If you give a bank $\$100$ the teller will flip a fair coin with sides labelled $\$50$ and $\$200$, and give you the result. Do you do this or keep your money?
This is how you're viewing the game show problem.
But the problem is more the case that: the teller has a coin with a side marked $\$100$ turned face up, and the bank guarantees that the other side is either labelled $\$50$ or $\$200$. If you give the bank $\$100$ the teller will turn that coin over and give you the result.
It's either one coin or the other, but how do you assess the probability of which?
Just looking into the box and finding $\$100$ will only tell you that the game show divided up a total either $\$150$ or $\$300$ into the two boxes. It won't tell you the probability of which. You can't determine the expected return of switching boxes.
To find: $\operatorname{E}(Y\mid X=100)$
You need: $\operatorname{E}(Y\mid X=100) = 50 \Pr(Y=50\mid X=100) + 200 \Pr(Y=200 \mid X=100)$
You only know: $ \Pr(X=100, Y=50 \mid X+Y=150)=\tfrac 12,
\\ \Pr(X=50, Y=100 \mid X+Y=150)=\tfrac 12,
\\ \Pr(X=100, Y=200 \mid X+Y=300)=\tfrac 12,
\\ \Pr(X=200, Y=100 \mid X+Y=300)=\tfrac 12$
You cannot get to there from that.
You had an equal ($50\%$) chance of picking the box containing either the smaller or larger amount. You picked the box containing $\$100$. This does not necessarily mean that there is an equal chance that $\$100$ was the smaller or larger amount. That depends entirely on how the game show selects the amount to distribute.
$$\Pr(X=z \mid X+Y=3z) \not\equiv \Pr(X+Y=3z \mid X=z)$$
@Bhoot No. You know, for instance:
$\begin{align}\Pr(Y=50\mid X=100) & = \Pr(X+Y=150\mid X=100)
\\ & = \tfrac{\Pr(X=100\mid X+Y=150) \Pr(X+Y=150)}{\Pr(X=100\mid X+Y=150) \Pr(X+Y=150) + \Pr(X=100\mid X+Y=300) \Pr(X+Y=300) }
\\ & = \frac{\tfrac 12 \Pr(X+Y=150)}{\tfrac 12 \Pr(X+Y=150) +\tfrac 12 \Pr(X+Y=300)}
\\ & = \frac{\Pr(X+Y=150)}{\Pr(X+Y=150) + \Pr(X+Y=300)}
\end{align}$
However, you don't know what $\Pr(X+Y=150)$ or $\Pr(X+Y=300)$ are.
The game show could choose the total amount in any way (and they are not restricted to using those two amounts). All you know is how the amount is divided and one of the divisions.
Basically, there are two independent random variables to this problem. $A$ the total amount, and $B$ the proportion of that amount in the chosen box.
We know that $B$ has a Bernoulii distribution. $(3B-1)\sim{\cal B}(1, \tfrac 12)$
Thus we can assess the expectation of the proportions of the amount in the boxes.
$$\Pr(B=b)=\begin{cases}\tfrac 12 & b\in\{\tfrac 13, \tfrac 23\}\\ 0 & \text{else} \end{cases}
\\\begin{align}\operatorname{E}(B) & =\frac 13\frac 12+\frac 23\frac 12 \\ & = \tfrac 12\end{align} $$
However, we know nothing of the distribution of $A$. So the condition $AB=\$100$ cannot inform us of the state of either $A$ or $B$.
$$\begin{align}
\Pr(B=b \mid AB=100)
& = \frac{\Pr(B=b \cap AB=100)}{\Pr(AB=100)}
\\ & = \frac{ \Pr(B=b)\Pr(A=\tfrac{100}{b}) }{ \Pr(B=\tfrac 13)\Pr(A=300)+\Pr(B=\tfrac 23)\Pr(A=150)}
\\ & = \frac{\Pr(A=\tfrac{100}b)}{\Pr(A=300)+\Pr(A=150)}
\\ \text{Likewise:}
\\ \Pr(A=a \mid AB=100) = & \frac{\Pr(B=\tfrac{100}a)}{\Pr(A=300)+\Pr(A=150)} & a\in \{150, 300\}
\end{align}$$
