Luck

Let's ask the question again in a mathematical manner: Was our professor of hepatology doing urgent endoscopy more frequently than that expected by chance?

The calculation should be done like this:

• Let's assume there are x endoscopists on the list, each sharing the same number of call.
• For a defined period of observation, there are n urgent endoscopy request, and each of them is independent from another.
• For each urgent endoscopy, therefore, the probability of a particular endoscopist to be called would be 1/x; let's call this probability p (in other words, p = 1/x)

For example, if we have 5 endoscopists sharing the duty, the probability of being called for each one in a particular case would be 20% (i.e. 0.2). If, as suggested by VW, there are 100 urgent requests in one year, each endoscopist would be expected to be called around 20 times. (The statistical jargon is: this is the mean value.)

Now, here come the critical step: How many times would be too many to be expected by chance (or random distribution) ?

Simple. The allocation of event (of urgent request) to independent categories (i.e. endoscopists) follow the binomial distribution, and the calculation goes like this:

• For that one year and 100 endoscopy request, the variance of allocation is n × p × (1-p)
• The standard deviation (SD) is the square root of the variance, and
• the upper limit of expectation (i.e. 95% confidence interval) is 1.96 (or 2) times SD.

In this case, therefore, the variance is 100 × 0.2 × (1 - 0.2), i.e. 16; the standard deviation is 4. In other words, the maximum number of request that you would expect to encounter by chance during one year is 20 + 2 × 4, i.e. 28.