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My explanation is correct and I can't say much more to convince you that I've already said.. yours is wrong in precisely why people are confused with the proble

My explanation is correct and I can't say much more to convince you that I've already said.. yours is wrong in precisely why people are confused with the problem and I'm kinda sorry you don't see it. Since you seem to be making some authority arguments just know that I have a PhD in ML / stats and was top 30 in my country in olympiad level competitive maths so I DO know what I do and in this case with absolute certainty. Look for the other guys answer if you want more details

baobabKoodaa 4y ago
> Since you seem to be making some authority arguments just know that I have a PhD in ML / stats and was top 30 in my country in olympiad level competitive maths so I DO know what I do and in this case with absolute certainty. Let's review what you're claiming with absolute certainty. This is the first claim that I'm contesting: > they say "one envelope contains 2A and the other A/2". When pointed out that they actually don't say this, you clarified your point: > the "writing" doesn't say it but the written formula at bullet point 7 says it. That's the mistake, the formula is wrong and does not describe reality. The second part of that quote is correct: bullet point 7 has the mistake, the formula is wrong and does not describe reality. The first part of that quote is incorrect: the written formula at bullet point 7 does not claim that one envelope contains 2A and the other envelope contains A/2. It claims that the first envelope contains A and the second envelope has a 50% probability of containing 2A and a 50% probability of containing A/2. You correctly identify where the mistake is, but you misidentify what the mistake is. If you read bullet point 6, it is very clear: "the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2". Notice that bullet point 6 does not claim "One envelope contains 2A and the other envelope contains A/2". To illustrate my point, consider a simple coin toss game. If the coin comes up heads, you win $2. If it comes up tails, you win $0.50. How might we construct the EV calculation for this coin toss? 0.5 * $2 + 0.5 * $0.50 Notice how both parts of the formula refer to the same coin. They don't refer to different coins. You have one coin that you flip, and depending on how that one coin lands, there is some probability that you get $2, and some probability that you get $0.50. If you looked at that formula, you wouldn't think that the first part refers to different coin that the second part. Similarly, here we have one envelope (similar to having one coin) and we have uncertainty about the what the envelope contains (similar to having uncertainty about how the coin will land). We have 50% chance that the envelope will contain 2A, and we have 50% chance that the envelope will contain A/2. Thus, we arrive at the (incorrect) formula: 0.5 * 2A + 0.5 * A/2 It's very obvious that both parts of the formula refer to the same envelope, just like in the coin toss formula both parts of the formula refer to the same coin. To be extremely clear, I am not claiming that the formula is correct. I am claiming that both parts of the formula refer to the same envelope, not to different envelopes. To be specific, I am refuting this claim that you made: > they say "one envelope contains 2*A and the other A/2". They don't say that. Furthermore, you make an additional incorrect claim: > The right formula is changing either the 0.5A by A or the 2A by A. Here you are just randomly changing the formula in order to make it output 0EV (which is the correct answer). Just because you get a correct answer does not mean your computation was correct. In this case it is incorrect, because A was defined badly. In order to fix the formula you would have to fix the definition of A. If you keep the incorrect definition of A and just shuffle symbols around until you get the correct result, your computation is still incorrect.
- davidguetta 4y ago
  > The first part of that quote is incorrect: the written formula at bullet point 7 does not claim that one envelope contains 2A and the other envelope contains A/2. It claims that the first envelope contains A and the second envelope has a 50% probability of containing 2A and a 50% probability of containing A/2. But it's wrong! Try with literally any A. > If you read bullet point 6, it is very clear: "the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2". Notice that bullet point 6 does not claim "One envelope contains 2A and the other envelope contains A/2". Sure but this bullet point is also dead wrong and possibly the start of the scam. If you say A is lowest value, Either it contain A with probability 1/2, or 2A. Otherwise you are changing the reality of A midway through the sentence. To meet you halfway, what you may actually want, or see in your head is the possibility that A (in the lowest value sense) may vary in the experiment, and try to compute the expectation of that new problem. But that's a different problem. And if you want to compute the expectation of this problem with A varying in a range you have to write an integral over the p(A) around the (corrected) equation 7. But the equation 7, even in this integral, need to use the same consistency for A: either you have A being the lowest envelope value and you only write 2A ever, or the highest value and then you only write 0.5A
  - baobabKoodaa 4y ago
    > But it's wrong! Try with literally any A. As I said, I agree that the formula is wrong. I disagree with you regarding where the mistake is and how to fix it. > Sure but this bullet point is also dead wrong and possibly the start of the scam. If you say A is lowest value, Either it contain A with probability 1/2, or 2A. I think maybe you have some typos here, because I don't understand what you're trying to say here. In any case our disagreement concerns whether the formula in Wikipedia refers to the same envelope in both parts of the formula, or different envelopes. You kept stating that it refers to different envelopes, and I wrote a long post to refute that. Now you came back saying that the bullet / formula / Wikipedia text is "wrong" and "possibly the start of the scam". Ok, sure, but the EV formula still clearly refers to the same envelope in both parts of the formula, despite the fact that you claimed otherwise with literally "absolute certainty". You were wrong about something that you claimed to know with "absolute certainty", so perhaps you should in the future adjust those estimates downwards with a bit of uncertainty added in? > Otherwise you are changing the reality of A midway through the sentence. Nope, A is defined as the "value of the firstly-chosen envelope" in the beginning of the sentence, and A is defined as the "value of the firstly-chosen envelope" in the end of the sentence. The definition for A does not change midway through the sentence. > But that's a different problem. And if you want to compute the expectation of this problem with A varying in a range you have to write an integral over the p(A) around the (corrected) equation 7. But the equation 7, even in this integral, need to use the same consistency for A: either you have A being the lowest envelope value and you only write 2A ever, or the highest value and then you only write 0.5A That's not the only way to compute the expectation for this problem. I actually ran some small simulations for this today, representing wagers on this problem. The simulations demonstrate the following claims: - Steps 1 through 7 in the Wikipedia page are correct, in terms of calculating the "expected value in the other envelope, relative to the value in the firstly-chosen envelope". The expected value for the switch, relative to the firstly-chosen envelope's value, is in fact positive (+25%). - At the same time, the "expected value for the switch in absolute terms" is zero. - These claims do not contradict each other. - The first error in Wikipedia's line of reasoning is step 8 that states "they stand to gain by swapping". This indicates that the player should try to maximize goal "expected value relative to firstly-chosen envelope", which is incorrect. The player should instead maximize goal "expected value in absolute terms". If you disagree with any of these, let's formulate our disagreement in the form of a wager that can be simulated in code.

