One in a Million V. One in a Billion

Relevant wiki: Binomial Distribution

Solving involves a calculation of probability . If a fair coin is flipped once, the odds of getting heads is $\frac{1}{2}$ . If the coin is flipped twice, then there are four possible outcomes, $\text{HH, HT, TH, HH}$ , with $\text{H}$ representing heads and $\text{T}$ representing tails. For this, there is a $\frac{1}{4}$ odds of getting two heads in a row, or $\frac{1}{2^2}$ . Similarly if a coin is flipped three times, there is a $\frac{1}{8}$ or $\frac{1}{2^3}$ chance of flipping three heads in a row.

So a generalized formula for the odds of flipping a fair coin all heads $n$ times in a row turns out to be $\frac{1}{2^n}$ .
In which case, if $n=20$ then $\frac{1}{2^{20}} = \frac{1}{1,048,576}$ .
If $n=30$ then $\frac{1}{2^30} = \frac{1}{1,073,741,824}$ .
Ten more times and the odds become one in a billion.

A quick answer using logarithms.

Because we are considering a Binomial Distribution $\text{Bi}_{n}(n,p)$ with parameters $n=20, p=\frac{1}{2}$ , we can check that indeed

$\begin{aligned} P(\text{Bi}_{20}=20) & ={20\choose20}\bigg(\frac{1}{2}\bigg)^{20} \\ & \approx\frac{1}{1000000}=10^{-6}.\end{aligned}$

Now we want to know when

$P(\text{Bi}_{n}=n)=\frac{1}{1000000000}=10^{-9},$

so we need to solve for $n$ :

$\begin{aligned} P(\text{Bi}_{n}=n) & ={n\choose n}\bigg(\frac{1}{2}\bigg)^n=2^{-n}=10^{-9} \\ & \implies 2^n=10^9 \\ & \implies \ln(2^n)=\ln(10^9) \\ & \implies n\ln2=9\ln10 \\ & \implies n=9\frac{\ln10}{\ln2}\approx\boxed{30}.\end{aligned}$

So the correct answer is ten times more .