Consider the powers of $2$:
$$ \begin{array}{rcl} 2^1 & = & 2\\ 2^2 & = & 4\\ 2^3 & = & 8\\ 2^4 & = & 16\\ 2^5 & = & 32\\ 2^6 & = & 64\\ 2^7 & = & 128\\ 2^8 & = & 256\\ 2^9 & = & 512\\ 2^{10} & = & 1024\\ \end{array} $$ $$\cdots$$
They always end in $2$, $4$, $6$ or $8$, but we know little about their beginnings
$$ \begin{array}{rcl} 2^1 & = & 2\\ 2^2 & = & 4\\ 2^3 & = & 8\\ 2^4 & = & 16\\ 2^5 & = & 32\\ 2^6 & = & 64\\ 2^7 & = & 128\\ 2^8 & = & 256\\ 2^9 & = & 512\\ 2^{10} & = & 1024\\ \end{array} $$ $$\cdots$$
They always end in $2$, $4$, $6$ or $8$, but we know little about their beginnings
 Is there a power of $2$ starting with $9$?
 Is there a power of $2$ starting with $9999\cdots 9999$ ($99$ nines)?
 When considering more and more powers of $2$, which is the average of powers starting with $9$ in the limit?
 There are powers of $2$ starting with $9$, for instance $2^{53} = 9007199254740992$

Now we can't proceed by direct inspection, because humongous numbers are involved... The right direction to prove that there are powers of $2$ starting with $9999\cdots 9999$ ($99$ nines) is to work with logarithms to the base $10$ (denoted $\log$): if some power starts with $9999\cdots 9999$
$$2^n=9999\cdots 9999\cdots\cdots\cdots$$
then
$$n\log 2=\log(2^n)=\log(9999\cdots 9999\cdots\cdots\cdots)=\log(9.999\cdots 9999\cdots\cdots\cdots)+m$$
where $m$ is the number of digits of $9999\cdots 9999\cdots\cdots\cdots$ minus $1$. Considering the fractional part,
$$\{n\log 2\}=\log(9.999\cdots\cdots\cdots)$$
that is, we're looking for some $n$ that satisfies
$$\log(9.999\cdots 9999)\leqslant\{n\log 2\}\lt\log(10)=1$$
and this is possible, because $\log 2$ is an irrational number (check it out! $2^q\neq 10^p$ for positive integers $p$ and $q$)! And this ensures that the fractional parts of $n\log 2$ are dense in the interval [0,1]: one may find arbitrarily close elements of this set to any given number $0\leqslant x\leqslant 1$. So we only have to wait for some $\{n\log 2\}$ to fall into $[\log(9.999\cdots 9999),1)$; we will have to wait a lot, sure, but there will be such powers
And with the same argument, we find out that there are powers of $2$ starting with any given finite sequence of numbers. Cool, isn't it?

Since in the limit the values $\{n\log 2\}$ are equally distributed in the interval $[0,1]$, the average of powers starting with $9$ is
$$\log 10\log 9\simeq 0.0458$$
That's quite interesting: there are powers of $2$ starting with any number, but bigger numbers appear less. Had you ever thought about it?