advanced expected value？ 

*David repeatedly flips a fair coin. Find the expected value of the total number of heads he will flip before flipping two consecutive tails.


I know that expected value is the sum of value multiplied by probability and all the probability sums up to $1$, however I find this somehow hard to approach so can anyone help me..
 A: Here is a combinatorial approach with an answer based upon  generating functions. We develop a generating function 
\begin{align*}
G(z)=\sum_{n=2}^\infty g_n z^n
\end{align*}
with $g_n$ counting the number of valid words of length $n$. These are binary words from a two character alphabet $V=\{T,H\}$ which have precisely one run of tails of length $2$ at the end of the word.

We can use $G(z)$ to calculate the   expectation value $E(Z)$ since
  \begin{align*}
E(Z)=\sum_{n=2}^\infty n \frac{g_n}{2^n}
=\frac{1}{2}\cdot\left.\frac{d}{dz}\left(G(z)\right)\right|_{z=\frac{1}{2}}\tag{1}
\end{align*}
We       observe
  
  
*
  
*The generating function $G(z)$ starts with $g_2z^2=z^2$ since the shortest valid word is $\color{blue}{TT}$.
  
*There is only one valid word with length $3$, namely $H\color{blue}{TT}$.
  
*Each valid word with length $n\geq 3$ ends with $H\color{blue}{TT}$.

The idea is to build valid words of length $n\geq 3$ by creating valid subwords of length $n-3$ and append $H\color{blue}{TT}$. In order to do so we start with a generating function for words of a two character alphabet $V=\{T,H\}$ which counts words with no consecutive equal characters at all.
These words are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information. (You might also find this answer helpful.)

The generating function $A(z)$ counting Smirnov words over a two character alphabet is according to the reference 
  \begin{align*}
  A(z)=\left(1-\frac{2z}{1+z}\right)^{-1}
  \end{align*}
The coefficient of $z^n$ of $A(z)$ gives the number of Smirnov words of length $n$, i.e. the number of words with no consecutive equal heads and tails.

Since there is no restriction to the distribution of heads, we can replace each character "H" in a Smirnov word by one or more "H"s, which means to replace
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots=\color{blue}{\frac{z}{1-z}}
\end{align*}
in the corresponding generating function $A(z)$.

Based upon $A(z)$ we obtain this way a generating function $B(z)$ with
  \begin{align*}
B(z)&=\left(1-\frac{z}{1+z}-\frac{\color{blue}{\frac{z}{1-z}}}{1+\color{blue}{\frac{z}{1-z}}}\right)^{-1}\\
&=\left(1-\frac{z}{1+z}-z\right)^{-1}\\
\end{align*}
The coefficient of $z^n$ of $B(z)$ gives the number of words of length $n$ with no consecutive equal tails.

Since  valid words  of length  $n\geq 3$ end up in $H\color{blue}{TT}$  we respect this fact by multiplying  $B(z)$ with $z^3$.  The special case, the word $TT$ which  does not end in $H\color{blue}{TT}$ is additionally to respect and we obtain the generating function $G(z)$ counting all valid words as
\begin{align*}
G(z)&=z^2+z^3B(z)\\
&=z^2+z^3\left(1-\frac{z}{1+z}-z\right)^{-1}\\
&=\frac{z^2}{1-z-z^2}\tag{2}\\
&=z^2+z^3+2z^4+3z^5+5z^6+\cdots
\end{align*}
Note that (2) is essentially the generating function of the Fibonacci numbers.   The coefficients give the number of valid words of length $n$
\begin{align*}
[z^2]G(z)=1:&\quad TT\\
[z^3]G(z)=1:&\quad HTT\\
[z^4]G(z)=2:&\quad HHTT,  THTT\\
[z^5]G(z)=3:&\quad HHHTT, HTHTT,  THHTT\\
[z^6]G(z)=5:&\quad HHHHTT, THHHTT, HTHHTT,  HHTHTT, THTHTT
\end{align*}

Finally,  we   obtain  from $G(z)$ according to (1) the expectation value
  \begin{align*}
\color{blue}{E(Z)}&=\frac{1}{2}\cdot\left.\frac{d}{dz}\left(G(z)\right)\right|_{z=\frac{1}{2}}\\
&=\frac{1}{2}\cdot\left.\frac{d}{dz}\left(\frac{z^2}{1-z-z^2}\right)\right|_{z=\frac{1}{2}}\\
&=\frac{1}{2}\cdot\left.\frac{z(2-z)}{(1-z-z^2)^2}\right|_{z=\frac{1}{2}}\\
&=\color{blue}{6}
\end{align*}

