Even if the questions were perfect (each question splitting the population in two exact halves, and all questions totally independent from each other) and therefore the algorithm would give each person a perfectly random number, the birthday paradox [1] tells us that even for just square(2^33)=~ 93k people we would have 50% probability of having a collision. To work we would need more bits. (Either that or create questions that are _not_ independent, so crafted in a way to make sure each person gets a different number)
No: the linked article requires that further questions split the previous groups each in half. The first question must split the group in half, the second question must split those two groups in half, and so on. When we hit the last bit, the final question is splitting groups of 2 people into the last bit, and each person has been assigned a unique number.
In this manner, 33 bits is enough, and the birthday problem is avoided. The page mentions all this.
[1] http://en.wikipedia.org/wiki/Birthday_problem