I wanted to consider some truly silly solution. But since taking only box A is out (and I can’t find a good reason for choosing box A, other than a vague argument based in irrationality along the lines that I’d rather not know if omniscience exists…), so I came up with this instead. I won't apologize for all the math-economics, but it might get dense.

Omega has been correct 100 times before, right? Fully intending to take both boxes, I’ll go to each of the 100 other people. There’re 4 categories of people. Let’s assume they aren’t bound by psychology and they’re risk-neutral, but they are bound by their beliefs.

1. Two-boxers who defend their decision do so on ground of “no backwards causality” (uh, what’s the smart-people term for that?). They don’t believe in Omega’s omniscience. There’s Q1 of these.

2. Two-boxers who regret their decision also concede to Omega’s near-perfect omniscience. There’re Q2 of these.

3. One-boxers who're happy also concede to Omega’s near-perfect omniscience. There’re Q3 of these.

4. One-boxers who regret foregoing $1000. They don’t believe in Omega’s omniscience. There’re Q4 of these.

I’ll offer groups 2 and 3 (believers in that I’ll only get 1000) to split my 1000 between them, in proportion to their bet, if they’re right. If they believe in Omega’s perfect predictive powers, they think there’s a 0% chance of me winning. Therefore, it’s a good bet for them. Expected profit = 1000/weight-0*(all their money)>0

Groups 1 and 4 are trickier. They think Omega has a P chance of being wrong about me. I’ll ask them to bet X=1001000P/((1-P)weight)-eps, where weight is a positive number >1 that’s a function of how many people donated how much. Explicitly defining weight(Q1, Q4, various money caps) is a medium-difficulty exercise for a beginning calculus student. If you insist, I’ll model it, but it will take me more time than I’d already spent on this. So, for a person in one of these groups, expected profit = -X(1-P)+1001000P/weight = eps > 0!

So what do I have now? (Should I pray to Bayes that my intuition be confirmed?) There’re two possible outcomes of taking both boxes.

1. Both are full. I give the 1001000 to groups 1 and 4, and collect Q21000+Q31000000 from groups 2 and 3, which is more than 1001000 if Q3>0 AND Q2>0, or if Q3>1. This outcome has potential for tremendous profit. Call this number PIE >> 1001000.

2. Only A is full. I split my 1000 between groups 2 and 3, and collect X1Q1+X4Q4 from groups 1 and 4. What are X1 and X4 again? X, the amount of money group 1 and group 4 bet, is unique for each group. I called group 1’s X X1, group 4’s X4.

I need to find the conditions when X1Q1+X4Q4 > 1000. So suppose I undermaximized my profit, and completely ignored the poor group 1 (their 1000 won’t make much difference either way). Then X=X4 becomes much simpler, X=1001000P/((1-P)Q4)-eps, and then they payoff I get is -Q4eps+1001000P/(1-P). P = 0.001 and Q4eps < $2 guarantee X1Q1+X4Q4 > X4Q4 > 1000.

That’s all well and good, but if P is low (under 0.5), I’m getting less than 1001000. What can I do? Hedge again! I would actually go to people of groups 1 and 4 again, except it’s getting too confusing, so let’s introduce a “bank” that has the same mentality as the people of groups 1 and 4 (that there’s a chance P that Omega will be wrong about me). Remember PIE? The bank estimates my chances of getting PIE at P. Let’s say if I don’t get PIE, I get 1000 (which is the lowest possible profit for outcome 2; otherwise it’s not worth making that bet). I ask the following sum from the bank: PIEP+1000(1-P) – eps. The bank makes a profit of eps > 0. Since PIE is a large number, my profit at the end is approximately PIEP+1000(1-P) > 1001000.

Note that I’d been trying to find the LOWER bound on this gambit. Actually plugging in numbers for P and Q’s easily yielded profits in the 5 mil to 50 mil range.