4 ms·

Yep so the way you've stated it there is my understanding of the problem, except for this bit: You choose an envelope, and find that it contains the number 8.3

by dools 13y ago

Yep so the way you've stated it there is my understanding of the problem, except for this bit: You choose an envelope, and find that it contains the number 8.3467394756297 You're actually given the chance to switch without having first seen the value in the envelope. If we forget the notion of switching and expected values for a second, I just want to focus on the very start of the statement where you say: Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists negates the other from existing, because in order for both to be possible you're introducing a 3rd value where only 2 existed at the start. If you ignore all the envelopes and switching and whatnot you can boil it down to this: I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero. So if A and B are envelopes, how does this change? If B is twice A how does it affect anything? All the other information provided in this paradox just serves to distract you from the fact that in order for you to have selected a value, and for the probability to exist that either half that value or twice that value is in the remaining envelope, you would have had to have 3 values present at the start of the exercise.

halisaurus 13y ago
Your A/B explanation misses the point that the selection doesn't reveal the relationship, only the absolute value. Once you reveal both options, the selection reveals the absolute value and the position in the set. Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?"
- dools 13y ago
  Your A/B explanation misses the point that the selection doesn't reveal the relationship, only the absolute value. The selection doesn't actually reveal anything. You're given the opportunity to switch prior to opening the envelope. Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?" A and B are just labels for the 2 options present at the start of the exercise. You can't assume you've selected B and then say that the remaining option is either A or C because only A and B existed to begin with (or B and C, whatever the labels are is irrelevant). To put it another way: if you select an envelope and attribute a dollar value to it (in the Wikipedia article this is $20) and then suppose it's either the higher or lower value, then compare the expected value of switching in each case, the things you're comparing have absolutely no relationship to each other. If you assume the value you chose was $20, and it's the lower value, then your initial values were $20 and $40, and you should switch. If you assume the value you chose was $20, and it's the higher value, then your initial values are $20 and $10 and you shouldn't switch. In either case the only relevant piece of information is whether or not you chose the higher or lower value and you can't include all 3 values ($10, $20 and $40) in an "expected value" calculation because only 2 values actually exist. Actually it becomes even more starkly obvious if you change the dollar value between each statement, eg: assume you chose $20 and it's the higher value. If you switch, you lose $10. But what if you chose $5,000 and it's the lower value? If you switch you'd gain $5,000 so you should always switch because you're only risking a 50/50 chance of losing $10 for a 50/50 chance of gaining $5,000. This is clearly ridiculous!
  - pencilcheck 13y ago
    I will try my best to explain this paradox. This is a paradox only if you act like a computer. So what do I mean by that? This problem simply means, that if you selected an envelope, and you know that there is a chance that the other envelope will have higher value than then one you selected, then by probability, you could conclude that you are better off swapping to the other envelope. However, the paradox starts after you switch envelope because now you can start this all over again and think, hey, even though I just swapped, but it is still possible that the other envelope has value higher than the one I just selected! To explain it in mathematical terms, when the value of the currently selected envelope is Z, the expected value of the other envelope will always be Z/2 * 0.5 + 2Z * 0.5 = 5/4Z higher than the one you selected. The trick is to think procedurally, instead of trying to be realistic. In other words, it is always statistically possible that the other envelope has higher value than the one you selected, therefore you should always swap.
    - dools 13y ago
      Right ... except that's not true :) I actually found a great article today on lesswrong.com which puts things far more elegantly than I ever could, but fundamentally you need to know something about how the values get into the envelope: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... I've submitted this as a fresh link on HN now. Hopefully we see some lively discussion there :)
dalai 13y ago
The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists negates the other from existing, because in order for both to be possible you're introducing a 3rd value where only 2 existed at the start. So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement? I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero. No it doesn't. According to the problem, both 2Z and Z/2 are possible and you don't have the information to decide either way. Think of a simple coin toss, before I show you the result it is 50-50 to be heads or tails, even though it is just one of the two.
- dools 13y ago
  So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement? Yes! Well, in some circumstances. Unless you know how the values were placed into the envelope to begin with, how could you possibly assign a probability to it? I just found this great link today: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... which basically resolves the problem I'm having by taking into account how the values get into the envelopes in the first place. If you have absolutely no knowledge of how the values got into the envelope in the first place then there's a 50/50 chance you got the higher value on your first pick, and a 50/50 chance you'll get the higher value on your second pick. You can't make a more informed decision unless you know how the values were initially chosen.
ColinWright 13y ago
> The issue as I see it is that the probability exists that the other > envelope contains Z/2 or 2Z, but not both - ie. the probability that > one of those exists negates the other from existing, because in order > for both to be possible you're introducing a 3rd value where only > 2 existed at the start. I can't make any sense of this at all. You choose an envelope and don't look at it. Instead, call the value inside Z. Under the conditions of the problem as stated, then other envelope now either contains Z/2, or it contains 2Z. You could have flipped a fair coin to decide which envelope to choose in the first place. Therefore there is a 50% chance the non-selected envelope contains Z/2, and a 50% chance the non-selected envelope contains 2Z. I don't understand where you think the impossible value C is being introduced. I don't understand where you think there is a third value.
- dools 13y ago
  Saying that I chose $20, and there's a 50% chance of there being $10 in the remaining envelope, and a 50% chance of there being $40, takes into account 3 amounts: $10, $20 and $40. However if you're presented with 2 envelopes, there are only 2 amounts to begin with. The only way you could assign a probability to what the second envelope contained would be if you started with 3 amounts then chose 2. For example I have a 5c coin, a 10c and a 20c coin. I pick the 10c coin and flip it: heads I pair it with the 5c coin and tails I pair it with the 20c coin, then I put each under a cup and invite you to choose. Now if you lift the cup and see 10c you know there is a 50% chance the other cup has 5c and a 50% chance it has 20c. You can take into account 3 amounts (ie. the one you chose and 2 possible remaining values) because the question starts off with 3 amounts. If your knowledge of the system commences with "here are 2 cups" you have no extra information -- you only have 2 amounts. The higher and the lower. You will pick the higher one 50% of the time on the first go, and 50% of the time if you switch.
  - ColinWright 13y ago
    > Saying that I chose $20, and there's a 50% chance of there > being $10 in the remaining envelope, and a 50% chance of there > being $40, takes into account 3 amounts: $10, $20 and $40. Right, but that's after you have opened the envelope. > However if you're presented with 2 envelopes, there are only > 2 amounts to begin with. I don't see why you think this is an issue, or relevant. > The only way you could assign a probability to what the second > envelope contained would be if you started with 3 amounts then > chose 2. That doesn't make sense. Suppose I flip a coin, and choose an envelope at random. Without opening it, suppose I designate the value it holds as Z. Now I know with probability 50% that the other envelope contains Z/2, and with 50% the other envelope contains 2Z. > For example I have a 5c coin, a 10c and a 20c coin. I pick the > 10c coin and flip it: heads I pair it with the 5c coin and tails > I pair it with the 20c coin, then I put each under a cup and > invite you to choose. > Now if you lift the cup and see 10c you know there is a 50% chance > the other cup has 5c and a 50% chance it has 20c. You can take into > account 3 amounts (ie. the one you chose and 2 possible remaining > values) because the question starts off with 3 amounts. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. > If your knowledge of the system commences with "here are 2 cups" > you have no extra information -- you only have 2 amounts. The > higher and the lower. You will pick the higher one 50% of the time > on the first go, and 50% of the time if you switch. Yes. So now, without lifting the cup, let's consider some statistics. Let's call the unknown amount under the cup I've chosen Z. The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). If I switch, my expected return is then 5Z/4, which is larger than Z. Hence, if given the option, I should switch, even though I don't know what the amount is that I've chosen. That is the paradox, which you still appear not to have resolved.

