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

Short question: If several people, all of whom have limited availability, need to meet, how far in the future will I need to schedule the meeting?

I was hoping there was a readily available answer to this, or a business-centric discussion of the problem, but I haven't found anything covering this particular problem (which seems to me a very common one!)

Here's a more precise statement of the problem conditions:

  1. There are a fixed number of meeting times per day (say 8).
  2. $N$ people need to meet, and all people must be available at the same time.
  3. Each person is available at a given time with some constant probability $p$ (i.e., if $p= \frac{1}{2}$, then each person is scheduled for $4$ meetings a day on average).
  4. The "expected wait time" I would define as the time where the probability is $\frac{1}{2}$ that there is an available meeting time.

After looking around some, I think this would be a poisson process, and for the trivial situation of $N=1$ then the probability of needing to wait a particular amount of time is $e^{-x}$, which integrated over time gives a nice neat an intuitive answer of $\frac{1}{p}$.

For $N>1$, I'm completely lost about how to proceed.

The underlying question I'm really interested in is how the expected wait time changes as $N$ changes. That is, is we add one more person to this meeting invite, how much longer are we going to need to wait until everyone is available at the same time?

share|cite|improve this question
up vote 0 down vote accepted

It is probably easiest to rescale the problem so that there is one meeting per day. It is certainly easiest for expository purposes.

Let us also assume that all people have the same probability $p$ of attending any meeting, and that the probabilities are independent. So the probability of all $N$ people being available at a given time is $p^N$. So the probability of having a meeting on exactly the $d$th day is $(1-p^N)^{d-1}p^n$. This is a Geometric Distribution, and it has mean $p^{-N}$. So we see that the effect of adding one more person extends the time til a meeting happens by a factor of $1/p$.

This is in a certain sense a discrete form of a poisson process with the Geometric Distribution being a discrete analog of the Exponential Distribution.

share|cite|improve this answer
Thank you! This is much simpler than the path I was trying to go down. The result is intuitive and easy to explain also. If I understand correctly, the probabilities are independent, meaning that my desired end result - waiting time when adding another person - can be stated in terms of the one person being added. i.e., adding in someone who is scheduled 3/4 of the time will extend the wait by 4x, while someone scheduled 1/4 of the time will extend it by 4/3. – evil otto Jun 3 '13 at 18:47
@evil Yes. If we have independence then the probability of missing an oportunity with one more person depends only on the probability of that last person. Independence is an often unjustified assumption to make in practice, and particularly so in this case, but dependence tricky to quantify,as we need to measure O(N^2) values as opposed to O(N). An assumption of independence is probably good enough for handwaving estimates. – deinst Jun 3 '13 at 19:50

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.