Cards in box - probability a given type is picked last I came out with a probability question which I find difficult to solve. I hope some kind souls can provide me with some ideas.
There is a box with four different types of cards, namely A, B, C, D. There are 7 A, 4 B, 3 C and 2 D. One starts to pick cards from the box. The card picked out is not put back into the box. I would like to calculate the probability for certain type the cards that is to be picked last.
For example, if the sequence of cards picked goes like AABABC, then D is identified instantly as the card to be picked last.
Can anyone provide a non-exhaustive method of calculating the probability of certain type of a cards to be picked last? Thank you!
Furthermore, it would be very nice of you to provide a generalized formula of evaluation.

The cards in the box are: 7 of type A, 4 of B, 3 of C and 2 of D.
Let, for example, $A=1,B=2,C=3$ represent the event of encountering the type A first, type B second, type C third, and type D last.  (We don't have to write the last, it's implicit.)  One such example is to draw cards in order $\mathbf A,A,\mathbf B, A, \mathbf C,B,A,C,\mathbf D, A...$
Clearly the probability of encountering A first is : $\mathsf P(A=1) =a/(a+b+c+d) =7/16$
Given that, the probability of encountering B second is: $\mathsf P(B=2 \mid A=1)=b/(b+c+d)= 4/9$
And likewise, $\mathsf P(C=3\mid A=1,B=2) = c/(c+d) = 3/5$
So $$\begin{align}
\mathsf P(A\!=\!1,B\!=\!2,C\!=\!3) &= \frac{abc}{(a+b+c+d)(b+c+d)(c+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(4+3+2)(3+2)}
\\
\mathsf P(A\!=\!1,B\!=\!3,C\!=\!2) &= \frac{abc}{(a+b+c+d)(b+c+d)(b+d)} &=\frac{7\times 4\times 3}{(7+4+3+2)(4+3+2)(4+2)}
\\
\mathsf P(A\!=\!2,B\!=\!1,C\!=\!3) &= \frac{abc}{(a+b+c+d)(a+c+d)(c+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+3+2)(3+2)}
\\
\mathsf P(A\!=\!2,B\!=\!3,C\!=\!1) &= \frac{abc}{(a+b+c+d)(a+b+d)(b+d)} &=\frac{7\times 4\times 3}{(7+4+3+2)(7+4+2)(4+2)}
\\
\mathsf P(A\!=\!3,B\!=\!1,C\!=\!2) &= \frac{abc}{(a+b+c+d)(a+c+d)(a+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+3+2)(7+2)}
\\
\mathsf P(A\!=\!3,B\!=\!2,C\!=\!1) &= \frac{abc}{(a+b+c+d)(a+b+d)(a+d)} &= \frac{7\times 4\times 3}{(7+4+3+2)(7+4+2)(7+2)}
\end{align}$$
Then $\mathsf P(D=4)$ is the sum of these six.  
$\begin{align}\mathsf P(D=4) & = \frac{7\times 4\times 3}{16}\times(\frac 1 {9\times 5}+\frac 1{ 12\times 5}+\frac 1{9 \times 6}+\frac 1{12 \times 9}+\frac 1{13\times 6}+\frac 1{13\times 9})
\\ & =\frac{721}{1560}\end{align}$
And so forth.
