Significance of Standard Deviation in Probability Let's say we have a set A of 10 games where the chance of you winning is $0.1$.
The probability of you winning at least one game in the set is $1 - 0.9^{10}$ while the expected number of times you win is $1$.
Now we increase the number of games in a set to $100$ (Set B), and the $p$ is changed to $0.01$. The expected number of times you win is still $1$, but the actual probability of you winning one game has changed ($1 - 0.99^{100}$).
My friend says that the difference in the two probabilities is due to standard deviation decreasing as the games increase, thus making the probability more "accurate". Is this true? I find that increasing the number of games while keeping $E$ as $1$ causes the probability of winning at least one game to tend towards ($1 - e^{-1}$). Why is this so?
And finally if this ($1 - e^{-1}$) is the most accurate probability in this series of sets, why isn't it the default answer for set A?
 A: Let $X_n$ be the random variable of the games. Then at the first setting $X$ is binomial distributed as $X_{10}\sim Bin(n^*,p^*)=Bin(10,0.1)$. The expected value is indeed $E(X_{10})=n\cdot p=10\cdot 0.1=1$ And 
$P(X\geq 1)=1-P(X=0)=1-\binom{10}{0}\cdot 0.1^0\cdot (1-0.1)^{10}=1-0.9^{10}$
In your second example $X_{100}$ is binomial distributes as $X_{100}\sim Bin(n^* \cdot 10,\frac{p^*}{10})=Bin(100,0.01)$ The expected value is still $E(X_{100})=1$.
This are just two different situations with two different distributions. You cannot say that one situation and it´s distribution is more exact than the other. You can hold $n\cdot p=\lambda$ constant and increase $n$ more and more. We will see how $X_n$ will be distributed if $n\to \infty$  
$$\lim_{n\to\infty}P(X_n=k)  =\lim_{n\to\infty}\frac{n!}{k!\,(n-k)!}\left(\frac{\lambda}{n}\right)^{k}\left(1-\frac{\lambda}{n}\right)^{n-k}$$
$$ =\lim_{n\to\infty}\left(\frac{\lambda^{k}}{k!}\right)\left(\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^{k}}\right)\left(1-\frac{\lambda}{n}\right)^{n}\left(1-\frac{\lambda}{n}\right)^{-k}$$
 $$ =\frac{\lambda^{k}}{k!}\cdot\lim_{n\to\infty}\underbrace{\left(\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+1}{n}\right)}_{\to1}\underbrace{\left(1-\frac{\lambda}{n}\right)^{n}}_{\to e^{-\lambda}}\underbrace{\left(1-\frac{\lambda}{n}\right)^{-k}}_{\to1}\\
 $$
$$  =\frac{\lambda^{k}\mathrm{e}^{-\lambda}}{k!}$$
Thus $\lim\limits_{n\to \infty} X_n$ is poisson distributed as $Z\sim Poi(\lambda)$ 
And $P(Z\geq 1)=1-P(Z=0)=1-\frac{\lambda^{0}\mathrm{e}^{-\lambda}}{0!}=1-\mathrm{e}^{-\lambda}$
But this only holds if $n\to \infty$. If n is not very large you have to use the binomial distribution, $X_n$ cannot approximated with the Poisson distribution.
