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Comment by Vvector | Hacker News Reader

Those odds are 4 quadrillion to one, if the numbers are picked randomly.

I think it's much better because the teacher doesn't have to match guesses to students. For example, for each student the odds are 30/100, roughly one in three. And any duplicates can be matched by a single guess.

I think odds should be 0.3 to the power of the amount of students.

E.g. teacher picking 1-30 and then each student has 0.3 odds of picking 1-30 or 31-100.

The issue is I think all it would take to beat the teacher is one unusual student.

That formula is missing something because for over 100 students the teacher can't lose. (they get over 100 guesses and there are only 100 numbers possible)

If the teacher does have 100 guesses then they wouldn't lose right.

Then it would be 1 to the power of 100.

I guess I should've clarified that the 0.3 refers to being able to choose 30 out of 100 numbers?

More precisely, if there are N students, the probability is (min(N,100)/100)^N. This is 1 for N ≥ 100. And the probability at N=30 is indeed a tiny 2e-16, which shows that the children's "random" picks were far from uniformly random.

(Incidentally, even with N=99 the probability is 0.37 ≈ 1/e, and the probability is lowest at N=37 ≈ 100/e. This is not a coincidence.)

Reverse it, and it becomes clear.

The teacher picks 30 numbers out of 100. Then each student (independently) picks one number. If random, that is 0.3 ^ 30. Obviously, the students are not picking random.

If I had to pick for the teacher:

multiples of 10: 10,20,30,40,50,60,70,80,90

double digits: 11,22,33,44,55,66,77,88,99

Not sure where to go next.

3 and 7 are most common digits people come up with with a random number 1-10

i would pick a 3-7, 30-37, 70-77, then some other fews from there like 1, 100, 50 etc

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