3 ms·

Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest a

by kmod 4y ago

Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.

pontus 4y ago
No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the expected value of both envelopes is infinite. Only after you look inside one of them does that collapse to two finite values that you can subtract.
- kmod 4y ago
  If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite
  - pontus 4y ago
    The point I'm making is that if you open the envelope you will always conclude that it's advantageous to switch regardless of what you see in that envelope. In the case where you see $1, you're guaranteed that switching is a good idea. If you see anything other than $1 you come to the same conclusion but you have to rely on a probabilistic argument using expected values. Now, if you will decide to switch regardless of what you see in the envelope, then why even bother opening it? But, if you don't open the envelope, then you're back to the original problem where the symmetry argument is valid.
    - kmod 4y ago
      Every envelope you open will have a higher expected value than the previous one. This is not paradoxical, it is a standard property of probability distributions with infinite means. I believe the problem you are trying to elicit is that you can almost show that either envelope is a priori better than the other. The problem with this is that the means are infinite so the math isn't sound. So there's not an actual contradiction; the strategy is to open either envelope and then switch to the other one. It's unintuitive but only because we're not used to reasoning about distributions with infinite means.
      - pontus 4y ago
        Right, I mentioned above that the resolution to this is that the expected value of both envelopes is infinite before you open anything and thus the difference is undefined. Once you open one, the difference becomes well defined. I still think it's quite unintuitive that one needs to open an envelope in order to regularize things this way. The fact that you will always decide to switch, regardless of what you see in the first envelope, yet you still have to open it is what I find paradoxical.
- deleted 4y ago
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by kmod 4y ago

pontus 4y ago

No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the expected value of both envelopes is infinite. Only after you look inside one of them does that collapse to two finite values that you can subtract.

kmod 4y ago
If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite
- pontus 4y ago
  The point I'm making is that if you open the envelope you will always conclude that it's advantageous to switch regardless of what you see in that envelope. In the case where you see $1, you're guaranteed that switching is a good idea. If you see anything other than $1 you come to the same conclusion but you have to rely on a probabilistic argument using expected values. Now, if you will decide to switch regardless of what you see in the envelope, then why even bother opening it? But, if you don't open the envelope, then you're back to the original problem where the symmetry argument is valid.
  - kmod 4y ago
    Every envelope you open will have a higher expected value than the previous one. This is not paradoxical, it is a standard property of probability distributions with infinite means. I believe the problem you are trying to elicit is that you can almost show that either envelope is a priori better than the other. The problem with this is that the means are infinite so the math isn't sound. So there's not an actual contradiction; the strategy is to open either envelope and then switch to the other one. It's unintuitive but only because we're not used to reasoning about distributions with infinite means.
    - pontus 4y ago
      Right, I mentioned above that the resolution to this is that the expected value of both envelopes is infinite before you open anything and thus the difference is undefined. Once you open one, the difference becomes well defined. I still think it's quite unintuitive that one needs to open an envelope in order to regularize things this way. The fact that you will always decide to switch, regardless of what you see in the first envelope, yet you still have to open it is what I find paradoxical.
deleted 4y ago
[deleted]