There's nothing wrong with series, though if one resistor fails open everything stops. (Kind of the point I know but inconvenient.) If in parallel one fails short the total impedance is still the average of the two, right? So twice as much current flows rather than thousands of times as much. Since the "safe" current range presently extends more than two orders of magnitude from the minimum useful value a factor of two increase isn't much.

Total parallel resistance is 1/( (1/R

_{1}) + (1/R

_{2}) + ... +(1/R

_{n}) ), not the average. With two 9k4 resistors(for 4k7 total), one shorting to .001Ω(I'm not sure if that's an accurate estimate, but I'll use it anyway) would result in a total of 0.0009999998936Ω. That'll be a [sarcasm]little[/sarcasm] more than twice the current