Calculating the possibility of having sequential numbers in randomly picked cards This is the question (3):

I am thinking  something like this: 
Answer will appeal when the probability of having no sequential numbers in pickings decrease below 50%.
So if I assume that I picked first number as 50, then possibility of second pick to be non sequential is:
$\frac{97}{99}$. 3rd is $\frac{94}{98}$, 4th is $\frac{91}{97}$and so on.
Therefore if when $\frac{97}{99}\times\frac{94}{98}\times\frac{91}{97} ...$ hits below 1/2, the iteration count + 1 is the answer. But I am assured that it is not the answer. So could you explain me what is the correct method to solve this problem?
 A: From $100$ balls, remove k of them, then $100-k$ balls remain.
$\uparrow\bullet\uparrow\bullet..... \uparrow\bullet \uparrow$
The $k$ removed balls can be inserted at any of the $(100-k+1)$ uparroows w/o being sequential,
and if you now number the balls, the $k$ inserted balls will always be non-sequential.
Thus P(none of the $k$ chosen numbers is sequential)$= \frac{\dbinom{101-k}{k}}{\dbinom{100}k}$, and we want this to be $<0.5$
$k = 9$ gives the first value below $0.5$

ALTERNATIVE EXPLANATION based on drawing rather than insertion
Here is an explanation in black and white to address multiple queries:
We shall illustrate with $n=7, k = 3$, chosen balls black, unchosen balls white.
There will be $4$ unchosen balls $\quad\circ\quad\circ\quad\circ\quad\circ$
The $3$ chosen black balls must have been drawn from any $3$ of the $5$ gaps including the two ends:
$\;-\circ\;-\circ\;-\circ\;-\circ-$
All such draws will have the black balls non-sequential
The numerator of $\binom53 = \binom{n-k+1}{k}$ thus represents all permissible drawals.
The denominator represents drawals of any $3$ balls.
