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Re:The German Tank Problem (2010) by justinsaccount on Friday November 06, 2015 @12:50PM (#10520776)
It's not "add a tenth":

> For example, let’s say 10 tanks were captured

It's "add 1/n" where n is the number you collected.

Think of this related problem. Numbers are randomly chosen from 1 of 2 ranges. One range is from 0-10, the other is 0-100. The numbers are:

4 3 1 8 5 2 7

Which range were the numbers chosen from? It could be either, but statistically the answer would be from 0-10, since if they were chosen from 0-100 the chances of all of the numbers being lower than 10 is very low.

Now, say you didn't know the ranges ahead of time, and the numbers were:

117, 232, 122, 173, 167, 12, 168, 204, 4, 229

Now, the upper bound can't be 100 or 200, since there is a 232. So it is at LEAST 232.

Which is more likely, that the upper bound is 256, 500 or 1000 ?

Intuitively you can assume that if it was 500+, you would expect at least one number in the 300-500 range.

The highest number should be 232 + some amount that is inline with all the other numbers. It could be exactly 232, but the chances that you happened to pick the highest number itself is also low.

Basically you are solving this problem:

If you were going to select 10 numbers at random from a range of 0 to N, how high should N be so that there is a high probability that the highest number you select is only 232.

Since a uniform distribution is assumed, you are basically just adding one more uniformly distributed data point.

Re:The German Tank Problem (2010) by JeremyHoward on Friday November 06, 2015 @06:49PM (#10522823)
It's actually "add m/n" where n is the number you collected and m is the highest sampled serial number.
Re:The German Tank Problem (2010) by SeriousM on Friday November 06, 2015 @03:47PM (#10521918)
I had the same question and your answer was pretty clear to me. Thank you!