Chance of getting a specific value from a combination with repetition

Question

Let's say I have a set of $m$ distinct elements and I create a new set of size $n$ ($n\geq m$) choosing those elements by a combination with repetition. Then, I randomly draw a value from this new set of $n$ elements. I wish to know the probability of getting a specific value from the original set.

Here's what I've got so far: I would have ${m+n-1}\choose{n}$ ways to make combinations with repetitions. I wish to know the probability that the new set contains exactly $k$ repetitions of a specific value of the original set. I believe that would be the ratio between the number of ways to choose with repetition $n-k$ (because $k$ values are already defined) from $m-1$ elements (excluding the specific one) and the total number of combinations, i.e., \begin{equation} P(k\times\text{specific value}) = \frac{{(m-1)+(n-k-1)}\choose{n-k}}{{m+n-1}\choose{n}} = \frac{{m+n-k-2}\choose{n-k}}{{m+n-1}\choose{n}}. \end{equation}

Now I would be able to partitionate the probability of choosing that specific value as \begin{equation} \text{Answer} = \sum_{k = 1}^n \overbrace{P(k\times\text{specific value})}^{\substack{\text{Chance of getting a combination} \\ \text{with } k \text{ repetitions of that value}}} \underbrace{\frac{k}{n}}_{\substack{\text{Chance of drawing} \\ \text{that value from} \\ \text{the new set} }} \end{equation}

But I am not able to work out this summation. It appears to have a neat result - and actually an even better way to express it. Intuitively, I understand that the answer should be $\frac{1}{m}$, like I am just randomly drawing the value from the original set of $m$ distinct elements, but I want to be able to show that this is the case after doing the combination with repetition. I would also be glad to find out if there are better methods than carrying out the sum above.

Answer 1

The question is, how to you prove $$ \binom{n+m-1}{n}^{-1}\sum_{k=0}^n \frac{k}n\binom{m+n-k-2}{n-k}\stackrel{?}=\frac1m $$ To do this, I will use another well-known identity. For any $a,b,r\in \mathbb N$,

$$\sum_{i=0}^r \binom{i}{a}\binom{r-i}{b}=\binom{r+1}{a+b+1}\tag{$*$}$$

For several proofs of $(*)$, see this question.

$$ \begin{align} \binom{n+m-1}{n}^{-1}\sum_{k=0}^n \frac{k}n\binom{m+n-k-2}{n-k} &=\frac1n\binom{n+m-1}{n}^{-1}\sum_{k=0}^n k\binom{m+n-k-2}{m-2} \\&=\frac1n\binom{n+m-1}{n}^{-1}\cdot \sum_{k=0}^n\binom{k}{1}\binom{(n+m-2)-k}{m-2} \\&\stackrel{*}=\frac1n\binom{n+m-1}{n}^{-1}\cdot \binom{n+m-1}{m} \\&=\frac1n\cdot \frac{n!(m-1)!}{(n+m-1)!}\cdot \frac{(n+m-1)!}{m!(n-1)!} \\&=\frac1m, \end{align} $$ as desired. In equation $\stackrel{*}=$, we have applied $(*)$ with $a\gets1,b\gets m-2,$ and $r\gets n+m-2$.