Law of Total Probability (Lecture 110 Harvard) I was watching Harvard STAT 110 lecture series on Youtube. As per lecture 10, given $$T=X+Y$$ I have the following question:

Why is $$P(T=t) = \sum P(T=t|X=x)\ P(X=x)$$ and not $$\sum P(T=t|X=x)\ P(X=x)+ \sum P(T=t|Y=y)\ P(Y=y)?$$       

 A: Presumably you meant $T=X+Y$, not $Y=X+Y$.

The cases for $X$ are exhaustive, since $X$ must take some value.

Analogously, the cases for $Y$ are exhaustive.

Thus, if you case it by values of $X$, you get
$$P(T=t) = \sum P(T=t|X=x)\,P(X=x)$$ 
and if you case it by values of $Y$, you get
$$P(T=t) = \sum P(T=t|Y=y)\,P(Y=y)$$
but if you sum both, you would get twice the correct answer.
A: Note that $T=X+Y$
so when $T=t,X=x$ you have $Y=t-x$, in other words,
\begin{align}
P(T=t)& = \sum_{x} \left[{P(T=t|X=x)\cdot P(X=x)}\right]\\
&=\sum_{x} \left[{P(X+Y=t|X=x)\cdot P(X=x)}\right]\\
&=\sum_{x} \left[{P(Y=t-x|X=x)\cdot P(X=x)}\right]
\end{align}
which means you have already unknowingly taken $Y$ into account!
A: Observe that for discrete random variables $X$ and $Y$: $$\{T=t\}=\bigcup_x\{T=t\wedge X=x\}=\bigcup_y\{T=t\wedge Y=y\}$$ where $x$ ranges over all values that are taken by $X$ and $y$ over all values that are taken by $Y$.
Then:
$$\sum_x P(T=t\mid X=x)P(X=x)+\sum_y P(T=t\mid Y=y)P(Y=y)=$$$$\sum_x P(T=t\wedge X=x)+\sum_y P(T=t\wedge Y=y)=P(T=t)+P(T=t)=2P(T=t)$$
A: @quasi's answer is an excellent rigorous answer, but if you'd like a more intuitive way to look at it: 
Suppose you want to get to a street corner that is north-east of your current location. You can either go one block north and one block east or go one block east and then one block north, but if you do both of them you'll overshoot your target. I.e. you can use either of the two sums and be valid, but you can't do both (i.e. add them together).
