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
 A: 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.
A: Assuming the die is fair, then the probability is still just $\frac{1}{6}$. Your formula $(\frac{5}{6})^{20}\cdot \frac{1}{6}$ is the probability of not getting a $6$ for $20$ consecutive throws and then getting a $6$ right after that.  So that would be the formula for $P(A \cap B)$, where $A$ is the event of not getting a $6$ for $20$ consecutive throws and $B$ the event for getting a $6$ on the $21$-st throw.  However, you are looking for the conditional probability $P(B|A)$, which given the independence of future throws and past throws (that is, given that the die has indeed no memory) is simply the same as $P(B)$, i.e $P(B|A) = P(B) = \frac{1}{6}$.
If this was a real life situation, however, then after not getting a $6$ for $20$ consecutive throws, I would form a slight suspicion that this may not be a fair die, and would suspect that the probability of getting a $6$ on the next throw is actually somewhat lower than $\frac{1}{6}$ ...
A: 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 ....
$\delta(20)=\frac{5^{20}}{1+5+5^2+...+5^{20}6}=\frac{5^{20}}{(5^{20}-1)/4+5^{20}6}$
