Originally Posted by

**jjmclell** That Bernoulli distribution was derived as follows:

You start at 0 on a number line, and for each step you can go one interval to the left or to the right. To reach any point, $\displaystyle m$, in $\displaystyle n$ steps, you have to take $\displaystyle a$ steps to the right and $\displaystyle b$ steps to the left. With that said:

$\displaystyle a + b = n$

$\displaystyle a - b = m$

Through substitution:

$\displaystyle a = (n + m)/2$

$\displaystyle b = (n - m)/2$

The total number of possible paths by which to arrive at $\displaystyle m$ in $\displaystyle n$ steps is:

$\displaystyle {n! \over a!b!} = {n! \over ((n + m)/2)!((n - m)/2)!} $

If we want to get to any point, $\displaystyle m$, in $\displaystyle n$ moves, then we know how many moves we must take to the right, $\displaystyle a$, and how many to the left, $\displaystyle b$. If $\displaystyle a$ represents successes and $\displaystyle b$ represents failures then:

$\displaystyle p(m,n) = ({1 \over 2})^a ({1 \over 2})^b {n!\over a!b!}$

Which equals:

$\displaystyle ({1 \over 2})^n {n! \over ((n + m)/2)!((n - m)/2)!}$

To reiterate my question, how is it that the distribution above can go from being a Bernoulli distribution to a Gaussian distribution as $\displaystyle n$ approaches infinity?...assuming that the above distribution is in fact a Bernoulli distribution and I'm not completely wrong about everything I just said.