Best estimate of a count from a single observation

Question

A company produces tickets that are individually numbered. The first ticket is numbered 1, the second ticket is numbered 2, etc...

You get a ticket numbered n. Knowing nothing else, we wish to estimate N, the total number of tickets produced. What is the best estimate of N? n would appear like a candidate although it would definitely be a biased estimate as it represents the lower bound. Maybe 2n? Maybe half-way between the lower and upper bounds (that would be ∞)? Does the question even make sense?

Answer 1

This sound a lot like the German-Tank-Problem: https://en.wikipedia.org/wiki/German_tank_problem

The minimum variance unbiased estimator is (see also https://web.archive.org/web/20140223104835/http://www.rsscse-edu.org.uk/tsj/wp-content/uploads/2011/03/johnson.pdf for more explanation):

$m(1 + k^{-1}) - 1$ with

$m$: Highest Oberved Number

$k$: Observations, in your case 1.

Answer 2

I don’t believe one ticket is enough information. You might get something silly, like n=1.

If you could get more tickets, then you could apply the central limit theorem.

Google “Random Number Between”, and here https://www.calculator.net/average-calculator.html is an average calculator.

I myself let there be 100 tickets, and got five randomly chosen groups of four numbers each, without replacement. The average of the averages of the five groups was 41. Two more groups, rose the average of the averages to 45.

As I get more random groups, the probability distribution of what my average-average may be becomes a normal distribution, centered at N/2*, and so I’d more and more often would get a number close to 50 as the average-of-averages, as I evaluate the averages of more random groups.

***(this is because all numbers from 0 to N are equally likely. If this were human height’s being evaluated from 0 to 7 feet, then it would centered at the average adult’s height of 5.5 feet, not the midway point of 3.5 feet, since even though 0 to 7 are all possible, adults are more likely to exist than kids.)

This would let you estimate N, if you had more, preferably much more, than 1 ticket.

With just one ticket, all assumptions of what N could be are equally likely.

Unless you know they made like, at least 1,000 tickets and no more than 10,000. If your random number n was below 1,000, that implies N is nearer to 1000 than it is to 10,000, since if N was close to 10,000 you would’ve been way, way more likely to have gotten a number bigger than 1000.