Probability of One Geometric Random Variable when Sum of Two is given This problem comes directly from MIT OCW 6.041 assignment #4, question #3.  The solution is given.  I think the solution is wrong.  The question is as follows:

Suppose that X and Y are independent, identically distributed, geometric random variables with parameter p. Show that
$P(X=i|X+Y=n)=\frac{1}{n-1}$, for $i$=1,2,...,$n$−1.

The solution given makes use of the fact that $X$ and $Y$ are independent, specifically the solution writes [Call this equation A]

$P(X=i∩X+Y=n)=P(X=i∩Y=n−i)=P(X=i)P(Y=n−i).$

This seems incorrect to me because if n and i are given, then $Y$ is completely dependent on those two values.  $Y$ has been implicitly specified.  So the distributions are not independent.  Further, the supposed solution to the problem is that all the $X=i$ are equally likely if we specify the sum of $X$ and $Y$, whereas prior to that sum $X$ follows a geometric distribution.  That sure seems like dependence to me.  Just using the law of multiplication the solution should have
$$
P(X=i∩X+Y=n)=P(X=i∩Y=n−i)=P(X=i)P(Y=n−i | X=i).
$$
$X$ is specified, one can't just throw away that dependence.  In other words, the quantity $n-i$ is not independent of $X$, so $Y$ cannot be independent of $X$ either, given the conditioning.
Here is my alternative solution.  I'm going to use Bayes' Rule:
$$
P(X=i|X+Y=n)= \frac{P(X+Y=n|X=i) P(X=i)}{\sum_{j=1}^{n-1} P(X+Y=n|X=j) P(X=j)}.
$$
The key, I think, is to recognize that $P(X+Y=n|X=j) = P(Y=n-j)$.  This is to say, that if $X$ is given to be $j$, then the distribution of $X+Y$ is just the distribution of $Y$.  This makes sense to me because I am adding a geometric random variable, $Y$, to a constant $j$, so the distribution will also be a geometric random variable.  So, the conditional probability comes out of the sum and cancels, leaving
$$
P(X=i|X+Y=n)= \frac{P(X=i)}{\sum_{j=1}^{n-1} P(X=j)} = \frac{p(1-p)^{i-1}}{1-(1-p)^{n-1}},\; \mathrm{for} \; i=1,2,...,n−1.
$$
All the conditional probabilities sum to 1, they are all non-negative, they are exclusive, and they are exhaustive.  This looks like a valid probability distribution to me.
Even the content of that final equation looks correct to me.  $X+Y=n$ is given, and we want to know what the probability of seeing any particular $X=i$ value is.  We know that $X$ is now limited to the interval $[1,n-1]$, because it is a geometric variable ($X \ge 1$) and we know that it is added to another geometric variable to get $n$, so $X \le n-1$.  So all I have to do is re-weight the probabilities I had before conditioning such that the now-allowed values sum to 1 and I should be done.
My question is: why is equation A used in the solution when every time I look at it, it appears wrong and I have a perfectly sensible solution that appears better in every way?  What am I missing about equation A?
Edit:
I'm going to answer the opposite of my question.  The mistake I made was pulling $P(X+Y=n|X=j) = P(Y=n-j)$ out of the sum in the denominator.  The identity is correct, but that probability literally depends on $j$ and I can't pull it out of the sum.  That was silly.  Plugging in the probabilities for the geometric variables I get the same answer as the answer key.
Furthermore, the identity I argue is true actually proves the identity I thought was untrue.  If I multiply both sides of my identity by $P(X=i)$ and use the multiplication rule to combine the conditional and marginal probabilities, I find $P(X=j∩X+Y=n) = P(Y=n-j)P(X=i)$, which I spent so many hours thinking was wrong.
It appears I made a mathematical error that led me to a specious answer which got me to believing a prior logical error.  What a ride.
 A: 
This seems incorrect to me because if n and i are given, then Y is completely dependent on those two values. Y has been implicitly specified.

No, the value for $Y$ in the event has been specified.  We do not yet know that that event has occurred; we still seek the probability for that.
$$\{\omega\in\Omega: X(\omega)=\imath, (X+Y)(\omega)=n\}=\{\omega\in\Omega: X(\omega)=\imath, Y(\omega)=n-\imath\}$$
In short, the event $X=\imath, X+Y=n$ and the event $X=\imath, Y=n-\imath$ are identical.

Next,  $X$ and $Y$ are independent random variables, so $\mathsf P(X=x,Y=y)=\mathsf P(X=x)\mathsf P(Y=y)$ for all values of $x$ and $y$, including constants: $\imath$ and $n-\imath$.
$$\mathsf P(X=\imath, Y=n-\imath)=\mathsf P(X=i)\,\mathsf P(Y=n-\imath)$$

Now we can seek the conditional probability using the above:
$$\begin{align}\mathsf P(X=\imath\mid X+Y=n)&=\dfrac{\mathsf P(X=\imath, X+Y=n)}{\mathsf P(X+Y=n)}&&\text{by definition}\\[1ex]&=\dfrac{\mathsf P(X=\imath, Y=n-\imath)}{\mathsf P(X+Y=n)}&&\text{by identicallity of the events}\\[1ex]&=\dfrac{\mathsf P(X=\imath)\,\mathsf P(Y=n-\imath)}{\mathsf P(X+Y=n)}&&\text{by independence of }X, Y\end{align}$$
A: I think the point you're getting confused about is that while $X$ and $Y$ are defined to be independent, once you condition on an event that involves both of them you are effectively moving into a universe where they do have a dependency.
So the expression $P(X = i \cap X + Y = n)$ is talking about two probability of two events both occurring, but $P(X = i | X + Y = n)$ is talking about the probability of one event occurring, conditional on another.
When rearranging the first probability into $P(X = i \cap Y = n - i)$, we're not conditioning on anything, and $n$ and $i$ are just a couple of parameters. Instead, we're re-writing the event in a way that covers the exact same area, but which is written as the intersection of two independent things.
Later, we then make use of the probability of that intersection of events to calculate the probability of the conditional event, but that's a separate step.
