4 ms·

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

• jmilloy 3y ago
  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 3y ago
    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 3y ago
      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)
      • mewpmewp2 3y ago
        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?
      • svat 3y ago
        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.)
  • Vvector 3y ago
    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 3y ago
      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