The following post is technical in nature. Skip it if you aren't into understanding electronics.
Lad, two corrections to your post.
First, a typical LED's maximum reverse voltage is about 5V, not 50V.
This data sheet for a typical LED illustrates this point. This is why we should provide a means to keep the reverse voltage low.
Second (and unimportant in this discussion), a conventional silicon diode typically drops closer to 1.0V at full load. The commonly mentioned .7V value is only appropriate when the diode is lightly loaded.
In addition (but seldom mentioned), the reverse voltage across an LED that is in series with a diode is not necessarily zero volts. The diode does not block all of the reverse voltage. Both the LED and the diode are reversed biased, and in this state they act as capacitors that are connected in series. The voltage divides INVERSELY according to the capacitances. If the 1N4007 has a larger capacitance than the LED, more reverse voltage will appear across the LED than the diode!
For the LED data sheet referenced above, the LED's capacitance at zero volts is typically 45 pF. The 1N4007 datasheet shows 15 pF for the diode at 4V reverse voltage. That would indicate that if we had 25V reverse voltage across the diode and LED (rectified 18V AC), the diode would have about 19V drop and the LED 6V.
If we change the diode to a 1N914 or 1N4148 (common small-signal diodes) with a capacitance of 4 pF, the LED will now have only 2V reverse voltage!
(The voltages in my examples were in error. Corrected 1/5.)
CONCLUSION: An LED is actually better protected by using a reverse diode in parallel with the LED rather than in series. The downside is that the power dissipated by the resistor doubles. The back-to-back paralleled LEDs provide this same protection.