user3377: Maybe this is a question for Electronics SE but I’m not a member there yet so hopefully someone will have an answer here.
I did an experiment with having some push-to-make buttons connected to my Uno. I copied the circuit from the aduino.cc website and all worked exactly as expected. Here is my schematic of the circuit as I have it now:

The pins are both held LOW by the Pulldown resistor until the switch is closed, when they go HIGH. 
All well and good. But I’ve just been looking at the circuit and something puzzles me. I can see a direct connection from Switch 1 to Pin 2 VIA the resistors and the Ground connection – and vice versa, S2 to P1. See my pathway in Red here:

Now, to be clear, this is NOT happening – but I just want to understand why not. Is it just that the 2 resistors in combination provide enough resistance to stop the opposite pin from going HIGH? In which case, this Pulldown resistor is actually serving 2 completely different functions in the circuit…?

user313: It helped me to draw an equivalent schematic to yours when S1 is closed and S2 is open. P2 is neither high nor low by itself; it is a high-impedance (which is why you need a pulldown or pullup resistor). It is pulled to ground by R2 and the fact that the junction of the two resistors is grounded. If you disconnected that junction from ground, P2 would see a 20K pullup, but as it is, the (left) end of R2 stays at ground, regardless of what you apply to the P1-R1 junction.

simulate this circuit – Schematic created using CircuitLab

user4143: An input pin is in what is known as high impedance mode. That is, to all intents and purposes, to the external circuit it looks like a massive resistor to ground.
This resistance can be in the millions of ohms.
So let's redraw your circuit with those input pins as big resistors:

simulate this circuit – Schematic created using CircuitLab
Now you can see how the schematic really looks (see how it's set up so +5V is at the top and GND is at the bottom - the [conventional] current flows then from top to bottom).
So with SW2 pushed (as shown) the current wants to flow from 5V (top) to GND (bottom). That goes through SW2, through R2, and down to ground. Some tiny amount also flows through R4, which represents the input pin.  Since by then the current is already at ground and has reached its destination there is no reason at all for it to then flow back up through R1, and back down to ground again through R3. For current to flow through R1 the voltage at IN1 would have to be negative since current will only flow from a higher to a lower voltage. And since it isn't negative, no current can flow.

user3917: Call the resistors R1 and R2, the negative terminal Ground, and the positive terminal V+.  The junction of R1 and R2 is connected to Ground.  This holds the resistor ends at the junction to zero volts relative to ground.  Ignoring P2 for the moment, the other end of R2 is open circuit to V+ when S2 is open.  Applying Ohm's law, we have a finite resistance in series with an infinite resistance, so zero volts across R2 when S2 is open.  Thus one sees that voltage across R2 is not affected by voltage changes across R1.
P2 as an input is not completely an open circuit; typically its effect is similar to a multi-gigaohms resistor to some voltage between Ground and V+.
Edit: 
More directly, observe that the voltage at the Ground ends of R1 and R2 is zero and remains at zero no matter what S1 and S2 do.  Thus, operation of S1 does not affect voltage at P2, and operation of S2 does not affect voltage at P1.

3377: <p>Maybe this is a question for Electronics SE but I’m not a member there yet so hopefully someone will have an answer here.</p>

<p>I did an experiment with having some push-to-make buttons connected to my Uno. I copied the circuit from the aduino.cc website and all worked exactly as expected. Here is my schematic of the circuit as I have it now:
<a href="https://i.stack.imgur.com/0q8Qp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0q8Qp.png" alt="Circuit"></a>
The pins are both held LOW by the Pulldown resistor until the switch is closed, when they go HIGH. 
All well and good. But I’ve just been looking at the circuit and something puzzles me. I can see a direct connection from Switch 1 to Pin 2 VIA the resistors and the Ground connection – and vice versa, S2 to P1. See my pathway in Red here:</p>

<p><a href="https://i.stack.imgur.com/GAHyx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GAHyx.jpg" alt="Alt Path"></a>
Now, to be clear, this is NOT happening – but I just want to understand why not. Is it just that the 2 resistors in combination provide enough resistance to stop the opposite pin from going HIGH? In which case, this Pulldown resistor is actually serving 2 completely different functions in the circuit…?</p>
 313: <p>It helped me to draw an equivalent schematic to yours when S1 is closed and S2 is open. P2 is neither high nor low by itself; it is a high-impedance (which is why you need a pulldown or pullup resistor). It is pulled to ground by R2 and the fact that the junction of the two resistors is grounded. If you disconnected that junction from ground, P2 would see a 20K pullup, but as it is, the (left) end of R2 stays at ground, regardless of what you apply to the P1-R1 junction.</p>



<p><img src="https://i.stack.imgur.com/BWuqg.png" alt="schematic"></p>



<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fBWuqg.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
 4143: <p>An input pin is in what is known as <em>high impedance</em> mode. That is, to all intents and purposes, to the external circuit it looks like a massive resistor to ground.</p>

<p>This resistance can be in the millions of ohms.</p>

<p>So let's redraw your circuit with those input pins as big resistors:</p>



<p><img src="https://i.stack.imgur.com/65fu6.png" alt="schematic"></p>



<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f65fu6.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>

<p>Now you can see how the schematic really looks (see how it's set up so +5V is at the top and GND is at the bottom - the [conventional] current flows then from top to bottom).</p>

<p>So with SW2 pushed (as shown) the current wants to flow from 5V (top) to GND (bottom). That goes through SW2, through R2, and down to ground. Some tiny amount also flows through R4, which represents the input pin.  Since by then the current is already at ground and has reached its destination there is no reason at all for it to then flow back up through R1, and back down to ground again through R3. For current to flow through R1 the voltage at IN1 would have to be negative since current will only flow from a higher to a lower voltage. And since it isn't negative, no current can flow.</p>
 3917: <p>Call the resistors R1 and R2, the negative terminal Ground, and the positive terminal V+.  The junction of R1 and R2 is connected to Ground.  This holds the resistor ends at the junction to zero volts relative to ground.  Ignoring P2 for the moment, the other end of R2 is open circuit to V+ when S2 is open.  Applying Ohm's law, we have a finite resistance in series with an infinite resistance, so zero volts across R2 when S2 is open.  Thus one sees that voltage across R2 is not affected by voltage changes across R1.</p>

<p>P2 as an input is not completely an open circuit; typically its effect is similar to a multi-gigaohms resistor to some voltage between Ground and V+.</p>

<p><em>Edit:</em> 
More directly, observe that the voltage at the Ground ends of R1 and R2 is zero and remains at zero no matter what S1 and S2 do.  Thus, operation of S1 does not affect voltage at P2, and operation of S2 does not affect voltage at P1.</p>
