided by the total number of boxes, or 2/4, or 1/2, or a 50% chance of a prize in the box we are considering. (You have also agreed this, previously.)

Therefore, remaining box 1, whatever one it is of the four original boxes, still has the same 50% chance that it always had of containing a prize, whether or not in actuality the two prizes are now providing us with more sunlight.

QED

I think that is the premise (at times) as well. But pkdk keeps moving the bar...and thus your (our) statements are sometimes wrong. Why can't there be 3 prizes in 4 boxes? The premise is there are two prizes in 4 boxes....and indeed there are 2....when there are a total of 3 (or 4) prizes.

I assume there can not be 0 or 1 prize.

I assume there can be 2,3 or 4 prizes.

pkdk, are there exactly 2 prizes in 4 boxes?

In the scenario we set up , we have four boxes with a prize in two of the boxes 2/4

We randomly shuffle the boxes and throw two boxes away leaving just two boxes.


The already agreed, is you all know that the two remaining boxes may have no prize in at all, or they could both have prizes in, or only one prize.


Some people, well most still think the answer of the chance of a prize is 1/2.

You did not answer my question

[ ] Yes, there are exactly two prizes. One is in one of the four boxes, the other is in a second of the four

[ ] No, there are two prizes. There may be three prizes. There may be four prizes.

Remember that time awhile ago when 2+2 was shutdown for an extended length of time due to outside forces.

I yearn for those halcyon days!

I did answer, neither of the above, there are exactly two prizes in two of four boxes in total.

Try it this way folks,

We have 4 boxes

[win][loss][win][loss]


We are going to shuffle the boxes and throw two away leaving:


[box1][box2]


Are the remaining boxes ''dead'' or ''alive''?


Now we already know and people have agreed, both boxes could be empty. So we cannot state 1/2 because we already know that could be untrue.
Quite oddly we could state a 50% chance, but we cannot state 1/2 although you would think it was the same.

Yea, the problem is that you do not know how fractions work.

1/2 = 0.5 = 50% = 2/4 = 50/100 = 244/488



So:

No, I get how fractions work. You are still not understanding what appears to be true is not necessarily true.


It appears to be 1/2 , but is not 1/2, the answer is 50% or ?/2 .