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I like Raymond Smullyan's version even better, it removes all probability issues and is so simple you feel it must be easy to figure out. The setup is the same

I like Raymond Smullyan's version even better, it removes all probability issues and is so simple you feel it must be easy to figure out. The setup is the same: two envelopes, once with twice as much money as the other, you pick one and before you open you are offered to switch. If you do switch you might either win money or lose money relative to what was in the envelope you chose first. Let's compare the possible gain to the possible loss (without worrying about the probability of having a gain or a loss): 1. The possible gain is equal to the possible loss. Proof: The two envelopes contain x and 2x, so you'd gain x if switching from x to 2x, and you'd lose x if switching from 2x to x. The possible gain and the possible loss are both equal to x. 2. The possible win is larger than the possible loss. Proof: Say the envelope you chose first has y. Then the other envelope has either y/2 or 2y. So if you gain money by switching you gain 2y - y = y, but if you lose in the switch, you lose y - y/2 = y/2. Clearly y > y/2. This version kept up at night for days when I first read it. I never figured it out --not that I had any chance of doing so, given that Smullyan couldn't figure it out either--, I just got used to not understanding it...

jeromec 16y ago
This doesn't eliminate probability, and it's really the same flaw, just with different wording. The flaw is still that the variable is undefined (or infinite probability) in one part of the word problem, but treated as defined in another. So, start with when you first chose the envelope which has y. At this point "y" is undefined, and can be the higher or lower amount. That's the key. Yet, you then continue to propose that by switching you gain 2y - y = y. That equation is not okay, because "y" could have been the higher amount, and the equation breaks down. Once you assert the equation holds true, you have thus defined the original value of "y" from that first envelope choice (i.e. "y" was lower). The same flaw surfaces by asserting your switch to lose equation is true, as well. Edit: BTW, the reason #1 works okay is because you define upfront the variable value ("x") as being either the higher or lower of the two possibilities, before constructing your equation(s).
- omaranto 16y ago
  You mention this doesn't eliminate probability but don't explain why you think that might be the case. I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). I don't find your argument convincing either, maybe it helps to remove the "undefined variable" you complain about: The first envelope chosen has some definite amount of money, let's say $10 (but the same argument can be adapted to any amount). Then the other envelope must have either $5 or $20. If it is $5, switching loses you $5. If it is $10, switching gains you $10. Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20?
  - jeromec 16y ago
    It doesn't eliminate probability because it's the exact same envelope experiment. ;) In other words, there is still a 50% probability of gaining or losing. My post above isn't an argument, it's an explanation. The "undefined variable" is established by your own proposal, i.e., in #2 you first choose an envelope which has an amount we call "y". At that point "y" is undefined, because you specifically explain it could be either the higher or lower amount. I think your confusion lies in not considering carefully enough the meaning of your wording. In other words, you're performing a play on words to get a specific result, akin to the "heads I win, tails you lose" word play. Let me try to explain further below: Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20? No. The other envelope contains the amount the person who prepared the envelopes placed in it. Try defining both values explicitly and this issue may become clearer. Imagine one envelope contains $10 and the other contains $5. Go through your above proposed equations using strictly those values. Remember, in a real experiment that would be completely valid.
  - Confusion 16y ago
    I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). The fact that he doesn't uses those words does not mean he doesn't use those concepts. The Smullyan version is simply less precisely stated and anyone solving it would first have to go through the trouble to convert the phrasing to incorporate the usual, well defined mathematical words. Solving it any other way is just handwaving.
Amnon 16y ago
Statement 1 isn't true. If you have 10$ in the envelope, and it's the smaller envelope, you gain 10$. But if it's the larger envelope you don't lose 10$. The fallacy is that "x" refers to different things in each case. Statement 2 is true but it isn't relevant, since you don't know the posterior probability given y. If the envelope contains the smaller amount of money then there is no way you can lose -- it's not a 50-50 chance.

jeromec 16y ago

This doesn't eliminate probability, and it's really the same flaw, just with different wording. The flaw is still that the variable is undefined (or infinite probability) in one part of the word problem, but treated as defined in another. So, start with when you first chose the envelope which has y. At this point "y" is undefined, and can be the higher or lower amount. That's the key. Yet, you then continue to propose that by switching you gain 2y - y = y. That equation is not okay, because "y" could have been the higher amount, and the equation breaks down. Once you assert the equation holds true, you have thus defined the original value of "y" from that first envelope choice (i.e. "y" was lower). The same flaw surfaces by asserting your switch to lose equation is true, as well. Edit: BTW, the reason #1 works okay is because you define upfront the variable value ("x") as being either the higher or lower of the two possibilities, before constructing your equation(s).

omaranto 16y ago
You mention this doesn't eliminate probability but don't explain why you think that might be the case. I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). I don't find your argument convincing either, maybe it helps to remove the "undefined variable" you complain about: The first envelope chosen has some definite amount of money, let's say $10 (but the same argument can be adapted to any amount). Then the other envelope must have either $5 or $20. If it is $5, switching loses you $5. If it is $10, switching gains you $10. Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20?
- jeromec 16y ago
  It doesn't eliminate probability because it's the exact same envelope experiment. ;) In other words, there is still a 50% probability of gaining or losing. My post above isn't an argument, it's an explanation. The "undefined variable" is established by your own proposal, i.e., in #2 you first choose an envelope which has an amount we call "y". At that point "y" is undefined, because you specifically explain it could be either the higher or lower amount. I think your confusion lies in not considering carefully enough the meaning of your wording. In other words, you're performing a play on words to get a specific result, akin to the "heads I win, tails you lose" word play. Let me try to explain further below: Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20? No. The other envelope contains the amount the person who prepared the envelopes placed in it. Try defining both values explicitly and this issue may become clearer. Imagine one envelope contains $10 and the other contains $5. Go through your above proposed equations using strictly those values. Remember, in a real experiment that would be completely valid.
- Confusion 16y ago
  I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). The fact that he doesn't uses those words does not mean he doesn't use those concepts. The Smullyan version is simply less precisely stated and anyone solving it would first have to go through the trouble to convert the phrasing to incorporate the usual, well defined mathematical words. Solving it any other way is just handwaving.

Amnon 16y ago

Statement 1 isn't true. If you have 10$ in the envelope, and it's the smaller envelope, you gain 10$. But if it's the larger envelope you don't lose 10$. The fallacy is that "x" refers to different things in each case. Statement 2 is true but it isn't relevant, since you don't know the posterior probability given y. If the envelope contains the smaller amount of money then there is no way you can lose -- it's not a 50-50 chance.