That's a lot less fun. Here goes:

As the picker:
- If I see an odd number, I switch (it's got to be the low).
- If I see a number over $333, I stay (it's got to be the high).

- If I see a number that's not a multiple of 4, it's probably the low, so I switch. (If it is the high, that means the stuffer put an odd number of bills in the other envelope, which is an odd strategy.)
- If I see a number that's over $166, I stay. (If it is the low, that means the stuffer put over $332 in the high envelope, which is an odd strategy.)

- If I see a multiple of $4 between $4 and $164, it's trickier. I pick X. The envelopes contain either (X and 2X) or (X/2 and X).
-- For (X and 2X) to be a good stuffing choice, 2X should be $164 or less (and thus 'ambiguous' whether it's the high or low value)...otherwise, if I had picked 2X instead of X, I'd be pretty sure it was the high value and stay. So if 2X is over $164, then X is probably not the low value.
-- For (X/2 and X) to be a good stuffing choice, X/2 should be a multiple of $4...otherwise, if I had picked X/2 instead of X, I'd be pretty sure it was the low value and switch. So if X is not a multiple of 8, it's probably not the high value.
-- So, if X is over $82, I stay. If X is not a multiple of $8, I swap.

- What if X is a multiple of $8 between $8 and $80? Here we go:
-- Again, the stuffing choices are (X and 2X) and (X/2 and X).
-- If it's (X and 2X), the stuffer would have to be worried about me choosing (2X) and thinking it might be either (X and 2X) or (2X and 4X). Thus 4X should be an ambiguous choice as well...so 4X must be under $164. So if X is a multiple of 8 over $41, X is probably not the low value, and I'll stay.
-- Likewise, if it's (X/2 and X), the stuffer would have to worried about me choosing (X/2) and considering the pair (X/4 and X/2). So X/4 should be a multiple of 4 as well. Thus, if X is not a multiple of 16, it's probably the low value, and I'll switch.
-- So, if X is over $41, I'll switch. If X is not a multiple of 16, I'll stay.

- So, what do I do with 16 and 32 (the multiples of 16 under $41)? It depends on my read of the stuffer:
-- If the stuffer (a) did not think out the problem or (b) thought out the problem only to the point of figuring out the (16, 32) is the most 'ambiguous' pairing, then I would stay on 32, swap on 16.
-- If the stuffer (a) figured out that (16, 32) is the most 'ambiguous' pair and (b) realizes that I, the picker, also would figure that out, then I would probably stay on 16, swap on 32...since he's probably counting on me to figure him for (16, 32) and make the appropriate move based on that.

- On the other hand, you could think about it this way: since I don't care about what happens to the picker, I just have to choose some value Y so that for all X > Y I'm equally happy and all X < Y I'm equally sad.
-- If X > Y, then I stay -- no regrets if it's high or low.
-- If X < Y but 2X > Y, then I swap -- I'm happy if I get 2X and would have been equally unhappy with either X or X/2.
-- If 2X < Y, it doesn't matter if I swap or not -- I'm not happy no matter what I do.
-- And, of course...this is FREE MONEY! Who cares if it's the most free money you could have gotten or not? It's FREE!

- The way NOT to look at this problem is this: I have X. If I swap, I'll either get 2X or X/2...so I'm risking X/2 to gain X with a 50% chance of winning. That's a 2:1 bet on even odds. Of course I'll swap! (Stupid paradoxes.)

As the stuffer:
- It depends on what I know of the picker.
-- Against a dunce I'd choose (16, 32).
-- Against anyone else, I'd probably go with (8, 16). If I get lucky and he chooses 16, he'll probably swap, since he reasoned that (16, 32) is a likely choice for me. If I get unlucky and he gets 8, well, maybe he'll get confused and outsmart himself by staying on 8. (I think this is better than choosing a combo like (19, 38), where you definitely lose on (19) and only probably win on (38).)

That's my mostly-muddled thinking on the problem.

Of course, depending on the exact interpretation of the rules, the stuffer's best strategy for X and 2X might be X = 0.
- 0 is an interger, after all...so both envelopes would be empty.
- If the rule is that you get paid off if the pick ends up with the least amount of money, then this is an automatic win.
- Even if not, if you really hate the picker, you can screw him over this way.

PP