Assume the envelope contents are distributed arbitrarily on [A,+infinity) where A is some large number. Let f(x)=1/x (the values for x<1 don't matter). Then the expected benefit of Thrun's algorithm is always 1/4, even though the difference in f(x) between the values of any two envelopes is less than 1/A. To convince yourself of that, work out the proof yourself or run a computational experiment:

from random import random
n = 100000
thrun = 0
chance = 0
for i in range(n):
smaller = 2+3*random() # replace this line with anything
selected,other = smaller,2*smaller
if random() > 0.5:
selected,other = other,selected
chance += (selected if (random() > 0.5) else other)
thrun += (selected if (random() > 1/selected) else other)
print (thrun-chance)/n