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The Two Envelopes Problem: the most boring paradox in the world

3 points by dools 13y ago

ColinWright 13y ago
The author appears not to understand the problem as presented - certainly the "solution" as given is inconsistent with my understanding of the problem. Consider your position. I present you with two envelopes, and tell you that each contains a number, and one of those numbers is twice the other. You choose an envelope, and find that it contains the number 8.3467394756297. You can choose to swap, if you like, and your objective is to maximize the expected value of the number in the envelope you're holding. Do you swap? Or not? Here's a line of reasoning. Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z, and it would appear to be 50:50 as to which it is. If you swap, my expected value is then (Z/2)0.5 + (2Z)0.5 = 5Z/4. That's bigger than Z, so you should swap. But that means you should always swap, which doesn't seem to make sense. Perhaps this version explains more clearly to you the paradox. Or not, maybe you still think it's all obvious. In that case, could you revise your explanation for this context, away from the $20 vs $10 or $40 issues?
- 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!
      - 13y ago

3 points by dools 13y ago

The author appears not to understand the problem as presented - certainly the "solution" as given is inconsistent with my understanding of the problem. Consider your position. I present you with two envelopes, and tell you that each contains a number, and one of those numbers is twice the other. You choose an envelope, and find that it contains the number 8.3467394756297. You can choose to swap, if you like, and your objective is to maximize the expected value of the number in the envelope you're holding. Do you swap? Or not? Here's a line of reasoning. Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z, and it would appear to be 50:50 as to which it is. If you swap, my expected value is then (Z/2)0.5 + (2Z)0.5 = 5Z/4. That's bigger than Z, so you should swap. But that means you should always swap, which doesn't seem to make sense. Perhaps this version explains more clearly to you the paradox. Or not, maybe you still think it's all obvious. In that case, could you revise your explanation for this context, away from the $20 vs $10 or $40 issues?

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!
    - 13y ago