Intuitively, and ignoring the constraint of the values being (A|2A) if I just assign a random number value to each envelope, does the order in which I have assi

Intuitively, and ignoring the constraint of the values being (A|2A) if I just assign a random number value to each envelope, does the order in which I have assigned the random numbers change the amount of information I have? I suspect not. However, I now have two random integers with no bits of information about either, where I have now removed information from the system (the relationship between the number of envelopes and the contents of one being twice the other), while leaving the states of the system intact. The envelope I in effect already have yields no bits of information because I didn't choose it, it is just the random (no information) one I start with, whereas the fact of the existence of the second envelope contains more information than my current envelope selection, and I have a decision to make about whether to switch. The null state of the system is me holding an envelope with no information, and addition of a second envelope is a new state of possibilities, where the information in this two envelope system is now non-zero. The decision with the most information is to switch. This additional information is independent of the implied desire for (A|2A), whereas the original problem is defined as the best possible decision is the one with the most information. Further, if I perform some operation with those two random integers (Za and Zb), the result has less "entropy" than the two random inputs in isolation, as by correlating them via a function I have produced some non-zero negentropy between them, say using xor, which is how I added information back into the system after removing it using the random numbers. I get this new number N composed of my two random numbers ( N = xor(Za, Zb) ), and I need to apply it to my choice of, not which envelope to select first (A|B) - but whether I should switch. The only way to get this number with information in it is to have assigned a random number to both envelopes and used the two bits of negative information to cancel each other - and therefore the only choice with non-zero information is the one which includes both envelopes - hence the decision to switch is the only one that yields the non-zero information outcome. Again, intuitively it seems related to how multiplying two negative integers yields a positive one, multiplying two entropy (negative information) sources yields negentropy (positive information). That could further imply that the information in the decision to switch is also based on whether the number of envelopes is even or odd, or maybe prime or composite? Sounding this out, it may also be related to graphs where the number of nodes/edges (where edges represent switching decisions and nodes envelopes) either forms a closed path or open one, and in the case of an even number of envelopes you get the max information by swtiching n-times, and in the case of odd ones, you switch n+1 times. Short version is, the choice with the highest number of bits of information in the two envelope case is to switch because it is predicated on there being more envelopes than the one you have started with.