compound interest when interest is a random variable Say you are investing, and you on average get $2$% interest per bet, with a standard deviation of $3$%.
How can I get, within a confidence interval, an average amount I will have after 100 bets, if interest compounds? 
 A: This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=\mu S_{t}\,dt+\sigma S_{t}\,dW_{t}$$
where $\mu=.02$, $\sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=\mu+\sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}\exp \left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_{t}\right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${\displaystyle f_{W_{t}}(x)={\frac {1}{\sqrt {2\pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}\exp \left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma x\right)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|\frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = \frac{\ln{\frac{y}{S_0}}-(\mu-\frac{\sigma^2}{2})t}{\sigma}$$
and 
$$\frac{dg^{-1}(x)}{dx} = \frac{1}{\sigma y}$$
Thus
$$f_{S_{t}}(x) = \frac{1}{\sigma y \sqrt{2\pi t}} \exp{\frac{-(\frac{\ln{\frac{y}{S_0}}-(\mu-\frac{\sigma^2}{2})t}{\sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $\mu=.02$, and $\sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $\approx 7.244S_0$ and standard deviation $\approx 2.176S_0$.
