Question: Consider two computers, one a bit-for-bit clone of the other. If I then perform any two arbitrary actions on machine A- like change the volume, then open a chrome browser, then do the same on computer B, but in reverse order, will the computers' state still be bit-for-bit identical (ram, disc, etc.)? The actions can of course also be more significant ones, like updating an application or driver.

Diving a bit deeper, I'm trying to figure out if the two machines...

* will *always* be in different states
* will *almost always* be in different states, but there are special cases that are identical
* can plausibly be either, but its impossible to tell without knowing the two exact actions

If this is a bad question - Downvotes without feedback don't help me figure out what is wrong with my question. Feedback is greatly appreciated.

Answer: 
Two computers doing absolutely the same thing as each other have absolutely no guarantee of being bit identical, let alone doing the same thing in different orders. In fact there is almost no way *for* two systems to be bit identical.

I say this because there are various encryption algorithms used at various points which all rely on cryptographically secure random number generation. What that means is that the algorithms are intended to be absolutely random and that two machines generating the random numbers should never hit the same sequence of numbers. Even given two perfectly identical machines if they both connect to an SSL website and use the same public key to begin a session-key exchange with a website, that session key absolutely will not be the same.

That's just one "minor" point where there is a big difference between otherwise "identical" machines.

On top of that modern systems employ [Address space layout randomization](https://en.wikipedia.org/wiki/Address_space_layout_randomization), which means that even running the exact same code two machines will have a different memory layout and therefore never be "bit identical".
These actions are not arbitrary so my answer is invalid.

Since not all math operations are commutative, it is easy to show the answer to this is NO. If I assume:
a = 1; b = 2, c = 3
The have one do

`c = a + b then c = c / b` (c = 1.5)

and the other do

`c = c / b then c = a + b` (c = 3)

You can see this clearly