The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number.
Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and guess that it is the larger one if and only if it shows a positive number.
Why does this work? If both numbers are positive, your strategy is equivalent to "pick a random envelope" and you're left with a uniformly random choice of envelope. If both numbers are negative, you're also left with a uniformly random choice of envelope. But if one is positive and the other is negative, you win 100% of the time.
What are the odds of the third case? Well every Gaussian has at least some mass on positive numbers and some mass on negative numbers. So there is some nonzero chance, even if you don't know what it is. So while the strategy might not always do better than a coin toss, some positive percent of the time you win every time. And so you get a distinct advantage over random guessing.
The problem is now how would one transfer such an approach to the original problem without the Gaussian assumption. Somehow the two problems are less different than they initially seem.