[mewpmewp2]

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.

[aidenn0]

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)

[svat]

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.)

[mewpmewp2]

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?