There are a few of you following this thread that seem genuinely interested but are completely missing the point of the original post. There is no trick here, no puzzle to be solved -- just a very easy to follow probabilistic analysis that leads to a surprisingly counter-intuitive result.

I seem to have led TheSnail astray by writing "Say I've randomly generated the number five and accordingly presented to you two checks, face down, labeled A and B, worth $4 and $8, respectively." That was intended only as an illustrative case. Please forget about it and take a fresh look at what I'm saying here.

You see, if you focus on the one case [A = 4, B = 8] you are considering only one possible event from the following ten-member "event-space":
[A = 1, B = 2] [A = 2, B = 4] [A = 4, B = 8] [A = 8, B = 16] [A = 16, B = 32]
[A = 2, B = 1] [A = 4, B = 2] [A = 8, B = 4] [A = 16, B = 8] [A = 32, B = 16]
if we can use this shorthand notation instead of talking about checks and dollars.

Now let's consider a sequence of "trials." In each trial I randomly choose to present to you for your choice one of these ten combinations (using a random number generator or a table of pseudo-random numbers) so that the probability of each combination is 1 in 10.

In any one trial you are presented with some combination [A = a, B = b] where, with equal probability, either a = 2b or a = b/2.

Then, over a succession of 100 trials, on average, each one of the ten possible combinations would occur ten times. Over those 100 trials, on average, A = 1 would occur 10 times; A = 32 would occur 10 times; and A = 2, A = 4, A =8, and A = 16 would each occur 20 times. (The last sentence remains true with B substituted for A.)

Clearly, in any given trial when both the values of A and B are unknown there is no advantage in choosing A or in choosing B.

However, if in a given trial you can learn the value of A, that new information changes the situation. For then
if A = 1, then B = 2 -- you should obviously choose B
if A = 2, then B = 1 or B = 4 with equal probability -- you should obviously choose B
if A = 4, then B = 2 or B = 8 with equal probability -- you should obviously choose B
if A = 8, then B = 4 or B = 16 with equal probability -- you should obviously choose B
if A = 16, then B = 8 or B = 32 with equal probability -- you should obviously choose B
if A = 32, then B = 16 -- you should obviously choose A

Note that if the value of A is revealed in a succession of trials, 90% of the time (whenever A ≠ 32) it's to your advantage to choose B.

Now consider a slightly larger example where the check values are bounded by 2¹⁰ = 1024 instead of 32; the same reasoning demonstrates that if the value of A is revealed in a succession of trials, 95% of the time (whenever A ≠ 1024) it's to your advantage to choose B.

In general, when the check values are bounded by 2^N the same reasoning demonstrates that if the value of A is revealed in a succession of trials, (100 - 50/N)% of the time (whenever A ≠ 2^N) it's to your advantage to choose B.

In the infinite case where there is no bound on the check values the same reasoning leads to the conclusion that it's always to your advantage to choose B, no matter what value of A is revealed.

Only in the infinite case is the result perplexing. Then one can ask: What's going on here? Why should the act of revealing the value of one check make it advisable to choose the other?