Confusion regarding conditional probability Let's take an example. I am throwing a fair dice many times. I want to get a $6$. Every time I throw the dice , probability of getting $6$ is $\frac{1}{6}$. Suppose that I haven't got a $6$ for $20$ consecutive times, then what is the probability of getting $6$ on $21^{st}$ roll? Is it $1/6$ or $(5/6)^{20}*1/6$ or anything else?  

I am confused because the dice doesn't know it did not get $6$ for so many times
No dice roll affects any future roll or previous roll - the die has no memory of its previous rolls when it rolls for you. That means, that if you do 20 rolls and they all flip up 6 (or all not 6, or 1 not 6 and 19 6s, etc.), then on your next roll there will always be a 1/6 chance of getting any of the 6 possible values.
Let's dnote $\delta(n)$ the probability of getting 6 after $n$ shots.

$\delta(0)=1/6$ obviously 

$\delta(1)=5/6*1/6=5/36$  consequently?? no! false! I will explain.

$\delta(1)$ is the chance to hit $6$ regarded failing all the other cases, because the sequence {$62$} is not taken as an option, the tosser should halt at a specific stage, when he makes his first trial. which means $\delta(1)=P(6\ after\ all\ failings)=\frac{combinations(succeeding\ a\ 2nd\ toss)}{combinations(All\ \ available\ tosses)}=\frac{5}{1+6*5}=5/31$

Extrapolating along the basis leads to ....

$\delta(20)=\frac{5^{20}}{1+5+5^2+...+5^{20}6}=\frac{5^{20}}{(5^{20}-1)/4+5^{20}6}$