I would say that the average number of attacks a FT can make is about three - that is, if the guy carrying survived incoming enemy fire.

The expected number of shots is 6.

The formula for expected number of events where an attempt depends upon the success of the previous attempt (like RoF or X#) is 1/(1-p) where p is the probability of a single successful event. 1/(1-p) is the sum of the infinite series 1 + p + p^2 + p^3 ...

With RoF you always have 1 guaranteed shot so is 1 + p + p^2 + ... = 1/(1-p)

RoF

[1]: 1/(1-1/6) = 6/5 = 1.2

[2]: 1/(1-1/3) = 6/4 = 1.5

[3]: 1/(1-1/2) = 6/3 = 2

[4]: 1/(1-2/3) = 6/2 = 3

[5]: 1/(1-5/6) = 6/1 = 6

[6]: 1/(1-1) = 1/0 = ±∞

OK, 1/0 is technically undefined but in practice is ±infinity depending whether you approach from the positive or negative side of 0.

x#: as per RoF.

x12: 1/(1-35/36) = 36/1 = 36

x11: 1/(1-33/36) = 36/3 = 12

x10: 1/(1-30/36) = 36/6 = 6

x9: 1/(1-26/36) = 36/10 = 3.6

x8: 1/(1-21/36) = 36/15 = 2.4

x7: 1/(1-15/36) = 36/21 = 1.714

That's regardless of opportunities for the use of a FT, sometimes the FT can't get there for various reasons but if you had unlimited opportunities ...

So 1/(1-p) is a very simple formula. Indeed RoF and x# are very similar except that RoF is 1D6 based and x# is 2D6 based and RoF is a success number while x# is fail based. If both were 1D6 or both were 2D6 based a RoF of [N] would be the equivalent of a x(N+1). An alternative way to look at it is 1/q where q is the probability of failure in a single event, as demonstrated by the 36/ (1, 3, 6, 10, 15, 21) progression in the x# table calculations.

EDIT: I messed up the difference of 1 between RoF (guaranteed 1st shot) and x# (with no guaranteed 1st shot). Fixed.

EDIT: I had it right 1st time, wrong 2nd time as Philippe pointed out.