How to compute an expected value in shorter ways (when taking all possibilities into account isn't plausible.) There is this question on which I have been spending a lot of time, trying to understand how to compute an expected value in a comprehensive way, as sorting out all the possibilities doesn't seem like that right thing to do, nor does it even seem possible.
The question states:

Dan tosses infinitely many standard, independent coins. The coins are tossed one by one. What is the expected number of tosses it will take Dan to arrive at two consecutive heads?

The answer says it is 6, and I didn't understand what to do. This is my attempt:

Firstly, I need to arrive at the first head. That for itself would have a geometric distribution with $\frac12$, which requires at least $2$ expected steps to be made.
Now, I either get another head and I am done, or I get tail, count one step, and then make 2 more expected steps.

Now I am in my 6th move and how do I know I am expected to arrive at head? Is that because last time I arrived at tail? I feel like I am in the right direction, but I don't fully comprehend the properties of expected value.
 A: Your English is fine, but just to confirm:  you are looking for E, the expected number of tosses before you have thrown HH. yes?
You have started well, but once you have tossed a few times the expectation may change.  Clearly, if your first toss is T, for example, you realize it will take longer than you thought (since part of your initial expectation relied on the possibility that you might throw HH to start!).
Let's follow your method:  If the first Toss is T (probability = $\frac 12$) then you are back to square 1, so your expectation from that state is E again, making your total expectation (along this route) E + 1.
If your first toss is H ((probability = $\frac 12$) then you throw again.  With probabilty $\frac 12$ you get H again and along this path it took you 2 tosses.  Thus we have a probability $\frac 14$ route to get HH in 2.  Of course you might throw a T for your second toss in which case you are back to the beginning, so we have a probability $\frac 14$ route to E + 2.  Putting this all together we see that: $$ E = \frac 14 (2)+ \frac 14 (E+2) + \frac 12 (E+1)$$ and it is easy to solve this to get E = 6.
