Conditional Probability lost card problem. Q : A card from a pack of 52 cards is lost. From the remaining cards of the
pack, two cards are drawn. What is the probability that they both are
diamonds?
My Approach : I calculated the probability of selecting 2 diamonds when A Diamond card is lost and added it to the probability of selecting 2 diamonds when a diamond card is not lost. 
P($\frac{D}{LD})$ + P($\frac{D}{NLD}$) $= \frac{^{12}C_2}{^{51}C_2} + \frac{^{13}C_2}{^{51}C_2}$
$=\frac{48}{425}$
But the answer given in the book is $\frac{1}{17}$. Where am I going wrong?
 A: We use your analysis, with some correction. Let $T$ be the event $2$ diamonds, and let $L$ be the event it was a diamond that was lost. Then
$$\Pr(T)=\Pr(T\mid L)\Pr(L)+\Pr(T\mid L^c)\Pr(L^c).$$
Note that $\Pr(L)=1/4$ and $\Pr(L^c)=3/4$.
Your expressions for $\Pr(T\mid L)$ and $\Pr(T\mid L^c)$ are correct. 
A: Alternatively, intuitively, it's like saying, instead of drawing the top two cards after shuffling, move the top one to the bottom and reveal card 2 and 3. What's the probability that these two are diamonds?
$$\frac{\binom{13}{2}}{\binom{52}{2}} = \frac{1}{17}.$$
This is symmetry.
A: I partitioned the cases and then used the Law of Total probability instead of using combinations.  Apologies if you need an approach that uses combinations.
The approach:
Let $P(X)$ be the probability of the lost card being a diamond ($=1/4$)
Let $P(Y)$ be the probability of the lost card not being a diamond ($=3/4$)
In the case of event $X$ (the lost card is a diamond), the probability of selecting 2 cards and having them both be a diamond under condition $X$ is then:
$P(2D | X) = (12/51)*(11/50)=(22/425)$
Under condition $Y$:
$P(2D | Y) = (13/51)*(12/50)=(26/425)$
With the four aforementioned probabilities, we can use the Law of Total probability to obtain the textbook's answer:
$P(2D) = P(2D | X)P(X) + P(2D | Y)P(Y) = (22/425)(1/4)+(26/425)(3/4) = 1/17$
(According to Wolfram Alpha).
