Geometric Distribution Let $X$ be a random variable with the geometric distribution.
Given $m <n $, what is $P( X > n| X > m)$?
 A: The geometric distribution is memoryless.
$$P(X=m+k)|X>m)=P(X=k)$$
The waiting time until the first head, given that you have already tossed the coin $m$ times without success, has the same distribution as the waiting time for the first head, if you are just starting to toss. The coin does not remember its previous history. This applies equally well to tossing dice, and other similar phenomena.
The result is essentially built into the definition of the geometric distribution, but it can also be reached through a computation. If the probability of success in any one trial is $p$, and the probability of failure is $q=1-p$, then the probability that $X>m$ is $q^m$, for the probability that the waiting time until the first success is greater than $m$ is simply the probability of $m$ successive failures. The probability that the waiting time is $m+k$ is $q^{m+k-1}p$. So using the usual formula for conditional probabilities, we find that the probability that $X=m+k$ given that $X>m$ is $q^{k-1}p$, precisely the same as the (unconditional) probability that the waiting time until first success is $k$. 
