Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to model a random variable which represents the number of failures before success in a repeated Bernoulli trial. I will conduct only utmost N trails and I am guaranteed of one success (only one success possible) if N trials are conducted. What is the appropriate distribution? If N can go to infinity I think geometric distribution models the situation, but in my case N is finite and is known in advance.


My situation is such that the random variable has the geometric distribution properties, in the sense that, the first trial is most likely to successful, the success rate decreasing as trials increases. But within N trials, I am sure of success.

share|cite|improve this question
up vote 2 down vote accepted

Suppose you throw a die repeatedly and the probability is $p$ that you get an ace on each trial. Then the probability distribution of the number of failures before the first success is a geometric distribution. Let $X$ be the number of failures before the first success. Then $\Pr(X = x) = (1-p)^x p$, for $x\in\{0,1,2,3,\ldots\}$.

But now let's look at the conditional probability distribution of $X$ given the event $A$ that exactly one success occurs in the first $N$ trials. We're looking for $$ \begin{align} \Pr(X = x \mid A) & = \frac{\Pr(X=x\ \&\ A)}{\Pr(A)} \\ \\ & = \frac{\overbrace{(1-p)\cdot (1-p)\cdot (1-p)\cdots (1-p)}^{x-1}\ \ p\ \ \overbrace{(1-p)\cdots(1-p)}^{N-x} }{\binom N 1 p^1(1-p)^{N-1}} \\ \\ & = \frac{1}{\binom N 1} = \frac 1 N. \end{align} $$ The bottom line is $1/N$, no matter what number $x$ is. It doesn't depend on $x$. In other words, the conditional distribution is a uniform distribution, regardless of the value of $p$, and notwithstanding the fact that the unconditional distribution is a geometric distribution.

share|cite|improve this answer

Assuming I've understood your question correctly, the distribution should be uniform over the N trials.

What you want is $P(k-1$ failures before $1^{st}$ success $|$ only 1 success in $N$ trials$)$. This is the same as $P($success at k^{th} location $|$ 1 success and $N-1$ failures in $N$ trials$)$.

P(success at $k^{th}$ location $|$ 1 success and $N-1$ failures) = P(success at $k^{th}$ location AND 1 success and $N-1$ failures)/P(1 success and $N-1$ failures).

the numerator is $p(1-p)^{N-1}$. The denominator is $Np(1-p)^{N-1}$.

This intuitively makes sense because if you guarantee exactly one success in N trials, there are N possible locations where you can put the successful trial and they are all equally likely.

share|cite|improve this answer
Thanks for the reply. But in my situation, the random variable has geometric properties and not uniform properties. I added an edit after seeing your answer. – suresh Mar 28 '12 at 0:54
@suresh - I assumed a geometric distribution in my calculations and I don't see any problems. The reason it comes out to be uniform is because you guarantee a single success in exactly N trials. That changes the situation from the standard geometric distribution where you have infinitely many tries allowed. – svenkatr Mar 28 '12 at 5:59
thanks svenkatr! – suresh Apr 4 '12 at 0:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.