I am going to take a shot at this one. Please feel free to correct any errors as you see fit.

First I solved to find the probability that H, E, or S would all occur at least once in the same 36-character line.

The probability of a specific letter occurring in an independent trial is 0.0384615384615385, therefore the probability of a specific letter not occurring in an independent trial is

1 - 0.0384615384615385 = 0.961538461538462

Therefore the probability of a specific letter not occurring in 36 independent trials is

0.961538461538462^36 = 0.243668721853164

This means that the chance a specific letter will occur at least once in 36 independent trials is

1 - 0.243668721853164 = 0.756331278146836

The chance of all 3 letters occurring in the same 36-character line is

0.756331278146836^3 = 0.432649477099284

Next we need to solve for the probability of either W or U occurring at least once in a 36-character line.

The probability that one of these two letters will occur in an independent trial is

2 * 0.0384615384615385 = 0.0769230769230769

The probability that neither of these two letters occurs in an independent trial is

1 - 0.0769230769230769 = 0.923076923076923

The probability that neither of these letters will occur at least once in 36 independent trials is

0.923076923076923^36 = 0.0560485232822473

Therefore the probability that one of these two letters will occur at least once in 36 independent trials is

1 - 0.0560485232822473 = 0.943951476717753

Therefore the probability that H, E, S and either U or W all occur at least once in a 36-character line is

0.432649477099284 * 0.943951476717753 = 0.408400112809032

Now we need to solve to find out the probability that each 36-character line in a 14 line Sonnet will contain H, E, S and either U or W. This is equal to the individual line probability of success to the power of trials or

0.408400112809032^14 = .00000359088872433288

Therefore the probability that at least one of the 14 lines of 36-characters will not contain an occurrence of H, E, S and either W or U at least once is

1 - .00000359088872433288 = 0.999996409111276

This means that in 154 trials the probability that none of these trials will contain 14 lines that all include the letters H, E, S and either U or W at least once is

0.999996409111276^154 = 0.999447155018702

This means that the probability that at least one of the 154 Sonnets has the letters H, E, S and either U or W occurring at least once in each of the 14 36-character lines is

1 - 0.999447155018702 = 0.000552844981298

*Note I did most of this in Excel and there may be some rounding errors