A: Excuse my brevity I'm on my phone.   Let $p$ be the probability of flipping a heads.   Let $N$ be the number of flips until you stop.   Let $Y_1$ be the number of flips until we get our first heads.   Note: we know the distribution of $Y_1$ to be geometric.   Let $Y_2$ be the number of flips until the second heads.   Clearly, for any realization where $Y_2=Y_1+1$ we have that $N=Y_2$. Now, by conditioning we have that $$\begin{align}\mathsf E(N) &= {\mathsf E(N\mid Y_2=Y_1+1)~\mathsf P(Y_2=Y_1+1) + \mathsf E(N\mid Y_2>Y_1+1)~\mathsf P(Y_2>Y_1+1)} \\ & = \mathsf E(Y_1+1)~p+(\mathsf E(Y_1+1)+\mathsf E(N))~(1-p)\end{align}$$ Then just solve the above equation.   The only unknown is $\mathsf E(N)$.   The second equality uses the fact that if $Y_2>Y_1+1$ then the ($Y_1+1$)st flip is a tails and so the process "resets" and the expected number of additional flips is as if you've never flipped the coin before, i.e., $\mathsf E(N)$.
A: $E(X)=(1(\frac{1}{2}+\frac{1}{2^2})+2(\frac{1}{2^2}+2C_1(\frac{1}{2^3})+\frac{1}{2^4})+3(\frac{1}{2^3}+3C_1(\frac{1}{2^4})+3C_2(\frac{1}{2^5})+3C_3(\frac{1}{2^6}))+.......)\frac{1}{2^2}
$
$= (\sum_{n=1}^\infty n(\frac{1}{2^n})(nC_0(\frac{1}{2^0})+nC_1(\frac{1}{2^1})+nC_2(\frac{1}{2^2})+nC_3(\frac{1}{2^3})+.....+nC_n(\frac{1}{2^n})))\frac{1}{2^2}$
$=(\sum_{n=1}^\infty n(\frac{1}{2^n})(\frac{3}{2^n}))\frac{1}{2^2}$
$=(\sum_{n=1}^\infty n(\frac{3}{4^n}))\frac{1}{2^2}$
$=3$
A: Let $T$ be the number of consecutive tails flipped in the past two flips. 
$T$ can take 3 values $\in \{ 0 , 1 ,  2 \}$.
We define $X$ to be the the number of heads and define its expected value recursively as follows: 
$\mathbb{E}(X|0) = \frac{1}{2} \cdot \{1 + \mathbb{E}(X|0) \} + \frac{1}{2} \cdot  \{0 + \mathbb{E}(X|1)\} $
$\mathbb{E}(X|1) = \frac{1}{2} \cdot \{0\} + \frac{1}{2}\cdot \{1 + \mathbb{E}(X|0) \} $
On Solving, we find $\mathbb{E}(X|0) = 3$
Hence, there are 3 expected heads before two consecutive tails.  
Detailed explanation: We begin at state "0", which means we don't have any tails in the last two flips. We can either go to state $1$ with $\frac{1}{2}$ probability by flipping a tails, adding essentially nothing to our expected value or we can flip a heads with probability $\frac{1}{2}$ and stay in the same state but we add $1$ to our expected value. 
Similarly, when we reach state one, we can flip a tails and the game ends, with probability $\frac{1}{2}$ (adding 0 to our expected value) and again with probabbility $\frac{1}{2}$ we can go into state $0$ since the we need to start all over again in order to get two consecutive tails, adding $1$ to the expected number of heads in the process. 
A: An interesting question indeed.
I have a very different approach. Let us instead make and ponder a much much much simpler one:
What is the expected number of flips required to achieve 2 consecutive tails?
Let us condition this expectation on a partition of the opening toss / tosses : H, TT, TH.
Notice that this is an "exhaustive and mutually exclusive" partition. By this I mean that only one but exactly one must happen. You will either get some Head followed by whatever. Tails Tails or Tails Heads. Notice I don't care what follows the Head. The probabilites for these events are respectively $ \frac 12, \frac 14  $ and $ \frac 14$
So we have E = $ \frac 12$ E|H  + $ \frac 14$  E|TT + $ \frac 14$ E|TH
We will now work through each of these different (easier) expected values.
Firstly E|TT = 2.

If we first tossed Tails Tails then we are done! It took only 2 tosses and we are finished.
Secondly E| H = E+1

If we toss a head first then we have just wasted a turn and are no closer to achieving our double tails. The game starts over again with a wasted toss. We don't yet know exactly what the value of E is but we know for sure if we tossed a heads first our expected value is the same as it was before just one more .
Thirdly E| TH = E+2

For the same logic as before instead we have wasted two tosses! The game is restarted and our previous two tosses have contributed nothing to getting us closer.
Filling this all in yields:


E = $ \frac 12$ E|H  + $ \frac 14$  E|TT + $ \frac 14$ E|TH


E = $ \frac 12$ E+1  + $ \frac 14$ 2 + $ \frac 14$ E +2


This is an easy equation we can solve and it gives E = 6.

Finally time for your original question . There are no more numbers now just an explanation with words. I want you to think about the Expected number of tails and the Expected number of heads. Is one larger than the other ? If I said to you , toss coins until the Yankees beat the Red Soxs or toss coins until the product of heads and tails is a cube number or toss coins until the difference between heads and tails is a prime greater than 15 Million . In those questions its simple, E(H) = E(T) surely. As long as the coin is fair you can give me ANY condition to toss coins until and E(H) = E(T) similar to the village that always wants boys problem. There is nothing you can do to make E(H) != E(T). Sure you want to keep tossing until we have ten thousand tails? E (H) = E(T) You want to toss coins until 2 consecutive tails? Still E(H) = E(T) .
So E(Tosses) = 6 if we toss until 2 consecutive tails. So obviously E(H) + E(T) = 6 as tosses are either Heads or Tails but again E(T) = E(H) so 2 E(H) = 6 and then E(H) = 3.
Oskar :)
