Here is my attempt: So denote by $P_i$ the probability of having $10$ heads in a row in $i$ tosses. So $P_{10}=1/2^{10}$.
For $i=10,...,20$, we can calculate the probability straight forward by saying: Probability that I get $10$ heads row and my heads start at $1$ plus probability that I get $10$ heads in a row and I start at $2$ plus, and so on. For instance, for $i=16$ we have $1/2^{10}+6/2^{11}=0.390625$ precisely what you got.
When $i>20$, then the counting becomes a little more messy, but I think its doable using recursion. Note that $P_{i+1}=P_i+$Probability that my $10$ heads in a row start at $i-9$ (let me denote this probability $Q_i$). So you know that $P_i$ is for the first couple of cases, note that $Q_i$ for $i\leq 20$ is just $1/2^{11}$ because we need a $T$ to go on the $i-10$ position and followed by $10$ heads, however, that does not work for say $i=21$ because we want more, we want the $11$th position to be a $T$, the following to be 10 $H$s, and the previouse ones not to be a string of $10$ heads because we are counting the strings whose starting point is at $12$. Thus, $Q_{21}=(1/2^{11})(1-P_{10})$. Hence, $P_{21}=P_{20}+Q_{21}$ to exactly what you have.
For $P_{22}=P_{21}+Q_{22}$ where $Q_{22}=(1/2^{11})(1-P_{11})$, I just checked and indeed I get the same result.
In general, you can build them recursively, and use $Q_i=(1/2^{11})(1-P_{i-11})$.
A little more clear (above was me thinking as I typed, so it turned out somewhat messy): You can denote $P_0,..,P_9$ all to be $0$, $P_{10}=1/2^{10}$, and $P_n=P_{n-1}+(1/2^{11})(1-P_{n-11})$.
For every $n\geqslant0$, let $p_n$ denote the probability that $n$ tails appear before any run of $10$ heads is completed. Then $p_0=1$ and, for every $n\geqslant0$, conditioning on the time when the $n$th tail appears, one gets $p_{n+1}=p_n\cdot(1-q)$, where $q$ is the probability that $10$ heads appear before the first next tail. Thus, $q=h^{10}$ and $p_n=(1-h^{10})^n$, where $h$ is the probability to get a head, here $h=\frac12$.
The probability to get $10$ heads in a row before flipping a total of $20$ tails is
$$
1-(1-2^{-10})^{20}\approx20\cdot2^{-10}\approx2\%.
$$
Refining only slightly this, one sees that the length of the longest run of heads before flipping a total of $n$ tails is of the order of $\log_2n$ when $n\to\infty$.
Best Answer
If you think that everybody flips simultaneously (presumably that was every 5 seconds, 24 hours a day) we can talk in terms of the number of flips. One of every 1024 will get 10 heads starting from the first flip. So it is almost certain that somebody will get it, and the first will be 45 seconds from the start of the experiment.