RoR -- What % loss before double?

[BruceZ]

I corrected an error in my computation of the mean fractional loss. Since P(f) is actually the probability of losing f or more, we need the density function -P'(f) to compute the expected value of f. This has the effect of multiplying the computed mean by -ln(ror), which for a ror of 0.1% makes the mean fractional loss 14.4%, which is closer to the median value of 10%. The changes are marked in red.