Changing Wattage on a Heat Pad?
You were generally correct in your figurings BUT calculated for new power loss of 50 Watts total. The heating pad would get 1/3 of that or about 16 Watts and the resistor about 34 Watts. As Rseries is added the current decreases (so total power drops) AND the portion available to the eatong pad drops - so power is related to I^2.

Throughout "=" === "~=" :-)

Assume 150 W , 120V heating pad.
Power = V x I so I = Power/V
I = 150/120 ~= 1.25A = 1250 mA.

How big is existing R?
Power = V^2/R  so R = V^2/Power
So Rpad = 120^2/150 = 96 Ohms.  

For 50W in pad you want new I to be smaller by a facti\or of 3^0.5 as power is to current squared.

So new Ipad = 1250mA/3^0.5 = 1250/1.732  = 720 mA.  

As I = V/R you want total new R to be 1.732 x  as large as before so added series R is 0.732 X existing R
Rseries = 0.732 x 96 ~=  70 Ohms.

Sanity check.
I = V/R = 120/(96+70) = 720 mA
Power = I^2 x R = (0.72)^2 x 96 = 50 Watts in heating pad. 
Power in 70 Ohm resistor = 36 Watts.

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Diode + Resistor.
As Spehro say - adding a series diode is an easy way to halve the power to about 75 Watts. You can then if wanted add a smaller series R to lower the power further.
From above (75/50)^0.5 = 1.22
Extra series R = Rpad x 0.22
= 96 R *0.22 = 21 Ohms.  (eg 20, 22, ...)
Power in 20 Ohms = 20/96 x 50W = 10W.
Rate resistors at ~= 50% of max rated power for good lifetimes.
2 x 10 Ohm 10W resistors on series or 2 x ~= 40R 10W resistors on parallel would be "about right".