In a coin tossing game with a fair coin, find expected tosses till we get HHH or THH.
Source: puzzledquant.com
My approach: This can be thought of as geometric distribution since trials are independent and p is fixed. So, probability of getting $HHH$ or $THH$ is $1/4 (=2 * 1/8)$. So expected tosses till we get either should be $4$. However, answer says expected tosses to get $HHH$ is $14$ and $THH$ is $8$. I am confused, how is this true?
The probability of getting $THH$ is much higher because the sequence of flips is continuous; that is, the trials are not independent.
If I have a chain of coin flips in which neither sequence has appeared, such as below:
$...HTHTT$
then it would be impossible for the sequence $HHH$ to appear before $THH$ appears.