by dools 13y ago

halisaurus 13y ago

Your A/B explanation misses the point that the selection doesn't reveal the relationship, only the absolute value. Once you reveal both options, the selection reveals the absolute value and the position in the set. Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?"

dools 13y ago
Your A/B explanation misses the point that the selection doesn't reveal the relationship, only the absolute value. The selection doesn't actually reveal anything. You're given the opportunity to switch prior to opening the envelope. Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?" A and B are just labels for the 2 options present at the start of the exercise. You can't assume you've selected B and then say that the remaining option is either A or C because only A and B existed to begin with (or B and C, whatever the labels are is irrelevant). To put it another way: if you select an envelope and attribute a dollar value to it (in the Wikipedia article this is $20) and then suppose it's either the higher or lower value, then compare the expected value of switching in each case, the things you're comparing have absolutely no relationship to each other. If you assume the value you chose was $20, and it's the lower value, then your initial values were $20 and $40, and you should switch. If you assume the value you chose was $20, and it's the higher value, then your initial values are $20 and $10 and you shouldn't switch. In either case the only relevant piece of information is whether or not you chose the higher or lower value and you can't include all 3 values ($10, $20 and $40) in an "expected value" calculation because only 2 values actually exist. Actually it becomes even more starkly obvious if you change the dollar value between each statement, eg: assume you chose $20 and it's the higher value. If you switch, you lose $10. But what if you chose $5,000 and it's the lower value? If you switch you'd gain $5,000 so you should always switch because you're only risking a 50/50 chance of losing $10 for a 50/50 chance of gaining $5,000. This is clearly ridiculous!
- pencilcheck 13y ago
  I will try my best to explain this paradox. This is a paradox only if you act like a computer. So what do I mean by that? This problem simply means, that if you selected an envelope, and you know that there is a chance that the other envelope will have higher value than then one you selected, then by probability, you could conclude that you are better off swapping to the other envelope. However, the paradox starts after you switch envelope because now you can start this all over again and think, hey, even though I just swapped, but it is still possible that the other envelope has value higher than the one I just selected! To explain it in mathematical terms, when the value of the currently selected envelope is Z, the expected value of the other envelope will always be Z/2 * 0.5 + 2Z * 0.5 = 5/4Z higher than the one you selected. The trick is to think procedurally, instead of trying to be realistic. In other words, it is always statistically possible that the other envelope has higher value than the one you selected, therefore you should always swap.
  - dools 13y ago
    Right ... except that's not true :) I actually found a great article today on lesswrong.com which puts things far more elegantly than I ever could, but fundamentally you need to know something about how the values get into the envelope: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... I've submitted this as a fresh link on HN now. Hopefully we see some lively discussion there :)

