Q:

Random range that iteratively multiplied by itself tends to 1

I was trying to emulate a random currency value, and first thing I thought was to start with value 1 and then iteratively add, say, Rand(-.1, .1), which is a function call that gives me a random number between -.1 and .1.  I then realized this could take my currency's value to less than zero, which would make no sense. 
So I figured every new value should be proportional to the old value, and tested multiplying the value times Rand(0, 2) iteratively.  My intuition behind it was that since is as likely for Rand(0, 2) to give a number < 1 than a number > 1, the value is as likely to increase as is to decrease.  But then for every test I ran the value dropped quickly to almost zero.  So I gave Rand(0, e) (e is the constant 2.718...) a try and it seemed to magically work (I also tried with Rand(0, 3) and it grew way above 1 every time).  
This was really just a wild guess, and my question is, was my guess right or is this just a coincidence, and if it was, why does it work?  

A:

Your guess is right :
$$ Y_n=\prod_{i=1}^n X_i$$ with $X_i$ independent random variable uniform on (a,b).  
Let $Z_n=\log(Y_n)$, then $$Z_n=\sum_{i=1}^n \log(X_i)$$
What is the distribution of $\log(X_i)$ ? 
$$\mathbf P(\log(X_i)<x)=\mathbf P(X_i<e^x)$$
So for any $x\in(\log(a),\log(b))$, $\mathbf P(\log(X_i)<x)=\frac{e^x-a}{b-a}$
Hence $\log(X_i)$ has a density function $f(x)=\frac{e^x}{b-a}$ if $x\in(log(a),log(b))$, else $f(x)=0$.
Hence $\mathbb E(\log(X_i))=\int_{\log(a)}^{\log(b)}\frac{x.e^x}{b-a}dx=\left[\frac{x.e^x}{b-a}\right]_{\log(a)}^{\log(b)}-\int_{\log(a)}^{\log(b)}\frac{e^x}{b-a}dx=\frac{b\log(b)-a\log(a)}{b-a}-1$
if $a=0$, $\mathbb E(\log(X_i))=\log(b)-1$. Note that variance of $\log(X_i)$ is defined and finite.  So we can use the central limit theorem to say that $Z_n$ will ultimately follow a normal distribution of mean $n.(\log(b)-1)$. 
Of course if $b=e$, then mean is equal to $0$, and $\lim\mathbf P(-{n}\epsilon<Z_n<{n}\epsilon)=1$. So for $b=e$ :
$$\forall\epsilon\lim_{n\rightarrow\infty}\mathbf P(e^{-\epsilon.{n}}<Y_n<e^{\epsilon.{n}})=1$$
Any other value of $b$ will make $Y_n$ quickly grows towards 0 or $+\infty$.
In particular, for $b=e$, $Y_n$ can grow quite large or small (due to the exponential factor), but it will eventually come back to 1. 
EDIT :
For what you want to do, it would be better to simulate the logarithm of the value.  Start from $v=0$, and at each step, add a random value between $-1$ and $1$. The value of your currency would be then $e^v$. It would be much more stable around $1$. 

