Ready?

So I've got for you a puzzle.

It's about darts.

It involves some higher dimensional geometry on a couple of the most famous numbers in math.

So I'll describe the game.

We want to hit us many bulls eyes as we can right now.

The bull's eyes, this tiny red thing.

But that's too small.

We're gonna start with the giant bull's eye that fills up the entirety of the board.

I have a proper compass.

I'm gonna go ahead and cut out a piece.

Just a piece of paper to indicate are enlarged Bull's eye, not the greatest circle I've ever drawn.

All you do is you try to hit the bull's eye each time we hit it.

Though based on where are shot is the bull's eye's gonna shrink.

So it's gonna get harder and harder as we go.

You're gonna hit the wall, eh?

I'm not that bad.

Okay.

Oh, my goodness.

Hang on.

That's amazing.

Thank you very much.

That is amazing.

And you're kind of rewarded for a good shot.

So let me tell you what the rule is for how we shrink it.

So, Steph one I'm going to draw a line from the center to where I hit and maybe for the future.

I'm just gonna call that distance H for That's the distance of my hit.

And then what I'm going to do is draw a cord of the circle that's perpendicular to this line.

That's our court, and our rule is that the length of this cord is the new diameter of the circle, which in this case, because I had a good shot, right, the shot was nice and close to the center.

I'm rewarded by the fact that it's not gonna shrink that much.

Is the center of the new circle where your dad hit or where the original straight question.

It's where the original center waas, so I'll draw a new circle where the original center was actually wish my shot was worse Now, because this is kind of too close to the edge.

I mean, you can kind of see it's it's barely smaller.

Okay, so that's just trimming that circle down of it.

So I'm just gonna be giving it a little trim is like getting your hair cut, you know, once every week or something like that.

And presumably this means that the next shot is gonna be similarly difficult.

It's actually fascinating, because when I animated, then it's like, Oh, boom, the computer does it.

And I'm not messing with the physical dartboard in any way.

I think I'll stick to the animated style for the sake of illustration.

I'm gonna try to make this the worst shot.

Oh, okay.

So if I do the algorithm again, like this court is a little bit shorter because we trimmed off the edges, right?

And so it'll also be just a tiny bit smaller so we can have could run that again.

But let me let me try to get a worse shot.

Can't even do it.

Consistency is only a virtue if you're not a screw up.

So this one's nice and far away from the center on.

Gonna be punished more because it's far away, so we'll do the same game.

We'll take this off.

We're gonna make some cuts.

Distances, H one.

And so the accord of the big circle perpendicular to that line Cord defines the new diameter, but it's kind of easier to think of as half the court is defining the new radius.

Certainly, if you're drawing with a compass.

So I'll set my new radius.

But this time, because I was closer to the edge of my quote.

Unquote Bullseye, That court ends up smaller, so I'm punished.

Which means that the new bull's eye is meaningful.

E smaller, smaller goals.

I play the same game.

I'm gonna set it up so that the center is at the center of the board.

Now I'm gonna actually try to hit closer to the center because I want more shots for a higher score.

And I didn't hit close to the center.

Yeah, you're gonna get punished for that.

So at this point, maybe you can already see how much that's gonna punish me.

The idea that it's close to the edge means that the cord defining the new radius he's gonna be quite a bit smaller.

Put it on the board.

Now we see how good you are.

So far, I've hit three.

This is for 1/4.

All right.

I missed Oh, dear.

Game over.

Game over.

So what's so your score for that?

Game three hit three.

I will say that would be the natural way to score it.

Based on what I want the answer to look like a the end.

Here, you get one point just for playing.

Okay, so you get one point just for playing and then one point for each bullseye that you hit.

So in this case, I got to score for you can say, maybe you're what we're counting is how many darts you throw in total.

So I threw four darts and the 4th 1 happened to be a mis.

Whatever you want to do to artificially add one, you'll see why we like that at the end.

I have to find a game.

I haven't defined the puzzle yet.

So for the puzzle, let me just draw our board again, or at least draw circle.

All right, so we're gonna have a random darts player.

So he throws it the board with some random distribution.

We all do.

But there's no skill hit is no.

There is absolutely no waiting for the skillet.

Could it's is likely to be anything.

This will be a very bad in fact.

So he throws it in a weird way.

We're gonna think of the square that encompasses the board.

And this is a little unrealistic because any real dart player.

Their distribution is probably rotational e symmetric, right?

I mean, why would it care about the diagonal directions?

But basically he hits it at a random point anywhere within a square encompassing whatever.

Our first circle, waas.

Okay, so maybe you imagine that the whole dartboard is a square, and then we'll be shrinking within that.

That's my ex co ordinates.

And maybe let's say that this happens to be a radius of one for the original circle, because why not?

The X coordinate for this person is always gonna be something between negative one and one, and it'll just be completely uniform within that.

So this would be the simplest thing to model on a computer.

To do it.

You just choose two different numbers, each one between negative one and one where every one of those is equally probable.

