gambling probability problem We are given a fair coin. We start out with 5 dollars. We keep tossing the coin. If the outcome is different than the previous one, we are awarded another 5 dollars. However, we do not get anything if the outcome is the same as the previous one. Let's say we toss the coin X times in the long run. How much do we expect to have in the end?

This problem may be solved using generating functions which is admittedly more powerful than strictly necessary in this case, but it does illustrate the method, which is worth knowing. Let $p_A(n)$ be a polynomial in $u$ such that $[u^k] p_A(n)$ is the probability of having a capital of $k$ dollars and the last coin came up heads and similarly $p_B(n)$ for tails. Then we have the following system of equations
$$ p_A(n) = \frac{1}{2} p_A(n-1) +  \frac{1}{2} u^5 p_B(n-1) \\
 p_B(n) = \frac{1}{2} u^5 p_A(n-1) +  \frac{1}{2} p_B(n-1)$$
with initial conditions
$$ p_A(0) = \frac{1}{2} u^5 \quad \text{and} \quad p_B(0) = \frac{1}{2} u^5.$$ 
Now introduce the generating functions
$$ P_A(z) = \sum_{n\ge 0} p_A(n) z^n  \quad \text{and} \quad 
P_B(z) = \sum_{n\ge 0} p_B(n) z^n.$$
Multiply the recurrences by $z^n:$
$$ p_A(n) z^n = z \frac{1}{2} z^{n-1} p_A(n-1) + z \frac{1}{2} u^5 z^{n-1} p_B(n-1) \\
 p_B(n) z^n = z \frac{1}{2} u^5 z^{n-1} p_A(n-1) + z \frac{1}{2} z^{n-1} p_B(n-1)$$
Sum over $n$ to obtain the following system of equations
$$ P_A(z) -  \frac{1}{2} u^5 = \frac{1}{2}z P_A(z) + \frac{1}{2}u^5 z P_B(z) \\
P_B(z)  -  \frac{1}{2} u^5 = \frac{1}{2}u^5 z P_A(z) + \frac{1}{2}z P_B(z).$$
The solution to this system is
$$ P_A(z)=-{\frac {{u}^{5}}{{u}^{5}z-2+z}}\\
P_B(z) =-{\frac {{u}^{5}}{{u}^{5}z-2+z}}.$$
The generating function for the expected capital after $n$ steps is then given by
$$ f(z) = \left( \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} =
-10\, \left( 2\,z-2 \right) ^{-1}+10\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}.$$
Now we have $$[z^n] f(z) = 5+5/2\,n,$$
which is precisely what was obtained above.
We can also compute the variance, which is where the generating functions start to shine. First compute the expected factorial moment $Q(Q-1)$ where $Q$ is the capital. This has generating function
$$ g(z) = \left( \frac{d}{du} \frac{d}{du} (P_A(z) + P_B(z)) \right)_{u=1} =
-40\, \left( 2\,z-2 \right) ^{-1}+140\,{\frac {z}{ \left( 2\,z-2 \right) ^{2}}}-100\,{\frac {{z}^{2}}{\left( 2\,z-2 \right) ^{3}}}
.$$ This gives $$[z^n] g(z) = 20+{\frac {115}{4}}\,n+{\frac {25}{4}}\,{n}^{2}.$$
The variance can now be obtained from 
$$Var[Q] = E[Q^2] - E[Q]^2 = E[Q(Q-1)] + E[Q] - E[Q]^2,$$
which gives
$$Var[Q] = {\frac {25}{4}}\,n.$$
We could in fact compute all factorial moments if desired.
