I liked this but as other commenters have pointed out it's purely binomial at this stage. The great thing about the central limit theorem is that it is more gen

I liked this but as other commenters have pointed out it's purely binomial at this stage. The great thing about the central limit theorem is that it is more general than just the limiting Binomial case. So, there's this thing called the cumulant-generating function. It's pretty much defined for any random variable X. If you want to get technical it is the logarithm of the Fourier transform of a probability density function f(x). You're on HN so you probably know the first two jargon words, "logarithm" and "Fourier transform". A "probability density" just means that f(x) dx is the probability for X to be in the interval (x, x + dx). The Fourier transform puts us into a "frequency space" indexed by some variable k, so we can write the CGF as some function c(k), or in other words: c(k) = ln[ ∫ dx f(x) exp(i k x) ] So for the "step left/right" variable which is -1 with probability 1/2 and +1 with probability 1/2, c(k) = ln[cos(k)]. (It can get a little messy when you ask what happens when cos(k) crosses 0 etc, but this function is infinitely-often differentiable on a disc centered at 0 which is all that we need.) It also turns out that since the sum of all the probability ∫ dx f(x) = 1, you can just evaluate this for k=0 as c(0) = ln[ 1 ] = 0. The cumulants are derivatives evaluated at k=0: c'(0) = i E[X] = i μ (where μ is the "expectation") c''(0) = - (E[X²] - E[X]²) = -σ² (where σ² is the "variance") c'''(0) = -i E[(x - μ)³] = -i σ³ γ (where γ is the "skewness") It is not very hard to prove that if I have a sum of two random variables X + Y = Z, then their CGFs add like cz(k) = cx(k) + cy(k). This is the Fundamental Awesomeness of Cumulants: cumulants are pseudo-linear; they're linear in independent random variables. It is also not hard to prove that if I have a scalar multiplication U = X / n, then cu(k) = cx(k / n). Combining these together, the mean M = (X1 + X2 + ... + Xn) / n of n identical and independent random variables looks like: cm(k) = n c(k / n) Now if you know calculus, you know the rest. Taylor-expand around k = 0 to find: cm(k) = i μ k − k²/2 * (σ²/n) − i k³/6 (σ³ γ / n²) + ... We see a geometric attenuation in this series, we keep dividing new terms by n. If we drop terms from c'' on we get a constant, M = μ. That's boring, so we keep one more term, to find: cm(k) ~= i E[X] − (Var[X] / n) k² / 2 This is extra-good if f(x) is symmetric about the mean (so that the skewness vanishes), but it is also pretty good even if the distribution is skewed. We can now invert all of the steps to return back to a probability density: you exponentiate, so you get exp(i μ k) exp(- k² σ²/ 2 n), then you inverse-Fourier-transform, which transforms the exp(i μ x) into a frequency offset x → x − μ and which transforms the Gaussian into another Gaussian, so you get a Gaussian centered at x = μ. In other words you get approximately a normal distribution with mean E[X] and variance Var[X] / n, with the error given by convolutions of higher-order terms, and the first correction disappearing when the distribution is symmetric about μ (or otherwise non-skew).