# 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?

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.

• I understand how the derivation comes about now, but this whole thing seems so logically strange. If you expect the same outcome in two different scenarios, both scenarios should logically be the same. But I guess you can't apply logic to maths! Thank you for your answer. – user148900 Apr 28 '18 at 15:31
• @user148900 The scenarios are different, only the expected value and the general distribution are the same. Suppose you have an urn with one $\color{red}{red}$ and one $\color{blue}{blue}$ ball. You draw a ball with repl. $n$ times. The probability that you draw $x$ red balls is $\binom{n}{x}\cdot 0.5^x\cdot 0.5^{n-x}$. The expected value is $0.5n$ Now suppose you have an urn with 1 red and 3 blue balls and You draw a ball with repl. $2n$ times. The expected value is still $\frac14\cdot 2n=0.5n$. The probability that you draw $x$ red balls is $\binom{2n}{x}\cdot 0.25^x\cdot 0.75^{2n-x}$. – callculus Apr 28 '18 at 15:58