dalai 13y ago

The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists negates the other from existing, because in order for both to be possible you're introducing a 3rd value where only 2 existed at the start. So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement? I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero. No it doesn't. According to the problem, both 2Z and Z/2 are possible and you don't have the information to decide either way. Think of a simple coin toss, before I show you the result it is 50-50 to be heads or tails, even though it is just one of the two.

dools 13y ago
So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement? Yes! Well, in some circumstances. Unless you know how the values were placed into the envelope to begin with, how could you possibly assign a probability to it? I just found this great link today: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... which basically resolves the problem I'm having by taking into account how the values get into the envelopes in the first place. If you have absolutely no knowledge of how the values got into the envelope in the first place then there's a 50/50 chance you got the higher value on your first pick, and a 50/50 chance you'll get the higher value on your second pick. You can't make a more informed decision unless you know how the values were initially chosen.

ColinWright 13y ago

> The issue as I see it is that the probability exists that the other > envelope contains Z/2 or 2Z, but not both - ie. the probability that > one of those exists negates the other from existing, because in order > for both to be possible you're introducing a 3rd value where only > 2 existed at the start. I can't make any sense of this at all. You choose an envelope and don't look at it. Instead, call the value inside Z. Under the conditions of the problem as stated, then other envelope now either contains Z/2, or it contains 2Z. You could have flipped a fair coin to decide which envelope to choose in the first place. Therefore there is a 50% chance the non-selected envelope contains Z/2, and a 50% chance the non-selected envelope contains 2Z. I don't understand where you think the impossible value C is being introduced. I don't understand where you think there is a third value.

dools 13y ago
Saying that I chose $20, and there's a 50% chance of there being $10 in the remaining envelope, and a 50% chance of there being $40, takes into account 3 amounts: $10, $20 and $40. However if you're presented with 2 envelopes, there are only 2 amounts to begin with. The only way you could assign a probability to what the second envelope contained would be if you started with 3 amounts then chose 2. For example I have a 5c coin, a 10c and a 20c coin. I pick the 10c coin and flip it: heads I pair it with the 5c coin and tails I pair it with the 20c coin, then I put each under a cup and invite you to choose. Now if you lift the cup and see 10c you know there is a 50% chance the other cup has 5c and a 50% chance it has 20c. You can take into account 3 amounts (ie. the one you chose and 2 possible remaining values) because the question starts off with 3 amounts. If your knowledge of the system commences with "here are 2 cups" you have no extra information -- you only have 2 amounts. The higher and the lower. You will pick the higher one 50% of the time on the first go, and 50% of the time if you switch.
- ColinWright 13y ago
  > Saying that I chose $20, and there's a 50% chance of there > being $10 in the remaining envelope, and a 50% chance of there > being $40, takes into account 3 amounts: $10, $20 and $40. Right, but that's after you have opened the envelope. > However if you're presented with 2 envelopes, there are only > 2 amounts to begin with. I don't see why you think this is an issue, or relevant. > The only way you could assign a probability to what the second > envelope contained would be if you started with 3 amounts then > chose 2. That doesn't make sense. Suppose I flip a coin, and choose an envelope at random. Without opening it, suppose I designate the value it holds as Z. Now I know with probability 50% that the other envelope contains Z/2, and with 50% the other envelope contains 2Z. > For example I have a 5c coin, a 10c and a 20c coin. I pick the > 10c coin and flip it: heads I pair it with the 5c coin and tails > I pair it with the 20c coin, then I put each under a cup and > invite you to choose. > Now if you lift the cup and see 10c you know there is a 50% chance > the other cup has 5c and a 50% chance it has 20c. You can take into > account 3 amounts (ie. the one you chose and 2 possible remaining > values) because the question starts off with 3 amounts. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. > If your knowledge of the system commences with "here are 2 cups" > you have no extra information -- you only have 2 amounts. The > higher and the lower. You will pick the higher one 50% of the time > on the first go, and 50% of the time if you switch. Yes. So now, without lifting the cup, let's consider some statistics. Let's call the unknown amount under the cup I've chosen Z. The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). If I switch, my expected return is then 5Z/4, which is larger than Z. Hence, if given the option, I should switch, even though I don't know what the amount is that I've chosen. That is the paradox, which you still appear not to have resolved.