[yor_on (OP)]

"For the box of gas, there aren't that many ways we could re-arrange the molecules to keep them in one corner, but there are many ways we could re-arrange them smoothly throughout the box. It is therefore very natural to evolve from a low-entropy state to a high-entropy state, simply because there are so many more high-entropy states to evolve to."

My question is if we consider a high entropy state at its final stage, is there really that many different states it could arrange itself in? Isn't it more like a 'bell shaped curve' where the probabilities will peak out somewhere in the middle of it time wise, if one could look at the process with the 'eye of a God'? Normal distribution

[Atomic-S]

Let us consider a box divided up into many small cells. We will toss objects in at random, the number of cells however being substantially larger than the number of objects. To simulate the event that all the objects will end up in one small portion of the box, we will select a certain few cells in advance to represent this portion. This would be like molecules spontaneously collecting in one small corner. This event corresponds also to the event that all cells in the rest of the box will remain empty. We wish to determine the probability that all the unselected boxes will remain empty. Given the way this problem is set up, we may regard the event that a cell does or does not receive a molecule, as equivalent to the event that, for that cell, a die that has n sides only 1 of which is colored, would when tossed end up with the colored side up (n being the ratio of the number of cells to the number of molecules, and in this problem n is considerably larger than 1). So what is the probability that, if all the, say, 100 sided dice that correspond to the unselected cells are randomly tossed, that none of them will turn with the colored side up? Well, it would be (99/100)^[however many unselected cells there are], which will be very close to zero.

[Geezer]

Let us consider how to format our posts so that others might understand them.