Question

A fair coin is tossed 9 times the probability that at least 5 consecutive heads occurs is

Answer 1

The whole possible results count 2^9=512. 
Then consider the first one as head and it as the beginning of the at least 5 consecutive heads, so the first 5 coins are all heads, left those 4 coins are random, in this condition the results count 2^4=16.
Thirdly consider the first one as tail and the second one is the same as the beginning head. the results will count 2^3=8.
Fourthly consider the second one as tail and the third one as the beginning head, as the first coin is also random, so the results count 2^3=8.
As there can be only one begining head, and the final beginning head is the fifth coin, and interestingly each condition the results all count 2^3.
So the answer is (2^4+4*2^3)/2^9=48/512.

PS: if the coin tossed more than 10 times, then there will be at least two beginning heads, and there will be also some inner amazing principle ; )