Probability riddle

[Beex]

But he was trying to not possess it. Justin a profitable way.

[unoduetre]

[mention=46027]Beex[/mention]

Maybe he didn't possess them but was possessed by them?

[Citizen]

So the answer was ghosts all along.

[unoduetre]

[mention=1400]Citizen[/mention]

Yup, the answer was d), but it was invisible, because it was a ghost answer (written by a ghostwriter).

[Moulders]

C and I'll tell you why -


This one always irritated me and it's seen as a paradox, but it's just one that arises through the vaguaries introduced through language.


The problem is the phrasing of that line "Now you are given an option to change your initial choice."

There is an apparent paradox introduced due to the difference in the size of the set. When you first made a choice the chances were 1 in 3. If you make another choice the value is 1 in 2 since one option had been taken away. The question is structured so that the choices are:

1) Stick with your original choice (1 in 3).
2) Change your answer. (1 in 2).

Which would make it appear that changing your answer yields a greater chance of being correct. This is not the case it's just the way the option was worded, there is no actual difference in probability.

You can verify this by the following:

Imagine a situation you picked A.
Option B is taken away, so the envelope with the money in either A or C.
If instead of giving the binary choice, of sticking with your initial guess and changing your guess the line was changed to 'Pick again' then no matter what you pick (A or C) the odds would be 1 in 2 for both.
Simply by asking the contestant to choose again you change the set size and even if you picked A again the choice would be 1 in 2.

Changing the odds simply by making a change to the question, with no practical change to your answer.

[lolin]

i'll admit to barely skimming the initial post but i'm pretty sure this is the monty hall problem which is solved

[unoduetre]

i'll admit to barely skimming the initial post but i'm pretty sure this is the monty hall problem which is solved

The wording is a bit different, so the outcome can be different. But you're right that it's based on the Monty Hall problem.

[unoduetre]

OK. So I will now reveal the answer in the next post.

[unoduetre]

And the answer is:

The question does not contain enough information to answer it properly.

There are circumstances in which any of the provided answers can be correct. So all of them are kind of correct.

So if you imagined a specific situation and chose the answer based on this situation, you were probably right for this situation. But there are more possible situations you have not considered.

I will now explain the circumstances by inserting the missing information to the question.

Some further explanation:

  Spoiler:  

It is important to notice the following. The whole story happened just once. So talking about probability doesn't make sense if by probability we mean a pattern which appears after repeating something.

To give you an example. If you want to check, if a coin is crooked, you can toss it many times and see if the number of heads and tails is more or less the same. But you can't really talk about probability (in this sense) if you can only toss the coin once.

But there is another way to think about the probability. It's the following one: "If exactly the same thing WERE repeated, what WOULD BE the outcome." If the coin is crooked, the outcome would be different from the outcome for a normal coin. But how do you know without repeating it? So you cannot have both. Either:
1) you KNOW that the coin is crooked, THEREFORE you know that the outcome WOULD BE as for a crooked coin, or
2) you DON'T KNOW whether the coin is crooked or not, THEREFORE to verify that, you MUST PERFORM the same toss MANY TIMES to discover it.

In this case we neither know whether the coin (the initial placement and the rule for opening of envelope C) was crooked or not nor we can repeat the same thing to make sure.

There are also other ways of thinking about probability and they might give different answers.

I'll now explain some circumstances (these are just examples, there can be more) for each option.

---

ANSWER a)

There are 3 closed envelopes with letters on them: A, B and C.

One of them contains a cheque for £1000. The other two contain nothing.

You are asked to choose one envelope.

After making the final decision, you will receive the contents of the envelope (if any).

You've chosen envelope A.

After making the choice, envelope C is opened and it is revealed that it contains nothing.

Now you are given an option to change your initial choice. This will be your final decision.

THE MISSING PART: The initial placement was at random (1 of 3). There was the following rule: "You will only be given an option to change the choice PROVIDED THAT your initial choice contains £1000."

You can either keep envelope A or switch and keep envelope B.

What should you do and why, assuming you want to maximize the probability that you win the cheque?

a) keep envelope A, because it maximizes the probability that you win the cheque
b) switch and keep envelope B, because it maximizes the probability that you win the cheque
c) it doesn't matter whether you switch or not, the probabilities that you win the cheque are the same for both choices, so you can do either

---

ANSWER b)

There are 3 closed envelopes with letters on them: A, B and C.

