Probability of 2 consecutive heads in 5 tosses of a fair coin I'm somehow getting the answer 1. Total number of possible events is $2^5$.

Now the number of favorable events:

2 consecutive heads can happen in following ways:
$$H\: H\: (H\:or \: T)\:\:(H\:or \: T)\:\:(H\:or \: T)$$ $$ (H\:or \: T)\:H\: H\:(H\:or \: T)\:(H\:or \: T)$$$$ (H\:or \: T)\:(H\:or \: T)\:H\: H\:(H\:or \: T)$$$$ (H\:or \: T)\:(H\:or \: T)\:(H\:or \: T)\:H\: H\:$$

Each of these $4$ possibilities can happen in $2^3$ ways corresponding to whether it's H or T in the remaining 3 tosses.

So, favorable number of events = $4\cdot 2^3$

Probability$=\frac{4\cdot 2^3}{2^5}=1$

What did I miss?
The possibilities you mention aren't mutually exclusive so you cannot just add them. $HHHHHH$ is in all of them e.g.

It might be easier to count the unfavorable ones instead?
The only ways to not get two consecutive heads are
$$\sf TTTTT\\ TTTTH, TTTHT, TTHTT, THTTT, HTTTT\\ TTHTH, THTTH,HTTTH, THTHT, HTTHT, HTHTT\\ HTHTH$$

So the probability you seek is $1-13/2^5$. 

That is $19/32$, or $0.59375$



The trouble with your method is the massive overcounting of common outcomes.

$$\require{cancel}\sf \phantom{[}H\phantom{\vert T]~[}H\phantom{\vert T]}~[H\cancel{\vert T}]~[H\vert T]~[H\vert T] \\ [H\cancel{\vert T}]~\phantom{[}H\phantom{\vert T]~[}H\phantom{\vert T]}~[H\vert T]~[H\vert T]$$