Poisson

You may notice I calculated by counting the total number of event during a period of time, while VW tried to focus on the frequency of a rare happening: being called 5 times in one night.

Of course the former method is generally preferable by statisticians because it considered all available information. (In their jargon, there is no inadvertent data reduction.) Nonetheless, let's look at the problem from the angle of our hepatology professor, and focus on the distribution of urgent request between call days rather than endoscopists.

In other words, during that year, what is the probability of being called 5 times a night?

Let's follow the simplified estimation and assume there are 300 days in a year. In other words, the average chance of being called in each day is one-third.

Now comes the critical question: What is the chance of being called twice?

No, it is not one-in-nine (1/3 × 1/3), as my friend tried to suggest. For uncommon events that happens this way, it follows the Poisson distribution, and the chance goes like this:

while p is the number of event you expect for that period, n is the actual event that happens, and e is the base of natural log. For things that we expect happening one in every three days (i.e. p equals one-third), the chance of having it twice in a single day is 0.0398 (i.e. 3.98%).

And, for n equals 5, the chance is 0.0025% (i.e. around once every 40000 days, or every 110 years).

So, VW, let's face it: You are the chosen one.