The voltage drop across EACH led is a function of the voltage drop across BOTH leds. If one led has a lower forward voltage than the other it will draw more current. Whether the other led limits that current depends on the forward voltage of the other led. Both will initially be off, then as the voltage rises , one of them will conduct , drawing more current. As the voltage increases further , the other will conduct. The first led cannot draw the current it needs to turn on completely because it is limited by the second led, which for that brief moment acts as a current limiter. It is not until the total voltage drop across BOTH leds is sufficient to turn on both leds that they both light up. One cannot light without the other. When VTotal = VForward-1 + VFoward-2 , then BOTH leds can conduct. Think of it like a handgun with TWO safetys. You can take Safety-1 OFF, but it still won't fire until you take the second Safety Off. When the total voltage = Vf1+Vf2, they both light, but when they do , the current will rise exponentially in both leds if the voltage further increases. It is impossible to know which of the two leds will self-destruct first. It's like a game of chance. You increase the voltage until they both turn on. If you increased it gradually enough you can hold the voltage at a safe point once they are both lit. If you continue to increase the voltage , one of them will suddenly get very bright and burn out. If you measure the forward voltage on both of them as soon as they both turn on, and note which one has the lower forward voltage , then you can reasonably assume it will be the first to self destruct with an increase in voltage. What is the minimum voltage for turn on and what is the maximum without damage, can only be determined empirically.