A brainteaser to solve before Easter
David Vose
Director of Vose Software. Risk Consultant. Advisor on quantitative IRM deployment.
Imagine you have a row of wicker baskets, and you are playing a hide-and-seek game with the Easter Bunny (EB).
EB is hiding in one of the baskets – silent, scarcely breathing, not even a twitch of its nose. You have no idea which basket EB is in. You are allowed to open a basket. IF EB is in the basket, you win. If EB isn’t there, you have to close the lid and EB will magically jump to a neighbouring basket at random - instantaneously.
Then you get to try again.
Suppose there was one basket, it would take just one attempt of course.
If there were two baskets, you open a basket and have a 50% chance of catching EB. Next try you open the same basket, because EB had to jump. So, on average it will take you 1.5 attempts to find EB.
If there were three baskets, you open the middle one and have a 1/3 chance of catching EB. If it's empty, you open the same middle basket again, because EB had to jump from whichever side to the middle. So, on average, it will take you 1 and 2/3 attempts.
But what if there were four baskets? What strategy would on average require the smallest number of attempts to catch EB?
Seeking God's will for me--One day at a time.
4 年I just use an ethernet to capture all my ether-bunnies!
Sr. System Safety Architect | Autonomous Vehicles at NVIDIA
4 年I think 2-2-2-3-3-2 guarantees EB capture.
Vice President, Quantitative Risk at Archer IRM
4 年22332 is the best I can do, so far. Brute force. 2.44 turns, but might be adding that up wrong
Vice President, Quantitative Risk at Archer IRM
4 年Is there a better strategy than the obvious one?
Risk Specialist / Part time Associate Professor Extraordinary at Stellenbosch University
4 年I will invite 3 friends and we will open the baskets at the same time. So, one turn.