Can you steal the most powerful wand in the wizarding world? - Dan Finkel
The fabled Mirzakhani wand is the most powerful magical item ever created. And that’s why the evil wizard Moldevort is planning to use it to conquer the world. You and Drumbledrore have finally discovered its hiding place in this cave. The wand is hidden by a system of 100 magical stones—including a glowing keystone—and 100 platforms. If the keystone is placed on the correct platform, the wand will be revealed. If placed incorrectly, the entire cave will collapse.
The keystone is immune to all magic, but the other stones aren't, meaning you can pick them up and cast a placement spell, and the platform that stone belongs on will glow. Place all 99 stones correctly, and the final platform must be the keystone’s correct resting place. You’re about to get started when one of Moldevort’s henchmen arrives and irreversibly seals a random stone to a random platform.
If you need to place a stone that belongs on a platform that's already occupied, your spell will make some random unoccupied platform glow instead. What are your odds of placing the keystone on the correct platform? Pause now to figure it out for yourself.
Answer in 3. Answer in 2. Answer in 1. Let’s imagine we knew everything about this situation. With perfect knowledge, we could label the stones 1 to 100, based on the order we plan to place them, and label the platforms they belong on in the same way. We’ll label the stone the henchman placed as 1, meaning it was supposed to go on platform 1, and the keystone as 100, belonging on platform 100.
Of course, we don’t know which platform is which, so the numbering of the platforms is actually invisible to us. There are three possibilities: one, that first stone was placed randomly onto its own platform, in which case, you’re guaranteed to succeed. Two, it was placed on the keystone’s platform, and you’re doomed to fail. But most likely—scenario three—it was placed somewhere else.
Suppose the henchman placed stone 1 on, say, platform 45. Then you’d place stone 2 on platform 2, 3 on 3, and so on, until you got to stone 45. Its platform being taken, a random platform would light up. And here, there are three possibilities: If it’s platform 1, you’ll win, because all of the remaining stones will go to the correct platforms. If platform 100 lights up, you lose, because the keystone’s spot will be taken. Any other platform, and you’re essentially back where you started, just with 54 remaining stones and one on the wrong platform.
In that scenario, let’s say the spell tells us to place stone 45 on platform 82. Then we place 46 to 81 correctly, and 82 at random. And here we reach the same three possibilities: pedestal 1, you win; pedestal 100, you lose; any other, you continue the process. In other words, you’re playing a game where you have equal chances to win and lose, and some chance to delay the decisive moment.
No matter how many times this process repeats, you’ll inevitably either place a stone on pedestal 1 or pedestal 100 before you reach the keystone. That’s all that determines whether you succeed or fail, and critically, the chances of those events are equal. This can be unintuitive, so let’s imagine another, similar game.
Say Drumbledrore magically generates numbers from 1 to 100. If it’s a 1, you win. If it’s 100, you lose. If it’s anything else, he picks again. Since the odds of winning by getting 1 are the same as losing by getting a 100, this is a game you’re just as likely to win as to lose. It might take a while, but the delays don’t give an advantage to getting a 1 before 100, or vice versa.
The same essential reasoning applies to our situation. You’re debating whether it’s worth risking a 50/50 chance of a cave-in when Drumbledrore reveals his secret weapon: a rare felush felucious potion, which grants extraordinary luck for a brief period of time. There’s a 1 in 100 chance the keystone’s platform was taken by the first stone and you’ve lost already, but otherwise, you’ve got even odds to win or lose. And right now, you’re feeling lucky.