[jmilloy]

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.

[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?

[Vvector]

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.

[me_me_me]

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