It seems to me that if you make a basic bayes net with utilities at the end. The choice with the higher expected utility is to one box. Say:
P(1,000,000 in box b and 10,000 in box a|I one box) = 99%
P(box b is empty and 10,000 in box a|I two box) = 99%
hence
P(box b is empty and 10,000 in box a|I one box) = 1%
P(1,000,000 in box b and 10,000 in box a|I two box) = 1%
So
If I one box i should expect 99%1,000,000+1%0 = 990,000
If I two box i should expect 99%10,000+1%1,010,000 = 20,000
Expected utility(I one box)/Expected utility(I two box) = 49.5, so I should one box by a land slide. This is assuming that omega has a 99% rate of true positive, and of true negative; it's more dramatic if we assume that omega is perfect. If P(1,000,000 in box b and 10,000 in box a|I one box) = P(box b is empty and 10,000 in box a|I two box) = 100%, then Expected utility(I one box)/Expected utility(I two box) = 100. If omega is perfect, by my calculation we should expect one boxing to be a 100 times more profitable than two boxing.

This is the sort of math I usually use to decide. Is this none-standard, did I make a mistake, or does this method produce stupid results elsewhere?