3 ms·

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 yo

by 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.

by kmod 4y ago

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.