<s>[INST] Coin Tossings probability I want to find the probability that in ten tossings a coin falls heads at least five times in succession. Is there any formula to compute this probability?
Answer provided is $\frac{7}{2^6}$
 [/INST] Corrected answer after Orangetree pointed out I forgot to take into account events were not mutually exclusive.
You need to think about how many different coin tossing sequences give at least $5$ consecutive heads, and how many coin tossing sequences there are in total, and then take the ratio of the two.
Clearly there is $2^{10}$ coin tossing sequences in total, since each of $10$ coins have $2$ possible outcomes.
To see how many sequences have $5$ consecutive heads consider the following. In any sequence
$$X-X-X-X-X-X-X-X-X-X$$
We can put the $5$ consecutive heads in $5$ places. This can be seen by putting it at the start of thee sequence, and then moving it over step by step.
$$H-H-H-H-H-X-X-X-X-X$$
$$X-H-H-H-H-H-X-X-X-X$$
$$X-X-H-H-H-H-H-X-X-X$$
$$X-X-X-H-H-H-H-H-X-X$$
$$X-X-X-X-H-H-H-H-H-X$$
$$X-X-X-X-X-H-H-H-H-H$$
For the first scenario, we have $2^5$ possible sequences since the remaining coins can be either heads or tails without affecting the result.
For the second scenario, we can only take $2^4$ since we need to fix the first coin as tails to avoid counting sequences more than once.
Similarly, for the third scenario we can count $2^4$ un-counted sequences, as we fix the coin preceeding the sequence as tails.. and so on
This gives total number of ways 
$$2^5 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4  = 2^4(2+1+1+1+1+1) = 2^4\times 7$$
Hence the probability should be
$$\frac{2^4 \times 7}{2^{10}} = \frac{7}{2^6}$$
To actually answer your question if a general formula exists, the reasoning above can be extended to give the probability of at least $m$ heads out of $n$ coin tosses as
$$\frac{2^{m-1}(2+(n-m))}{2^n} = \frac{2+n-m}{2^{n-m+1}}$$
</s>