If the game is iterated, so that you can accumulate data about the sample space and its probability distribution, then it's an interesting but completely different problem. We can still talk about strategies in the non-iterative case.
The three choices are:
- Never Switch,
- Always Switch, and
- Sometimes Switch.
There is some
evidence that Sometimes Switch increases your expected gain. (I've played around a tiny bit with this, and the results were all over the place but almost always positive. I don't really know how to simulate this.) That's interesting but not quite the "puzzle" here.
If you were trying to find a strategy that is nearly indistinguishable from Never Switch over a large number of trials (not iterated, just repeated), then it would be Always Switch. If, for instance, you fix the value of X, so that there is a single pair of envelopes used over and over again, then we're just talking about coin flips. A Never Switch strategy might result in HTHHTHTT... while Always Switch would be THTTHTHH... and it couldn't be more obvious they'll end up equivalent.
So then the
puzzle is what to do about the Always Switch argument, which appears to show that given any value for an envelope you can expect the other envelope to be worth 1/4 more, so over a large number of trials you should realize a gain by always switching. This is patently false, so the puzzle is to figure out what's wrong with the argument.
The closest thing I have to answer is this:
If you define the values of the envelopes as
X and
2X for some unknown
X, you're fine.
If you want instead to use variables for the value of the envelope you selected,
Y, and the value of the one you didn't,
U, you can do that. But if you want to
eliminate one of them, so you can calculate an expectation only in terms of
Y or
U, you
have to know which one is larger. (Note that there is no ambiguity with
X and
2X.) Which one is larger you do not and cannot know.