baobabKoodaa 4y ago

> Since you seem to be making some authority arguments just know that I have a PhD in ML / stats and was top 30 in my country in olympiad level competitive maths so I DO know what I do and in this case with absolute certainty. Let's review what you're claiming with absolute certainty. This is the first claim that I'm contesting: > they say "one envelope contains 2A and the other A/2". When pointed out that they actually don't say this, you clarified your point: > the "writing" doesn't say it but the written formula at bullet point 7 says it. That's the mistake, the formula is wrong and does not describe reality. The second part of that quote is correct: bullet point 7 has the mistake, the formula is wrong and does not describe reality. The first part of that quote is incorrect: the written formula at bullet point 7 does not claim that one envelope contains 2A and the other envelope contains A/2. It claims that the first envelope contains A and the second envelope has a 50% probability of containing 2A and a 50% probability of containing A/2. You correctly identify where the mistake is, but you misidentify what the mistake is. If you read bullet point 6, it is very clear: "the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2". Notice that bullet point 6 does not claim "One envelope contains 2A and the other envelope contains A/2". To illustrate my point, consider a simple coin toss game. If the coin comes up heads, you win $2. If it comes up tails, you win $0.50. How might we construct the EV calculation for this coin toss? 0.5 * $2 + 0.5 * $0.50 Notice how both parts of the formula refer to the same coin. They don't refer to different coins. You have one coin that you flip, and depending on how that one coin lands, there is some probability that you get $2, and some probability that you get $0.50. If you looked at that formula, you wouldn't think that the first part refers to different coin that the second part. Similarly, here we have one envelope (similar to having one coin) and we have uncertainty about the what the envelope contains (similar to having uncertainty about how the coin will land). We have 50% chance that the envelope will contain 2A, and we have 50% chance that the envelope will contain A/2. Thus, we arrive at the (incorrect) formula: 0.5 * 2A + 0.5 * A/2 It's very obvious that both parts of the formula refer to the same envelope, just like in the coin toss formula both parts of the formula refer to the same coin. To be extremely clear, I am not claiming that the formula is correct. I am claiming that both parts of the formula refer to the same envelope, not to different envelopes. To be specific, I am refuting this claim that you made: > they say "one envelope contains 2*A and the other A/2". They don't say that. Furthermore, you make an additional incorrect claim: > The right formula is changing either the 0.5A by A or the 2A by A. Here you are just randomly changing the formula in order to make it output 0EV (which is the correct answer). Just because you get a correct answer does not mean your computation was correct. In this case it is incorrect, because A was defined badly. In order to fix the formula you would have to fix the definition of A. If you keep the incorrect definition of A and just shuffle symbols around until you get the correct result, your computation is still incorrect.

