Oh, I understand now. Even if we don't know how it's distributed, if it's the top among 9 choices with the same variance that puts it in the 80th percentile for specialness, and signal and noise contribute to that equally. So it's likely to be in the 80th percentile of noise.

It might have been clearer if you'd instead made the boxes actually contain coins normally distributed about 40 with variance 15 and B=30, and made an alternative of 50/1, since you'd have been holding yourself to more proper unbiased generation of the numbers and still, in all likelihood, come up with a highest-labeled box that contained less than the sure thing. You have to basically divide your distance from the norm by the ratio of specialness you expect to get from signal and noise. The "all 45" thing just makes it feel like a trick.