One of them must be a possible value of X. Not necessarily both.

Sorry, still not following you. One of what?

Fraud.

Sorry, yes, I mean Y = 10 → . X = 5 ∨ X = 10.

You know that at least one of 5 and 10 are possible values of X. Do you know that both are?

I would say yes I do. I feel a discussion over the meaning of 'possible' may be approaching, but let's see where this goes.

I just don't understand how you know that.

To me, to say 'B is possible' means 'I would not be astonished if I discovered B to be the case'.

If I have 10 and somebody opens the other envelope, I will be astonished neither if it contains 5 nor if it contains 20. I would however be astonished - or think the games mistress has deceived me - if it contained 100, or any other number except 5 or 20.

That's more answer than I'm asking for. What in the rules of the game or in the knowledge you have acquired (i.e., that Y = 10) assures you that the games mistress includes both 5 and 10 in the sample space for X? Maybe a tenner is the smallest bill she has.

It's not about the games mistress's sample space. She doesn't have one. She has two amounts that she has always intended to put into the envelopes. It's about my sample space, which consists of the events that I cannot rule out based on my knowledge to date.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events.

I must to bed. Cheers.

Dormez bien, mon ami.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events. — andrewk

Based on your knowledge you know that the sample space has to follow the algebraic form of [X,2X],and you know by the definition of a sample space, that your sample space can only contain possible outcomes (or at least I hope you do).

Therefore your sample space could never be [5,20], as it does not follow the form of [X, 2X]. Your sample space could be [5,10] or [10,20] but there is no way you can mathematically justify a sample space of [5,20] as it is not consistent with [X,2X].

I have no idea what a Frequentist would do. — andrewk

Flip a coin.

A simple random sample, which is what Classical statistics is based on, is a subset of a population which is obtained by a process which gives all sets of n distinct items in a population an equal chance of being selected. There are two n distinct items, so they would flip a coin.

Both Classical and Bayesian approach uses a random process to make selections from a sample space. Bayesian methods combine their prior with random samples from a sample distribution to create posterior distributions. Then they make their assessments off the posterior distributions.

Based on your knowledge you know that the sample space has to follow the algebraic form of [X,2X],and you know by the definition of a sample space, that your sample space can only contain possible outcomes (or at least I hope you do).

Therefore your sample space could never be [5,20], as it does not follow the form of [X, 2X]. Your sample space could be [5,10] or [10,20] but there is no way you can mathematically justify a sample space of [5,20] as it is not consistent with [X,2X]. — Jeremiah

[X, 2X] is our sample space for both envelopes, but given that we know that there's £10 in our envelope, what's our sample space for the other envelope?

What if there are two people and each time both are randomly given an envelope. Is it then beneficial for both to switch the envelopes? So each of them gets more than half of the money on average.

What if these two people then put the money back together and distribute it in new envelopes? Do they by switching the envelopes again keep infinitely increasing the amount of money?

You just said it yourself the sample space for both is X and 2X.

It went over this right here:

https://thephilosophyforum.com/discussion/comment/192854

Also, it should be made clear that an uninformative prior is a Classical apporach. A Baysian approach with an uninformative prior is philosophically the same apporach as a Classical apporach. Baysian divergence is when it introduces an informative prior.

You just said it yourself the sample space for both is X and 2X. — Jeremiah

Yes, and then I asked you what the sample space is for the other envelope, given that your envelope contains £10.

You read it too fast and understood too little. I very clearly and fully addressed your concern.

What if there are two people and each time both are randomly given an envelope. Is it then beneficial for both to switch the envelopes? — BlueBanana

I have flipped a coin and know the result. I will let you bet on it with a 2:1 payout if you win. However, I decide how much you have to bet, and only after you tell me your guess (although you're free to back out if the bet is too high).

Should you bet?

You read it too fast and understood too little. I very clearly and fully addressed your concern. — Jeremiah

Clearly not clearly enough, as I didn't understand it. What is the sample space of the other envelope?

I have flipped a coin and know the result. I will let you bet on it with a 2:1 payout if you win. However, I decide how much you have to bet, and only after you tell me your guess.

Should you bet? — Michael

I don't understand how this is supposed to be comparable or relevant to the envelope paradox. Are you the person opening the other envelope or the person organizing the test?

Both people know the amount in their envelope. Let's say they can both choose the same envelope but can't communicate with each other. What should they choose? Always choosing the envelope of the other person can't be beneficial unless A+B≠B+A.

I don't understand how this is supposed to be comparable or relevant to the envelope paradox. — BlueBanana

The point is that it's an advantageous bet, even though I will ask you to bet £20 if you guess wrong and another person £10 if he guesses right, so that the average payout is £0.

As you've stated it, it's not an advantageous bet if you have to bet twice as much when you lose.

Let's say you put the two envelopes on a table, and randomly choose to put one on the left side and the other on the right side. You can each time open the one on the left and then switch to the right one. If this is beneficial, you can just choose the right one each time without checking the amount on the left envelope.

If that's not absurd enough, you can do the same reasoning for the left envelope.

As you've stated it, it's not an advantageous bet if you have to bet twice as much when you lose. — BlueBanana

I offer you a fixed amount of £20 whether you pick the winner or the loser. It's just that if you pick the loser then I will offer someone else who picks the winner a £10 bet.

Can the participant be certain they're the person getting the fair chance? The person betting £10 will always win and the person betting £40 will always lose.

Everyone gets a fair chance. What’s the difference between me offering you a £20 bet regardless of your choice and me offering you a £20 bet only if you pick the loser? The different bets is just about balancing my books. For the participants it’s still just a 50% chance with a 2:1 payout.

You can't balance your books with £20 bets with 50% chance of 2:1 payout and £10 bets with certain wins so clarify the system in your example.

It's just that if you pick the loser then I will offer someone else who picks the winner a £10 bet. — Michael

And if I pick the winner you lose £40, so I assume you also offer someone who picks the loser a £40 bet, right?

For the participants it’s still just a 50% chance with a 2:1 payout. — Michael

True, but notice that you can't make profit out of betting in this system. Taking part in it still isn't profitable.

True, but notice that you can't make profit out of betting in this system. Taking part in it still isn't profitable. — BlueBanana

You keep thinking about it as what would happen if you play repeatedly. I'm just talking about playing a single game. There's a 50% chance of winning and the payout is 2:1. It doesn't matter that my potential winnings are offset by someone else losing after betting twice as much or that my potential losses are offset by someone else winning after betting half as much. We can play the game and not even consider anybody else being involved.

There's a 50% chance of winning and the payout is 2:1. — Michael

Only if you're the person betting £20. And that's fine as a thought experiment of its own but it doesn't relate to the envelope paradox. You don't know whether the amount of money in your envelope is the amount of money that has 50% chance of winning with 2:1 payout.