Am I drawing more current if I'm giving multiple paths from supply to ground?
\$ Ohm's Law: U = R \times I \$.


Print it big, frame it and hang it on your den's wall. It's the basis of lots of circuit calculations.
Here we're interested in the current, so we rearrange the equation:  


  \$ I = \dfrac{U}{R} \$


That means that a higher voltage across the resistor results in a higher current, while a higher resistor value will give a lower current. All resistances are \$10k\Omega\$ and the voltage across each resistor is also the same \$9V\$, so for each resistor 


  \$ I_R = \dfrac{9V}{10k\Omega} = 900\mu A\$


and the total current in your left schematic is \$4 \times 900\mu A = 3.6mA\$, or 4 times the current of your right schematic.  

This higher current will drain your battery four times as fast.



For extra credit:
Any number of parallel resistors can be replaced by a single one. We know the voltage across it, that's \$9V\$, and the total current, that's \$3.6mA\$, then, again according to Ohm's Law


  \$ R_{EQUIVALENT} = \dfrac{V}{I} = \dfrac{9V}{3.6mA} = 2500\Omega\$


This means that 4 identical resistors in parallel can be replaced by a single one \$1/4\$ their value, and more general \$N\$ identical resistors in parallel can be replaced by a single resistor \$1/N\$ their value.