The >50% strategy is actually for a different problem, but I'll share it anyways because it's pretty cool.

In this problem, there are two envelopes. One has more money than the other, but you don't know what the amounts are. You're given an envelope and can see how much money is inside. After doing so, you can either keep this envelope or switch with the other one. You can only do this once.

Given that you don't know the values, it's strange that you can ensure a >50% chance you pick the better envelope. The trick is that you choose whether to switch randomly, in a way that is more likely to switch when you have the smaller envelope.

For simplicity, let the envelope values be A and B, where A < B and both are positive integers. Use the following switching scheme: look at the amount of money in your envelope, let's say that's M. Flip a fair coin M times. If you get all heads, switch envelopes. Otherwise, keep envelopes.

The chance you get the envelope with B is 1/2 * (1 - 1/2^B) + 1/2 * (1/2^A) = 1/2 + 1/2^{A+1} - 1/2^{B+1}. Since A < B, this is very slightly larger than 1/2.

What we're basically doing here is getting a sample from the geometric distribution, the number of trials needed before reaching an ending condition. In this scheme, it's # of flips until first tails. If the sample (the # of flips) is >= M we switch, and otherwise we don't. To generalize this to real numbers, you can use a continuous probability distribution, such as the normal distribution. All you need is P(sample >= A) > P(sample >= B) for A < B, which is always true as long as your distribution has non-zero weight over the range [0, infinity)