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 henchmen 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.