davidguetta 4y ago
> The first part of that quote is incorrect: the written formula at bullet point 7 does not claim that one envelope contains 2A and the other envelope contains A/2. It claims that the first envelope contains A and the second envelope has a 50% probability of containing 2A and a 50% probability of containing A/2. But it's wrong! Try with literally any A. > If you read bullet point 6, it is very clear: "the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2". Notice that bullet point 6 does not claim "One envelope contains 2A and the other envelope contains A/2". Sure but this bullet point is also dead wrong and possibly the start of the scam. If you say A is lowest value, Either it contain A with probability 1/2, or 2A. Otherwise you are changing the reality of A midway through the sentence. To meet you halfway, what you may actually want, or see in your head is the possibility that A (in the lowest value sense) may vary in the experiment, and try to compute the expectation of that new problem. But that's a different problem. And if you want to compute the expectation of this problem with A varying in a range you have to write an integral over the p(A) around the (corrected) equation 7. But the equation 7, even in this integral, need to use the same consistency for A: either you have A being the lowest envelope value and you only write 2A ever, or the highest value and then you only write 0.5A
- baobabKoodaa 4y ago
  > But it's wrong! Try with literally any A. As I said, I agree that the formula is wrong. I disagree with you regarding where the mistake is and how to fix it. > Sure but this bullet point is also dead wrong and possibly the start of the scam. If you say A is lowest value, Either it contain A with probability 1/2, or 2A. I think maybe you have some typos here, because I don't understand what you're trying to say here. In any case our disagreement concerns whether the formula in Wikipedia refers to the same envelope in both parts of the formula, or different envelopes. You kept stating that it refers to different envelopes, and I wrote a long post to refute that. Now you came back saying that the bullet / formula / Wikipedia text is "wrong" and "possibly the start of the scam". Ok, sure, but the EV formula still clearly refers to the same envelope in both parts of the formula, despite the fact that you claimed otherwise with literally "absolute certainty". You were wrong about something that you claimed to know with "absolute certainty", so perhaps you should in the future adjust those estimates downwards with a bit of uncertainty added in? > Otherwise you are changing the reality of A midway through the sentence. Nope, A is defined as the "value of the firstly-chosen envelope" in the beginning of the sentence, and A is defined as the "value of the firstly-chosen envelope" in the end of the sentence. The definition for A does not change midway through the sentence. > But that's a different problem. And if you want to compute the expectation of this problem with A varying in a range you have to write an integral over the p(A) around the (corrected) equation 7. But the equation 7, even in this integral, need to use the same consistency for A: either you have A being the lowest envelope value and you only write 2A ever, or the highest value and then you only write 0.5A That's not the only way to compute the expectation for this problem. I actually ran some small simulations for this today, representing wagers on this problem. The simulations demonstrate the following claims: - Steps 1 through 7 in the Wikipedia page are correct, in terms of calculating the "expected value in the other envelope, relative to the value in the firstly-chosen envelope". The expected value for the switch, relative to the firstly-chosen envelope's value, is in fact positive (+25%). - At the same time, the "expected value for the switch in absolute terms" is zero. - These claims do not contradict each other. - The first error in Wikipedia's line of reasoning is step 8 that states "they stand to gain by swapping". This indicates that the player should try to maximize goal "expected value relative to firstly-chosen envelope", which is incorrect. The player should instead maximize goal "expected value in absolute terms". If you disagree with any of these, let's formulate our disagreement in the form of a wager that can be simulated in code.