Logic question

[ASL_v2.0]

@Danger_Man

If the first envelope is fixed to 10, and the second envelope is set randomly to either 5 or 20, then the statement on both envelopes will be true.

@glee

The ‘incorrect expectation value calculations’ arise from not realising that the two values are x and 2x; but people think there are somehow three values; x, 2x and x/2!

I guess it comes down to, are the envelopes provided in pairs, having been sorted into sets of two that satisfy the written conditions before being distributed, or are they essentially sets (perhaps even open sets) of three or more, where only the first envelope is provided initially (and presumably not at random), and then the second envelope is provided on request, either at random (but from a prior pool of envelopes that will comply with the written conditions) or by a thinking agent?

Consider…

Scenario one:

1. Game master (G) provides Player (P) 2 envelopes that G knows will comply with the written statements, one containing X and the other 2X.
2. P reads the first envelope and decides to switch to the second, because all P knows is that the envelope contains a value Y. P does not know if Y = X, or if Y = 2X. But the second envelope could be either Y/2, or 2Y, so P reasons the EV of the second envelope is 1.25Y
3. Player then reads the second envelope.
4. At this point, P should conclude that there is no difference in EV as, with better information, P can conclude that one envelope (doesn’t know which) must contain Z (an unknown lowest value corresponding to the unknown X) and the other must contain 2Z. Each envelope has an equal probability of it. So the expected value of each is 1.5Z.

Scenario two:

1. G provides P with a single envelope that G knows contains 2X. G also has, at G’s disposal, some additional number of envelopes behind their back that G knows contains either X or 4X.
2. P reads the first envelope, and ponders their choice, perhaps assuming that it doesn’t make sense for G to have any more than two envelopes behind G’s back.
   (a) But of course, G could have any combination or number of envelopes behind their back. They could have a hundred envelopes, of which only one contains 4X and the other 99 contain X, or vice versa. Or, heck, they could have any number of envelopes containing X or 4X (from zero to an essentially infinite number), as long as G has at least one envelope containing X or 4X, G was careful to provide P an envelope containing 2X, and there isn’t a single envelope envelope containing a value other than either X or 4X.
3. But assuming—as I think is reasonable—that this variation of the game would be limited to three envelopes, X, 2X, and 4X, or at the very least to an equal number of X and 4X envelopes and with G always being careful to hand P a 2X envelope to start, the EV of the second envelope is… 2.5X (or 1.25Y if you want to go that way, but you don’t really need to, because in this game P can deduce they have been handed the middle-value envelope to start with).

And I’m sure there are more scenarios you could come up with, depending on what is assumed about G and the conditions under which they provide envelopes.