[irchans]

Picking any number Y before opening the first envelope and switching whenever the envelope contains less than Y has a higher expectation than random as long as there is any chance that the stuffer can put $x and $z into envelopes where x < Y < z. This is the best "solution" to this problem that I have seen in the past.

Here is an interesting alternate solution. Open the first envelope. Say it contains d$. Generate a random number x between 0 and 1. If x < Exp(-d), then switch. This alternate strategy is guaranteed to have a higher expectation than the random strategy for any stuffer probability distribution.