> if I have one 3 OHM and one 1 OHM, both of which are 100W, and I wire them in series, I end up with 4 OHMS at 200w ?
No. 4 ohms 133 watts.
But take your assumption and work backwards. 200 Watts in 4 ohms is, ummm, screw square-root formulas. 24 volts in 4 ohms is 6 Amps. 24 Volts times 6 Amps is 144 Watts, too low. 32 Volts in 4 ohms is 8 Amps, 32V*8A is 256 Watts, too high. 28V in 4 ohms is 7 Amps, 28V*7A= 196 Watts, close enough.
When you put resistors in series, the voltage divides by the resistor ratio. If you put 4V across 3+1 ohms, you find 3V across the 3 ohm and 1V across the 1 ohm.
So if we put 28V across 3+1 ohms we will get 21V across the 3 ohms and 7V across the 1 ohm. (We could instead have noted that 7 Amps times 3 Ohms is 21 Volts.)
And 21V in 3 ohms or 7 Amps gives 21V*7A= 147 Watts in the 3 ohm resistor. Which is significantly in excess of the 100W rating. The fact that the 1 ohm is loafing at 47 Watts, half of its rating, will not save the 3 Ohm working way over its rating.
And when the 3 Ohm is working AT 100 Watts, the 1 Ohm will take 1/3rd of that power, 33 Watts. The maximum safe value for the combination is 133.33 Watts.
There are very few tube amps which can cook a 133 Watt resistor.
Most tube amps also have a 8 ohm tap. Your 3+3+1+1 or 8 Ohm array can take 266.7 Watts. If you have a tube amp which'll cook a 250 Watt resistor, post pictures!
If you are doing solid-state: it is nice to find 16 ohm resistors in large Watts and at low prices. You parallel two for 8 and four for 16. A typical transistor amp may be rated 150W in 8, 100W in 16, 250 Watts in 4. If it had zero power supply sag, it would be 100W in 16, 200W in 8, 400W in 4; but sag happens. If you can find 16 ohms which can eat the 16-ohm power of the amplifier, two or four in parallel will be able to eat the 8 and 4 ohm powers.