prove that $13/42\sum_{x=0}^{29} (x+1) \dfrac{{29\choose x}}{{41\choose x}} = 43/14$ 
Prove that $13/42\sum_{x=0}^{29} (x+1) \dfrac{{29\choose x}}{{41\choose x}} = 43/14$.

If I were to guess a generalization, I'd say that for positive integers $m,n,$ with $m\leq n-1$ we have $\sum_{x=0}^m (x+1) \dfrac{{m\choose x}}{{n-1\choose x}} \dfrac{n-m}{n} = \dfrac{n+1}{n-m+1}.$ The latter sum can be described in probabilistic terms as the expected value of a certain random variable X, described as follows. Suppose we have a deck of $m$ red cards and $n-m>0$ blue cards and all cards are distinct. We shuffle the deck randomly so that each permutation is equally likely to occur. We then select the top card and remove it until we get a blue card. Let X be the number of cards we removed from the top.
The expected value calculation follows directly from the definition; the probability of drawing $x+1$ cards is $m(m-1)\cdots (m-x+1) (n-m) (n-x-1)!/n!,$ as there are $m(m-1)\cdots (m-x+1) $ ways to draw the first x red cards, $n-m$ ways to select the $(x+1)$th blue card, and $(n-x-1)!$ ways to arrange the remaining cards, and there are of course n! total permutations. Using the definition of expected value, we can easily obtain the required sum.

However evaluating the sum is another issue in itself, and I'm unsure of any good ways to make progress. It seems intuitive that the expected value would be $\dfrac{m}{n-m+1} + 1,$ but I'm not sure how to formally justify this.

I'm not sure if there are any useful binomial identities for proving the stated equality. I know some tricks involving binomial coefficients that involve integration and differentiation, but they don't seem useful for this problem (e.g. one can evaluate $\sum_{x=0}^n {n\choose x} \dfrac{1}{x+1}$ easily using integration and the resulting sum would be $\dfrac{2^{n+1}-1}{n+1}$).
 A: Your generalization is correct.
Generalization. For $0\le m\leq n-1$,  we have $\displaystyle\sum_{x=0}^m (x+1) \dfrac{{m\choose x}}{{n-1\choose x}} \dfrac{n-m}{n} = \dfrac{n+1}{n-m+1}.$
Proof.
$$\text{LHS}=\frac{m!(n-m)!}{n!}\sum_{x=0}^m (x+1) {n-1-x\choose n-m-1}$$
$$\begin{aligned}
&\quad\sum_{x=0}^m (x+1) {n-1-x\choose n-m-1}
=\sum_{x=0}^m\left(\sum_{k=0}^x1\right){n-1-x\choose n-m-1}\\
&=\sum_{k=0}^m\sum_{x=k}^m{n-1-x\choose n-m-1}
=\sum_{k=0}^m\sum_{i=n-m-1}^{n-1-k}{i\choose n-m-1}\\
&=\sum_{k=0}^m{n-k\choose n-m}
=\sum_{i=n-m}^n{i\choose n-m}\\
&={n+1\choose n-m+1},
\end{aligned}$$
where each time we eliminate the summation index $i$, we are applying Christmas stocking identity. (Isn't today Christmas?!) (As shown in another answer, the above equality can be proved by double counting, too).
So $$\text{LHS}=\frac{m!(n-m)!}{n!}{n+1\choose n-m+1}=\text{RHS}.$$
A: Note that
$$
\sum_{x=0}^m (x+1) \frac{\binom{m}{x}}{\binom{n-1}{x}}\frac{n-m}{n}=\sum_{x=0}^m \frac{\binom{m}{x}\binom{n-m}{1}}{\binom{n}{x+1}}=\sum_{x=0}^m \frac{\binom{x+1}{1}\binom{n-x-1}{n-m-1}}{\binom{n}{m}}=\frac{\binom{n+1}{n-m+1}}{\binom{n}{m}}=\frac{n+1}{n-m+1}.
$$
The identity used to obtain the fourth sum from the third one is
$$
\sum_{x=0}^{a}\binom{x}{a}\binom{b-x}{c}=\binom{b+1}{a+c+1}.
$$
This can be shown combinatorially as follows. Any choice of $a+c+1$ elements out of $b+1$, is a choice of the $(a+1)$-st element (call it $x+1$), followed by choices of $a$ elements from the $x$ elements before it and $c$ elements from the $b-x$ elements after it.
A: After conversion of the problem into red and blue cards,
we can get an answer using linearity of expectation,
which operates even when the variables are not independent.
There are $n-m$ blue cards, and $m$ red ones.
Let $X_i$ be an indicator random variable that assumes a value of $1$ if the $i^{th}$ red card is ahead of the first blue one, and $0$ otherwise.
Consider the $i^{th}\;$ red card in conjunction  with the $n-m$ blue ones.
Since each red card is equally likely to be ahead of the first blue one,
$P(X_i)  =\frac1{n-m+1}$
Now the expectation of an indicator random variable is nothing but the probability of the event it indicates, so $\Bbb E[X_i]=\frac1{n-m+1}$
And by linearity of expectation,
we have  $\Bbb E[X] = \frac{m}{n-m+1}\;$ cards drawn before the first blue card
The answer you intuited is the number of cards drawn including the first blue card