One of them contains a cheque for £1000. The other two contain nothing.

You are asked to choose one envelope.

After making the final decision, you will receive the contents of the envelope (if any).

You've chosen envelope A.

After making the choice, envelope C is opened and it is revealed that it contains nothing.

Now you are given an option to change your initial choice. This will be your final decision.

THE MISSING PART: The initial placement was at random (1 of 3). There was the following rule: "You will only be given an option to change the choice PROVIDED THAT your initial choice doesn't contain £1000."

You can either keep envelope A or switch and keep envelope B.

What should you do and why, assuming you want to maximize the probability that you win the cheque?

a) keep envelope A, because it maximizes the probability that you win the cheque
b) switch and keep envelope B, because it maximizes the probability that you win the cheque
c) it doesn't matter whether you switch or not, the probabilities that you win the cheque are the same for both choices, so you can do either

---

ANSWER c)

There are 3 closed envelopes with letters on them: A, B and C.

One of them contains a cheque for £1000. The other two contain nothing.

You are asked to choose one envelope.

After making the final decision, you will receive the contents of the envelope (if any).

You've chosen envelope A.

After making the choice, envelope C is opened and it is revealed that it contains nothing.

Now you are given an option to change your initial choice. This will be your final decision.

THE MISSING PART: The initial placement was at random (1 of 3). There was the following rule: "After putting money into one of the envelopes, one of the two remaining envelopes was chosen randomly and remembered (the no-choice-envelope). You will only be given an option to change your choice PROVIDED THAT your initial choice was different from the no-choice-envelope." Envelope A wasn't the no-choice-envelope and that's why you were given the option to change.

You can either keep envelope A or switch and keep envelope B.

What should you do and why, assuming you want to maximize the probability that you win the cheque?

a) keep envelope A, because it maximizes the probability that you win the cheque
b) switch and keep envelope B, because it maximizes the probability that you win the cheque
c) it doesn't matter whether you switch or not, the probabilities that you win the cheque are the same for both choices, so you can do either

On the other hand, if you've chosen the no-choice envelope in the first place, the story would've been like this:

There are 3 closed envelopes with letters on them: A, B and C.

One of them contains a cheque for £1000. The other two contain nothing.

You are asked to choose one envelope.

After making the final decision, you will receive the contents of the envelope (if any).

You've chosen envelope A.

You are asked, if this is your final choice and you confirm it.

The envelope A is opened and it contains nothing.

---

Now you can see that the situation described in the question can match all of these 3. There is not enough information to decide which one it is, because it was just a single event. If you can repeat the same thing again and gain, you can see which of the situations (or maybe a situation different from these 3) it is.

Example of other elements of the rules:

"£1000 is always placed inside envelope A…"
"£1000 is always placed inside envelope B…"
"…You will always be given the option to change. The revealed envelope will always be an empty one."
"…You will always be given the option to change. The revealed envelope will be chosen at random from the remaining two and it may or may not contain £1000."

[unoduetre]

This is also similar to multiple outs and magician's choice.

https://en.wikipedia.org/wiki/Equivocation_(magic)

Let me explain it to you, so you can learn some illusion.

You hide three envelopes in different places. One (containing letter A) behind the door, one (containing letter B) under the bed, one (containing letter C) inside the book on the shelf.

Now you ask someone to write down one of three letters A, B or C on a piece of paper. Depending what has been written, you ask the person to check either behind the door, or under the bed, or inside the book.

The audience is under illusion that there was only one envelope containing the answer in the first place.

An example of multiple outs from: https://secrets-explained.com/basic-tec ... tiple-outs

You put three pieces of paper on a table, each with a number from 1 to 3 written on it. You hand your spectator a folded piece of paper and ask him to put it on any of the three pieces of paper already on the table. He puts it on 3. You then ask him to unfold the paper. The writing on it says: "You will select 3."

However, if he selects 2, you ask him to turn over the paper he has selected. On the back side of it, you have written: "You will select this paper." The other two papers have nothing written on the back side.

If the spectator selects 1, you ignore the papers. Instead, you ask him to reach under his seat. There he finds your prediction: "You will select 1".

