# how do I model probable time until simultaneous availability?

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?

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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 $1/p$.