Consider the equation $$\frac{kc-b}{b-c}=\sum_{k\leq s\leq n-1}\frac{(n-k)\cdot\ldots\cdot(n-s)}{k\cdot\ldots\cdot s}(\frac{t}{1-t})^{s-k+1},$$ where $t\in(0,1)$, $n\geq 3$, and $2\leq k\leq n-1$. The term $t$ represents a probability and $n,k,s\in\mathbb{Z}$.
Let's assume that $b=10$ and $c=4$. We have that $k=[b/c]+1$ where $[b/c]$ is the greatest integer less than $b/c$.
I am not sure how to do the computations of equation $(1)$. Which are the terms that we are adding?
For example, if $n=4$, then do we get $$\frac{1}{3}=\sum_{3\leq s\leq 4-1}\frac{(4-3)}{3}(\frac{t}{1-t})^{s-3+1}=\frac{1}{3}\frac{t}{1-t}\implies 1-t=t\implies t=1/2$$?
If $n=5$, then do we get $$\frac{1}{3}=\sum_{3\leq s\leq 5-1}\frac{(5-3)\cdot(5-4)}{3\cdot 4}(\frac{t}{1-t})^{s-3+1}=\frac{(5-3)\cdot(5-4)}{3\cdot 4}(\frac{t}{1-t})^{3-3+1}+\frac{(5-3)\cdot(5-4)}{3\cdot 4}(\frac{t}{1-t})^{4-3+1}=\frac{1}{6}(\frac{t}{1-t}+(\frac{t}{1-t})^2)\implies 2=\frac{t}{1-t}+(\frac{t}{1-t})^2$$?
The last leads to $t=1/2,2$ which doesn't make any sense.
I'd appreciate any help.
When $n = 5$ and $k = 3 $, there are two terms in the sum: one for $s=3$ and one for $s = 4$. The product $(n-k)\cdot...\cdot (n-s) = \prod_{j=k}^{s}(n-j)$ has just one factor when $s = 3$: $n-3 = 5 - 3$; it has two factors when $s = 4$: $(n-3)\cdot(n-4)$. Similarly the product in the denominator has one factor when $s=3$, two factors when $s = 4$. Putting it all together, I get
$$ \frac{5 -3}{3}\frac{t}{1-t} + \frac{(5-3)\cdot(5-4)}{3\cdot 4}\left(\frac{t}{1-t}\right)^2 $$
I hope that makes things clearer. However, you will still have a quadratic equation with two solutions, one of which will be bigger than 1.