Finding expected value of remaining piece A father has a pie made for his two sons. Eating more than half of the pie will give indigestion to anyone. While he is away, the older son helps himself to a piece of the pie. The younger son then comes and has a piece of what is left by the brother. Assume that the size of each of the two pieces eaten by the sons is random and uniformly distributed over what is currently available.
  
  What is the expected size of the remaining piece give that no son has indigestion?  


After a while of thinking this is what I did. 
Let $\theta_1$ be the total angle corresponding to the amount of pie that son $1$ eats. Similarly, let $\theta_2$ be the total angle corresponding to the amount of pie for son $2$. 

Let $X$ be the remaining piece. So, 
$$
X = 2\pi - \theta_1 - \theta_2 \\
\mathbb{E}(X) = 2\pi - \mathbb{E}(\theta_1) - \mathbb{E}(\theta_2).  
$$
So after a while of thinking, given that no son has indigestion, it feels as if son 1 and son 2 are independent, since they eat less than $\pi$... 

So $\theta_1 \sim \operatorname{Unif}(0,\pi)$ and also $\theta_2\sim \operatorname{Unif}(0,\pi)$. Then $\mathbb{E}(X) = \pi$.  

Is this correct?
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{} }$As is pointed out by @AndreaL, conditional probability should be used to calculate to desired probability.

Suppose the elder son eats an angle of $X_1$ and the younger son eats an angle of $X_2$. Translating to probabilistic notation, there is $X_1 \sim U(0, 2π)$ and $X_2 \mid X_1 \sim U(0, 2π - X_1)$, and what needs to be found is $E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)$.

The joint density of $(X_1, X_2)$ is$$
f_{X_1, X_2}(x_1, x_2) = f_{X_2 \mid X_1}(x_2 \mid x_1) f_{X_1}(x_1) = \frac{1}{2π(2π - x_1)}, \quad \forall x_1, x_2 &gt; 0, x_1 + x_2 \leqslant 2π
$$
thus\begin{align*}
&amp;\peq P(X_1 \leqslant π, X_2 \leqslant π)\\
&amp;= \iint\limits_{0 &lt; x_1, x_2 \leqslant π} \frac{1}{2π(2π - x_1)} \,\d x_1\d x_2 = \frac{1}{2π} \int_0^π \int_0^π \frac{1}{2π - x_1} \,\d x_2\d x_1\\
&amp;= \frac{1}{2π} \int_0^π \frac{π}{2π - x_1} \,\d x_1 = \frac{\ln 2}{2},
\end{align*}
and\begin{align*}
&amp;\peq E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\
&amp;= 2π - E(X_1 + X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\
&amp;= 2π - \iint\limits_{0 &lt; x_1, x_2 \leqslant π} (x_1 + x_2)·\frac{f_{X_1, X_2}(x_1, x_2)}{P(X_1 \leqslant π, X_2 \leqslant π)} \,\d x_1\d x_2\\
&amp;= 2π - \frac{1}{π\ln 2} \int_0^π \int_0^π \frac{x_1 + x_2}{2π - x_1} \,\d x_2 \d x_1\\
&amp;= 2π - \frac{1}{π\ln 2} \int_0^π \frac{π x_1 + \frac{π^2}{2}}{2π - x_1} \,\d x_1\\
&amp;= 2π - \frac{1}{π\ln 2}·\frac{1}{2} (5 \ln 2 - 2) π^2 = \frac{2 - \ln 2}{2\ln 2} π.
\end{align*}
This is a complicated situation, but the question happens to greatly simply the process of answering it. The question states that no son has indigestion.

Therefore, neither son ate half the pie or more. (amount either son ate: $0 \le eaten \lt .5$)

Therefore, there is always some pie left. (amount left: $0 \lt remaining \le 1$)

Since the portions both sons ate are uniformly random and independent (since neither son eats more than half), in the expected case we just say that they both ate the middle value (halfway between 0 and .5, or .25). And since two sons ate, .25 + .25 = .5, half the pie is gone.

Another way to look at it is to see that the amount of pie left is uniformly random, and deduce that it is .5.

So while it is true that in the situation the amount the second son eats is highly dependent on the amount the first son eats, the amount they eat is independent for this specific question only.

So yes, you are correct.