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A simple model for the beginning of the universe: $n$ particles are created. Every particle can be either matter or antimatter, with probability $\frac{1}{2}$. After the particles are created, matter and antimatter "cancel each other out" until particles of only one type remains, in a 1:1 ratio. The question: For a given amount of particles $k$, what is the probability that this amount will remain?

If X is the number of matter particles, we have that $K=|2X-n|$, so call $Y=2X-n$. Since X is binomially distributed, $E[X]=n/2$ and $V[X]=n/4$ and so $E[Y]=0$ (as expected...) and $V[Y]=2^2V[X]=n$. Using Chebyshev we get $P(K>k)\le\frac{n}{k^2}$.

This seems to imply that the amount of particles in the beginning of the universe (in our simple model) should have been at least quadratic of the amount of particles existing today. However, I feel this result is weak and can still be improved, maybe using a variant of Chernoff I'm not thinking about? My "gut feeling" is that as $n$ grows larger, the system behaves in a less chaotic way and the probability of "almost zero" values of K increases, maybe even to 1. Am I correct in my guess, and if so - how to prove it?

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Yes, you're right that $$\lim_{n\to\infty} P(M=k+\frac{n}{2} | M~Binomial(n,\frac{1}{2})) = 0$$ for any specific k. But as long as $k+\frac{n}{2}$ is an integer in the range 0 to n, then k is non-zero for any finite n. Yes, for observed k, your best estimate of N is going to be proportional to {k^2}

Note that although, using your notation, E[Y]=0, that does not mean that E[k]=0. Indeed, E[k] is not zero. You need to be careful about what you're denoting as "matter" rather than anti-matter. If matter is just defined to be whatever type of particle is left over after all the cancelling out, so that k is your number of particles of matter, then E[K] is non-zero. I think $E[k] = 0.4\sqrt{n}$

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The number of matter particles is distributed $M = \mathrm{Bin}(n,1/2)$. If $M \geq n-M$ then we will end up with $M-(n-M) = 2M-n$ matter particles. Similarly, if $M \leq n-M$ then we will end up with $(n-M)-M = n-2M$ antimatter particles. So the quantity you're interested in is $|2M-n| = 2|M-n/2|$.

If $n$ is large, then the number of matter particles approaches a normal distribution, by the central limit theorem. In particular, the probability that $|2M-n| \leq C\sqrt{n}$ approaches some constant which depends only on $C$, and can be readily calculated by considering the mean and variance of $M$.

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