Let $X$ be a random variable with the geometric distribution.
Given $m
Geometric Distribution
1 Answers
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 $k$.
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0How do you mean memoryless? – 2011-04-18
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0Oh I see, so the answer in my case is just P(X=n) – 2011-04-18
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0@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 $P(X>n-m)$. – 2011-04-18
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0I'm really confused :( Maybe I don't understand the basic definition of geometric distributions??? – 2011-04-18
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0@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. – 2011-04-18
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0@Cherie de Coning: What I wrote in the comment above is potentially ambiguous. I am using the definition that $X$ is the number of trials until the first success, *including* in the count the trial on which that success occurred. Your instructor or book may not include in the count the trial on which the first success occurred. It makes a tiny difference in the formulas, but nothing changes fundamentally. – 2011-04-18
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0Yeah, I am not given the number of trials – 2011-04-19
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0@Cherie de Coning: Naturally you are not given the number of trials. But suppose we toss a fair die until we first toss a $6$, and then stop. Let $X$ be the total number of tosses, and let $Y$ be the number of tosses strictly before the first $6$. So for example if your sequence of tosses yields $1$, $3$, $3$, $6$ then $X=4$ and $Y=3$. Some people say only $X$ has geometric distribution, some people say only $Y$ does. Most people accept both as geometric, and check which definition the writer is using. In your problem, it happens not to matter, since the answer is $q^{n-m}$ in any case. – 2011-04-20
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0Thank you, I think I get it now. – 2011-04-26