Works out to be a bit more average damage than axes; between 1/12 and 1/4 of a point, depending on die size. (Huh: interestingly, it's 1/4 of a point more damage with a d4, 1/6th of a point more with a d6, etc.)
Actually, if I've done my math right, it should actually be 1/3 of a point for a d4 and 1/5 of a point for a d6. In general, for a d

, the average value of vorpal would be 1/(n - 1) more than the average value of brutal 1.
Math stuff below:
[SBLOCK]Let the average value of a Brutal 1 d

be B

.
B

= (2 + 3 + ... + n)/(n - 1)
= [1 + 1 + 1 + 2 + ... + 1 + (n - 1)]/(n - 1)
The expression in the numerator is the sum from 1 to (n - 1), plus (n - 1), so:
B

= [1 + 2 + ... + (n - 1) + (n - 1)]/(n - 1)
Hence, using the standard expression for the sum from 1 to N, N x (N + 1)/2, and substituting (n -1) for N, we get:
B

= [(n - 1) x n/2 + (n - 1)]/(n - 1)
= (n - 1) x (n/2 + 1)/(n - 1)
= (n + 2)/2
Let the average value of a Vorpal d

be V

.
When the dice roll is a maximum, you add n and roll the die again, so the average value of a maximum roll is n + V

.
V

= [1 + 2 + ... + n + V

]/n
Using the same standard expression for the sum from 1 to N, we get:
V

= [n x (n + 1)/2 + V

]/n
n x V

= n x (n + 1)/2 + V

n x V

- V

= n x (n + 1)/2
(n - 1) x V

= n x (n + 1)/2
V

= n x (n + 1)/[2 x (n - 1)]
Subtracting,
V

- B

= n x (n + 1)/[2 x (n - 1)] - (n + 2)/2
= n x (n + 1)/[2 x (n - 1)] - (n + 2) x (n - 1)/[2 x (n - 1)]
= [n x (n + 1) - (n +2) x (n - 1)]/[2 x (n - 1)]
= [n^2 + n - (n^2 + 2n - n - 2)]/[2 x (n - 1)]
= (n^2 + n - n^2 - 2n + n + 2)/[2 x (n - 1)]
= 2/[2 x (n - 1)]
= 1/(n - 1)[/SBLOCK]