## Washrooms in a plane

2010-04-08

Most people have some familiarity with the following situation: you’re in the plane, already crammed into your seat and you need to go to the washroom. But, at least in the planes I’ve been, the washroom seems to be occupied all the time. So you go, stay in line, get in, and when you get out the line is still there – other people waiting to take your place.

What is the actual probability to have the washroom occupied during your flight? This should be relatively easy to answer. Suppose we have N people in a plane with Nw washrooms. We want to know P(all washrooms are occupied) and we will denote it with P:

$P = \prod_{k = 1}^{Nw} P(\text{washroom } k \text{ is occupied}) \quad (1)$

We can safely multiply in the previous relation since the probability that a washroom is occupied is independent of other washrooms being occupied or not.

To calculate the probability that a washroom is occupied we need to know in average how many times a man uses the washroom and for how long.

We can infer the first value by knowing the normal volume of urination and the volume of the human bladder. Let’s call this Np and based on the previous links it can be anywhere from 3 to 6.

For the second parameter I couldn’t find any … informative links so let’s just call this tw and give it a reasonable value of 3 minutes.

The probability that a man will be in the washroom is:

$P_{hp} = \frac{Np \cdot tw}{60 \cdot 24} = \frac{Np \cdot tw}{1440} \approx 0.0083 \quad (2)$

Of course you can try your own calculations using your preferred values for Np and tw. Also I had friends arguing that my whole argument is not valid since they know people who stay more or less than my assumed values. Of course I’m aiming here for an “average” probability, and if you want to know exactly the value for your particular flight/home/work place, you can experimentally find the above probability per person and plug it in the equations below.

If we have N people in the same place and one washroom, the probability that the washroom is occupied will be:

$P(\text{washroom is occupied}) = 1 - P(\text{nobody is in the washroom})$

$P(\text{washroom is occupied}) = 1 - (1 - P_{hp})^N$

since P(nobody is in the washroom) equals P(person 1 is not in the washroom) and P(person 2 is not in the washroom) and … and P(person N is not in the washroom) which are independent events and thus can be multiplied together.

Going back to equation (1) we can write now:

$P = \prod_{k = 1}^{Nw} [1 - (1 - P_{hp})^{(N - k + 1)}]$

And to finish with a mental image here is a plot of the P(all washrooms are occupied) for different values of N and Nw. Considering that a reasonable plane capacity is around or above N = 200 you can see for yourself that there’s a good reason the washroom is almost always occupied.
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