[learnstats2]

This is a really interesting thought. Assuming you can only add two atoms at random, there are four options:

odd + odd = even (odd reduced by 2, evens increased by 1)

odd + even = odd (evens reduced by 1)

even + odd = odd (evens reduced by 1)

even + even = even (evens reduced by 1!)

When the evens outnumber the odds, odd+odd will be rare, so the percentage of evens will tend to decrease until it reaches 50%.

No matter many of each type you start with, the equilibrium position is 50-50% odd and even.

However: there may be an exception to this. If all the odds are in one place, then odd-odd may be more likely than random. If you have all evens in one place, you can't get back to having odds again - you can't (ever) increase the number of odds that already exist to balance things out, you can only decrease the evens. 0-100% odd and even could also be an equilibrium position.