OK, there is no easy way. If you need to figure these things out, you need to know how to do it.
You start with the voltage that the decoder puts out for the lights. 12vdc(?), then the voltage of the light bulbs that you want to use, 1.5vdc in your case. Now you have to figure out how much voltage to drop across the resistor, so 12vdc minus 1.5vdc = 10.5vdc.
Now you have to know the current that your light bulb will draw when lit by 1.5vdc. This should be provided by the maker of the bulb and noted on the package, in most cases. *** If you can't get that information, everything from here on out is all guesswork. *** So let's assume for this lesson that the current is 10ma. (10x10 to the -3) Thats 0.01 amps.
Now we have to use a formula. R=E/I That is the Resistance needed equals the Voltage (E) divided by the Current (I) of the bulb.
So 10.5 (The voltage you want the resistor to drop) divided by 0.01 (The current of the bulb) equals 1,050 Ohms of resistance.
Because 1,050 is not a common value, a 1Kohm resistor is close, BUT, it is too small by 50 ohms and you could burn out the bulb, so you should go higher in value, which could be 1.1Kohm which is 1,100 ohms. Depending on where you get the resistor, you may have to go a little higher.
Now comes the power rating of the resistor, or the Wattage. Again a formula. P=IxE. That's Power (in watts) = the Current of the bulb (I) times the Voltage (E) the resistor is dropping. So, P= 0.01 x 10.5 which is 0.105watts. So a 1/8 watt resistor would work, but it will get warm because it is close to the 1/10 watt the resistor will dissipate. So to be safe, a 1/4 watt, or even a 1/2 watt would be better.
You will need one resistor per bulb, wired to the white / yellow wires. Since the blue wire is common to each bulb, don't put the resistors on the blue wire. Resistors have no polarity to worry about.
If you wanted to figure one resistor for two bulbs, you add the current of each bulb together (assuming the bulbs are wired in parallel) and use it in place of 0.01 or (I), and recalculate. You will also have to recalculate the Power or Watts, because you are using more current through the resistor. You CAN do this!!
I know this is technical, and you might be able to find a chart somewhere that will tell you the same thing, but if you can get you head around this, you will learn more.
And this works with LED's too.
Elmer.
The above is my opinion, from an active and experienced Model Railroader in N scale and HO since 1961.
(Modeling Freelance, Eastern US, HO scale, in 1962, with NCE DCC for locomotive control and a stand alone LocoNet for block detection and signals.) http://waynes-trains.com/ at home, and N scale at the Club.