The trick here is that there is no way you could be wrong. Whatever the selection is, you have a way of revealing your prediction.

An example of magician's choice from: https://secrets-explained.com/basic-tec ... n-s-choice

Let's look at a simple example. You have three objects: A, B and C. You want your spectator to pick the object A. So you tell him to select two objects and then you act according to his choice. If he picks B and C, you say you will eliminate these two and the remaining object (A) is his decision. However, if he selects A and B, you eliminate C and ask him to select one of the remaining two objects. If he selects A, that is his selected object. If he selects B, you eliminate it and the remaining (A) is his chosen object. I doesn't matter what he selects, you will always end up with A.

[unoduetre]

C and I'll tell you why -


This one always irritated me and it's seen as a paradox, but it's just one that arises through the vaguaries introduced through language.


The problem is the phrasing of that line "Now you are given an option to change your initial choice."

There is an apparent paradox introduced due to the difference in the size of the set. When you first made a choice the chances were 1 in 3. If you make another choice the value is 1 in 2 since one option had been taken away. The question is structured so that the choices are:

1) Stick with your original choice (1 in 3).
2) Change your answer. (1 in 2).

Which would make it appear that changing your answer yields a greater chance of being correct. This is not the case it's just the way the option was worded, there is no actual difference in probability.

You can verify this by the following:

Imagine a situation you picked A.
Option B is taken away, so the envelope with the money in either A or C.
If instead of giving the binary choice, of sticking with your initial guess and changing your guess the line was changed to 'Pick again' then no matter what you pick (A or C) the odds would be 1 in 2 for both.
Simply by asking the contestant to choose again you change the set size and even if you picked A again the choice would be 1 in 2.

Changing the odds simply by making a change to the question, with no practical change to your answer.

The "pick again" you mentioned is identical to "stick or switch". The situation doesn't change.

Assuming the rule was "The initial placement was at random between the 3 envelopes. You will be always given an option to switch and the revealed envelope will always be an empty one."

"Change your answer" is not 1 in 2. It's 2 in 3.

It's not your choice which sets the probability. It's the initial placement of £1000 (into which envelope to put it). The initial placement is tacitly assumed to be random and 1 in 3. The situation would be different, if £1000 were always in envelope A for example or always in envelope B, or more likely to be in A than B. So there is also not enough information about the initial placement in the question, and not only the rule when to give the option to switch. Nothing really can be said based on the information provided in the question only.

But let's say we make all of these assumptions:
- The initial placement was fully at random between 3 possibilities.
- Regardles of your initial choice, you will always get an option to switch and the revealed envelope will always be an empty one.

Your initial probability to choose correctly is 1/3. Your initial probability to choose incorrectly is 2/3. After the reveal, it's still the same. If you don't switch you still have 1/3 chance. If you switch you now have 2/3. This is because one of the envelopes has been eliminated and you can only switch to the remaining one. So it's not 1/2.

Imagine a similar scenario, but now instead of revealing the empty envelope, you can either keep envelope A or switch and take both envelope B and C. Which one would you chose? I assume you'd agree that the first option has the probability of success 1/3 and the second option has the probability of success 2/3. Does it change anything, if one of the envelopes is revealed? No. You effectively take both, but it happens you know which one is an empty one for sure and you don't know about the other one.

The probability is in the selection method, not in the envelopes themselves. The whole procedure works like a crooked coin with 1/3 of chance for one side and 2/3 for the other side. This is because one of the envelopes is eliminated during the "toss", so there are only "two sides" left, but they have different probabilities. It only appears to be a symmetric choice, but it isn't. The envelopes are symmetric, but the choice procedure isn't.

Or to explain it differently.

Imagine you have a triangular bar like this one. Let's ignore the triangles at the top and at the bottom.

The attachment Triangular-Bar.PNG is no longer available

The bar's sides are painted in red, green and blue.

You are asked to choose a colour.

You choose red.

The green side is now painted in blue, so you have 1 red side and 2 blue sides.

Would you switch to blue or not?

You now only have 2 colours, so is the probability to win 1/2? No.

[unoduetre]

[mention=1400]Citizen[/mention]

You were closest to the right answer, but unfortunately you got arrested.