3 ms·

There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then plac

by pontus 4y ago

There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it. Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This happens with probability 0.2*0.8^4 ~ 8.192%. Alternatively 8 was generated and the envelope you have is already the largest one. This happens with probability 0.2*0.8^3 ~ 10.24%. Given that you see 16 already, the relative probability becomes 4/9 vs 5/9. In other words the chance that you have the largest envelope already is 5/9. The expected value of the other envelope is then 5/9*8 + 4/9*32 = 18.67 > 16 so you should switch. In fact, try this for different initial envelopes and you'll see that you should always switch. But, if you should always switch, why even look inside the envelope? But, then you're back at the symmetry / no information argument.

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
    [deleted]
judahmeek 4y ago
So the real predictor is the probability. Once the probability gets around 7/9*.5X + 2/9*4X, then switching is no longer attractive.
amalcon 4y ago
This is similar to the St. Petersburg problem. The way I satisfy myself with that answer is to note that it's kind of a spherical cow problem. Nobody in the real world can afford to run this game, since it requires an infinite amount of money -- even a currency issuer would expect to just make their currency worthless by running it, and an offering of physical resources instead of currency would violate thermodynamics in all sorts of ways. Spherical cow problems often end in counterintuitive results.

by pontus 4y ago

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
  [deleted]

judahmeek 4y ago

So the real predictor is the probability. Once the probability gets around 7/9*.5X + 2/9*4X, then switching is no longer attractive.

amalcon 4y ago

This is similar to the St. Petersburg problem. The way I satisfy myself with that answer is to note that it's kind of a spherical cow problem. Nobody in the real world can afford to run this game, since it requires an infinite amount of money -- even a currency issuer would expect to just make their currency worthless by running it, and an offering of physical resources instead of currency would violate thermodynamics in all sorts of ways. Spherical cow problems often end in counterintuitive results.