Two Eggs 1

You are in a 100-story building and have two identical eggs. An egg will survive a fall from below a certain floor number, $$X$$ , but will break if dropped from floor $$X$$ or above. Your task is to determine the least number of egg drops required to accurately find $$X$$ in the worst-case scenario. What is the minimum number of drops needed to determine this floor $$X$$ ?

Answer

Suppose that we have a strategy with a maximum number of $$N$$ drops. For the first drop of the first egg, we try the $$N$$ th floor. If the egg breaks, we then start with the second egg from the first floor and incrementally increase the floor number by one until the second egg breaks. In this case, at most, there are $$N-1$$ floors to test. However, if the first egg dropped from the $$N$$ th floor does not break, then we have $$N-1$$ drops left. We next go to the $$(2N - 1)$$ th floor. If the egg doesn't break again, we proceed to the $$(3N - 3)$$ th floor with $$N-2$$ drops left, and so on. This pattern continues until we find the critical floor or cover all 100 floors. Mathematically, the number of floors covered with a maximum of $$N$$ drops is given by the sum

N + (N-1) + (N-2) + \ldots + 1.

This sum must be at least 100 to ensure we can test every floor. Simplifying, we get

\frac{N(N+1)}{2} \geq 100.

Solving this for $$N$$ gives us $$N = 14$$ . Therefore, the minimum number of drops required in the worst-case scenario to determine the exact floor $$X$$ is 14.