So every point in this whole square is equally probable, and that square doesn't shrink.

It's only the circle that shrinks.

And the question is, what's the expected score?

Okay, so we want to know he expected score for this thrower.

So grant, after you throw your first stop and the circle shrinks, the square doesn't shrink with it.

It does not drink no.

So the distribution for the thrower is the same.

You might imagine the dartboard itself, which is fixed a square and the bull's eye is shrinking because otherwise it's gonna be the same probability for each shot on.

Some viewers might be able to guess what that probability is, and it wouldn't be as interesting a problem.

Another way to think about the expected score.

Imagine this random player plays 1000 games, right?

Just what's his average score for those 1000 games?

Another way to think about it is you say it's gonna be the probability that the score is equal to one times one for that score, plus the probability.

Because, remember, the lowest score you can get is one, because you get that one point just for playing the probability that the score is equal to two times two, plus the probability that the score is equal to three times three and so on.

So what we need to figure out is all of these probabilities in principle, you could have a perfect game, right?

If you hit in the center every single time and it never shrinks, you're hitting a perfect game.

Um, what we'll find is that that's there's a probability of zero that you hit, Ah, Bullseye infinitely many times.

But it's also interesting.

How quickly does it shrink?

Would a score of 10 be absurd?

Or would that be maybe expected?

What is the probability off getting an exact bull's eye?

Because an exact bull's eyes, like presumably an infant decimal.

The small point is probability.

Zero.

Yeah, So, like one of those paradoxical facts about probability is, you can have events that are possible.

It just probability.

Zero right.

It's possible to hit in exact bull's eye.

It's just probability.

Zero.

You know, if you choose a random number from the number line, it's possible that it's rational.

But it's probability.

Zero.

And this.

A lot of people kind of try to come to philosophical terms with, like, how could it be possible?

But probability?

Zero.

The probability would have to be slightly bigger, but it's just one of those things of math, all right, so any time you're solving a puzzle that's hard and this this is hard, you see if you can ask an easier version of it or some easy sub problem within it.

What's the easiest thing you can get a foothold in?

And maybe in this case, the easiest question to start that's not totally trivial is what's the probability of hitting the first shot?

And this isn't that your score is exactly equal to one.

If you hit a first shot, your score is greater than or equal to one or no, actually, because you do get that point for free.

This is the probability that your score is greater than one.

Actually, maybe you can tell me, right?

What's the probability that on this first shot with this random distribution within the square, it hits somewhere inside this hole?

Sir, I'm imagining we we need the area of the circle in the area of the square, the area of the circle.

This is pi R squared, and that radius starts off as one.

So that ends up just being pie.

And the area of the square is equal to well, the square side length is two times that radius, right, so it's two by two.

So it's four, which means the probability that you hit that first bullseye, that your score will be greater than one because you get that one point free just for playing is gonna be the area of that circle divided by the area of that square ground.

What happens if the dart lands on the line?

That's probability.

Zero.

Don't worry about it.

It won't mess with the problem.

So now Ah, much, much harder.

Question is, okay, you've hit that first shot.

What's the probability that you hit at least two?

And this is where it starts to become interesting, because this depends on where your first shot Waas.

I'm going to start writing a little bit of a chart where I'm going to keep track of the radius of each bull's eye, and I'm gonna keep track of just the hit length and each point.

And so the radius starts off is one and that first hit.

It's a it's a random point.

Let's just give it a name for now.

Let's say it goes there.

That's H zero so the h zero and by the Pythagorean theorem, if we want to describe what this is in terms of X and Y, which will be helpful for later on to be thinking of it.

In terms of that, it's the square root of X squared.

That's why I squared.

If those are the coordinates of your first shot and because we're gonna have other shots, I'm also gonna give these guys a sub script.

So now it's a geometry question asked, What's the new radius?

Maybe we call this our zeroth radius night, and I want to know what's the next radius after that initial shot.

And if we remember a rule, they say, you drive the cord and the court was defined to be perpendicular to that.

So when you know that it's defined to be perpendicular, you know that you're probably gonna use that fact in your solution to the puzzle.

So the new Radius is this length here, right?

It's half of that court.

So what we might do and I recognized all sort of be drawing over myself of it here because we have a right triangle here, and that's what defined it.

We're going to use protection the Pythagorean theorem, because this was our old radius radius one.

That's the new iPod news.

So we know that the old high part news one squared is equal to one of the legs, which was h not squared, plus the new radius squared Plus are one squared.

So what that means for what are one is is It's the square root of one squared minus h not squared.

And remember, the only way that you're even getting to this point is this.

Each knot was less than one if you got that 1st 1 in the bull's eye, so this will be nice and well defined.

No imaginary numbers or anything like that.

Similarly, Now we're gonna have a new shot.

It's determined by two new random coordinates.

Whatever the coordinates of your second shot, I happen to be where the coordinates of the next shot.

Maybe I should say.