My $6$ friends and I tried buying tickets to a popular event. Everyone who wanted a ticket got a random number and if your number is less than or equal the number of tickets available, you can buy a ticket.
None of us $7$ was selected to buy tickets. How can I estimate the probability of successfully buying a ticket? Using maximum likelihood on a binomial model the answer is $0$, but I know tickets were sold, so it can't be $0$.
The underlying statistical model is $X_1,\ldots,X_n\sim Bernoulli(p)$ iid ($X_i=1$ iff friend $i$ gets a ticket; $n=7$), i.e. $Y_n=\sum X_i\sim Binomial(n,p)$. You wish to estimate $p$, having observed $Y_n=0$ Here are two approaches.
The maximum likelihood estimate is indeed 0. You could construct a confidence interval. A rule of thumb in this situation is the rule of three: a 95% confidence interval is $[0, \frac3n]$.
You can perform inference in a Bayesian fashion. Your prior for $p$ might be a $Beta(a,b)$, for example Jeffreys' prior is $Beta(\frac12,\frac12)$. Your posterior distribution for $p$ is then $Beta(a+Y_n, b+n-Y_n)$. With Jeffreys' prior and your observations, the posterior is therefore $p|Y_n\sim Beta(0.5, 7.5)$.
If you insist on a point estimate, you can choose between the MAP (posterior mode), which is at 0 like the maximum likelihood estimate, or the posterior mean, which is at $\frac{a+Y_n}{a+b+n}=0.0625$. Of course, the estimate is different with a different prior.
You might also want to read about the sunrise problem, which is related.
