# Geometric Distribution

Let X be a random variable with the geometric distribution. For m<n, what is P( X > n| X > m)?

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$$P(X=m+k)|X>m)=P(X=k)$$ 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 $k$.
@Cherie de Coning: Not exactly. The probability that $X=m+1$ given that $X>m$ is $P(X=1)$, the probability that $X=m+2$ given that $X>m$ is $P(X=2)$, and so on. So the probability that $X>n$ given that $X>m$ is $P(X>n-m)$. –  André Nicolas Apr 18 '11 at 5:23
@Cherie de Coning: There may be excellent reason for your confusion, since there are two closely related distributions that are called the geometric distribution, and your instructor may be using the other definition. The version I used gives the total number of trials before the first success. The other one gives the total number of failures before the first success. Of course these are simply related, the second is just the first minus $1$. If your instructor is using the second definition, it should be straightforward to adapt what I wrote. –  André Nicolas Apr 18 '11 at 13:27