Calculating a sum which includes binomial coeffeicients

Question

Some formula for calculating the probablity that the difference between the number $6$ and the average of accidentally selected $100$ points among $10000$ points which are distributed in the interval $[2,8]$ with equal distances, is not more than $0.05$; lead to this sum:

$$\sum_{r=493701}^{506199}\sum_{k=0}^{100}(-1)^k\frac{\binom{100}{k}\binom{r-10001k+99}{99}}{\binom{r+99}{r}}$$

It seems that if $r-10001k+99$ is negative then $\binom{r-10001k+99}{99}$ is to be considered zero. btw, it may not be important if there is a calculator or a way that can deal with the sum; then I will correct the details. I tried to calculate it in Wolframalpha with this command:

Sum[Sum[((-1)^k)*Binomial[100,k]*Binomial[r-10001*k+99,99]/Binomial[r+99,r],{k,1,100}],{r,493701,506199}]

But it didn't show me any result. Online Sage seems to be out of access for my internet connection. Is there any other way to calculate or estimate the result of these iterated sums?

Also I tested gap with:

sign:= function(n)         
        if n < 0 then
           return 0;
        elif n = 0 then
           return 0;
        else
           return 1;
        fi;
    end;

 Sum([493701..506199],r->Sum([0..100],k->((-1)^k)*Binomial(100,k)*Binomial(r-10001*k+99,99)*sign(r-10001*k+99)/Binomial(r+99,r)));

and the result:

<integer 714...349 (28661 digits)>/<integer 547...000 (28670 digits)>

---

Edit: Clément Guérin's answer suggests to write:

$$\sum_{r=493701}^{506199}\sum_{k=0}^{100}(-1)^k\frac{\binom{100}{k}\binom{r-10001k+99}{99}}{\binom{r+99}{r}} = \sum_{k=0}^{100}(-1)^k\binom{100}{k}\sum_{r=493701}^{506199}\frac{\binom{r-10001k+99}{99}}{\binom{r+99}{r}} = \sum_{k=0}^{100}(-1)^k\binom{100}{k}\sum_{r=493701}^{506199}\prod_{s=1}^{99}\left(1-\frac{10001k}{r+s} \right) $$ But I couldn't give this to Wolfram. In fact, one should calculate $$ \sum_{k=0}^{100}(-1)^k\binom{100}{k}\sum_{r=493701}^{506199}\prod_{s=1}^{99}\left(1-\frac{10001k}{r+s} \right)\frac{sgn(r-10001k+99)+1}{2}.$$

Answer 1

First (setting r'=r+99) :

$$\frac{\binom{r-10001k+99}{99}}{\binom{r+99}{r}}=\prod_{i=0}^{98}(1-\frac{10001k}{r'-i})$$

Hence :

$$\frac{\binom{r-10001k+99}{99}}{\binom{r+99}{r}}=1+\sum_{s=1}^{99}(-1)^s\sum_{0\leq i_1<...<i_s\leq 98}\frac{(10001k)^s}{(r'-i_1)...(r'-i_s)}$$

Now your sum is equal to :

$$\sum_{r=493701}^{506199}\sum_{k=0}^{100}(-1)^k\frac{\binom{100}{k}\binom{r-10001k+99}{99}}{\binom{r+99}{r}}=$$

$$\sum_{r=493701}^{506199}\sum_{k=0}^{100}(-1)^k\binom{100}{k}+\sum_{r=493701}^{506199}\sum_{s=1}^{99}(-1)^s\sum_{0\leq i_1<...<i_s\leq 98}\frac{10001^s}{(r'-i_1)...(r'-i_s)}\sum_{k=0}^{100}(-1)^k\binom{100}{k}k^s$$

Now I claim that for any $s<n$ we have that‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌:

$$\sum_{k=0}^n(-1)^k\begin{pmatrix}n\\k\end{pmatrix}k^s=0 $$

I won't prove it generally, for $s=0$ this is the binomial formula.

For $s=1$ :

$$\sum_{k=0}^nkx^k\begin{pmatrix}n\\k\end{pmatrix}=x\times \frac{d}{dx}(x+1)^n=nx(x+1)^{n-1}$$

So this is true. For $s=2$ :

$$\sum_{k=0}^nk^2x^k\begin{pmatrix}n\\k\end{pmatrix}=x\times \frac{d}{dx}x\times \frac{d}{dx}(x+1)^n=x\frac{d}{dx}nx(x+1)^{n-1}=xn(x+1)^{n-1}+x^2n(n-1)(x+1)^{n-2}$$

And so on...

Finally this gives that the your probability is null(!). Although I cannot se where is the error, I might be wrong (given that you are looking for a probability), anyway I think there is a way here to help you compute the sum.