Found the following in another thread researching this. The formulas shown and the data for a 1157 bulb when worked out show that a 6 ohm 30 watt resistor should be correct to duplicate the load of a single 1157 bulb. Your 6 ohm 50 watt actually should be ideal for the intended function theoretically.
The resistance of a light bulb will vary depending on whether it's illuminated or not. That said, if you know the voltage drop across the bulb (nominally, 12 v) and the current draw (which I don't seem to have in front of me) you can calculate the resistance of a fully illuminated bulb using Ohm's law--
R=E/I where R is resistance, E is voltage, and I is current, in amps.
So if the bulb draws 1 amp (which would be the case for a 12 watt bulb), the resistance is
R = 12 / 1 = 12 ohms. Anything close would work.
In your case, you're using two bulbs in series, so the drop is 6 volts across each bulb. Assuming the current drain is the same, you calculate resistance the same way; it'd be 6 ohms.
Know also that you'll need a resistor with sufficient wattage to dissipate the heat. Wattage is
P=I^2 * R (that's I squared times R)
In this example, W = (1* 1) * 12 = 12 watts, so a fifteen--watt resistor would work fine.
So, anybody know the wattage of an 1157? that's all that's missing here.