Tossing 2 coins, distribution We are tossing 2 fair coins once. Then for the next toss we are only tossing the coin(s) that came up heads before.
Let $X$ be the total number of heads. 
The question is $EX$ and the distribution of $X$
I tried to calculate the expected values for small $X=x$-es but it gets really complicated soon.
 A: Start by considering each coin in isolation, and the question becomes easier. Let $Y$ denote the number of heads for the first coin, and $Z$ denote the number of heads for the second coin, so $X=Y+Z$. $Y$ and $Z$ are identically distributed so let's just consider $Y$. 
First, we know that if $Y=y$, then the first coin must have come up heads exactly $y$ times, and then tails once. The probability of $y$ heads in a row is $\left ( \frac{1}{2} \right )^{y}$, and the probability of getting tails after that is $\frac{1}{2} $. Thus:
$P(Y=y) = \left ( \frac{1}{2} \right )^{(y+1)}$
To calculate the expected value of $Y$, we sum $y \cdot P(Y=y)$ over all values of $Y$, from zero to infinity:
\begin{align} 
E[Y] &= \sum_{y=0}^{\infty} y \cdot P(Y = y)\\
&= \sum_{y=0}^{\infty} y \cdot \left ( \frac{1}{2} \right )^{(y+1)} \\
&= 1
\end{align}
The expectation of the sum of two random variables is the sum of their expectations, so $E[X] = E[Y+Z] = E[Y] + E[Z] = 2$.
How do we use $P(Y)$ to get $P(X)$? Here's an example: Say $X=2$. Then we know there are three possibilities: (1) $Y=2$ and $Z=0$, (2) $Y=1$ and $Z=1$, or (3) $Y=0$ and $Z=1$. Since $Y$ and $Z$ are independent, we have: 
\begin{align}
P(X=2) &= \left( \left( \frac{1}{2} \right)^{3}\cdot \frac{1}{2} \right ) + \left( \left( \frac{1}{2} \right)^{2}\cdot \left(\frac{1}{2} \right)^2 \right) + \left( \frac{1}{2} \cdot \left( \frac{1}{2} \right)^{3} \right )\\
&= 3 \cdot \left(  \frac{1}{2} \right)^{4}
\end{align}
This example gives the intuition that maybe $P(X=x) = (x+1) \cdot \left(  \frac{1}{2} \right)^{(x+2)}$. It is true for $X=0$: both heads have to come up tails on the first flip, and the probability of that occurring is $\frac{1}{4} = (0+1) \cdot \left(  \frac{1}{2} \right)^{(0+2)}$. 
It should be simple to show by induction that this is true for all values of $X$. Here is a sketch. First note that if $X=x$ there are $x+1$ possible combinations of $Y$ and $Z$ values that can produce $y + z = x$. Each value of $Y$ corresponds to a unique series of heads and a tail (and likewise for $Z$). If we iterate, and ask what values of $Y$ and $Z$ could give $y' + z' = x+1$, we can start with our original set of possible combinations of $Y$ and $Z$ values, and just add an extra head to the start of each run for the first coin, which would multiply the probability of each combination by $\frac{1}{2}$. That is, we set $y'= y+1$ and $z'=z$. Then we need to add one new term to the sum, to account for the case where $y=0$, and $z'=x+1$. 
