2. 80% of the people +1 will be correct.


1. This is what we would call an ill-posed problem, because we have no information as to how the number of people in the room is selected. We can work this problem if we assume that this number is chosen randomly to lie between 0 and some N, with all numbers equally likely (uniform distribution). Now, the equating of a uniform distribution with total lack of knowledge is a topic of philosophical debate, and has led philosophers to some rather strange conclusions since Plato. Nevertheless, here is the result of that assumpion in this case:


From the definition of conditional probability:


P(A|B) = P(AB)/P(B) = P(B|A)P(A)/P(B)


where P(A|B) means the probability of A given B, and P(AB) means the probability of A AND B.


Let A be 25 people or less

Let B be you are 5th


P(A) = 25/N

P(B) = 1/N(1/5 + 1/6 + 1/7 + ... + 1/N)


that is, the probability that there are any number of people n in the room is 1/N, and the probability that you are 5th for that n is 1/n.


P(B|A) = 1/25(1/5 + 1/6 + 1/7 + ... + 1/25)


P(A|B) = sum[n=5 to 25](1/n)/sum[n=5 to N](1/n)


Now I ask:


3. If you had to guess how many people were in the room, what guess would have the highest probability of being correct.


4. What guess would on average come the closest.


5. What guess would on average minimize the mean squared error (average of the square of the error).


Make the uniform assumption for 3-5.