Bertrand's Paradox (with 3blue1brown) - Numberphile

Featuring Grant Sanderson, creator of 3blue1brown.
More links & stuff in full description below ↓↓↓

NUMBERPHILE

Videos by Brady Haran

Special thanks to our friend Jeff for the accommodation and filming space.

[hamgelato]

It must be nice to collab with Grant since he did his own animation

[NANO]

4:52 for anyone who was also confused why the area was not pi/4:
He meant "The inner circle has an area of 1/4 [of the outer circle area]".😅

[collardgreen]

Every other person: This is because of my bad drawing.

Grant: Thats a "function" of my Bad Drawing.
7:22

[kesim]

A triple cheers for bringing this important example up and making it known to the greater public! And so wonderfully rendered and explained, too!

[KillerMZE]

This ended in such a cliffhanger, never felt so compelled to see the extra footage

[alexshih]

I think this paradox is a perfect example of how slippery a lot of concepts in probability theory can be. Even Erdos got the Monty Hall problem wrong.

[nomekop]

This is a beautiful video. The style of numberphile mixed with Grant and his animations makes for a wonderful combination

[drakoz]

It's fascinating: each method is a different way of seeing the topology of the circle. The first is the classic S1, the second is the disc filling in S1, and the last is the disk as [0, 1]×S1. (In particular, quotienting out the S1 part).

My intuition tells me that the first and second should be different, but the second and third should be the same. Interesting!

[nileshpandey]

Back to back videos by Grant. He just released one video on his channel 3b1b. What an amazing day!

[jaredparkes]

Any point outside the circle will have two tangent lines to the circle; the intersection points of these lines to the circle will uniquely define a chord of the circle, thus any point outside the circle uniquely defines a chord. There is a finite region around the circle (bounded by a circle of radius 2r centred at a origin) where the chord defined by any point within will be shorter than s; any point outside this region defines a chord that is longer than s. The probability that a randomly selected point outside the circle will fall within the region where the chord would be shorter than s is 0; therefore the probability that a randomly selected chord will be longer than s is 1.


Select a random point on the circle, then construct a ray line from the centre of the circle which passes through the point on the circle which is 90 degrees clockwise from the selected point. A randomly selected point on the ray line, along with the selected point on the circle will uniquely define a line which intersects the circle at two points, which defines a chord. If the selected point on the line is within r/sqrt(3) of the origin, then the chord will longer than s. The probability that the selected point on the line is within r/sqrt(3) is 0; therefore the probability that a randomly selected chord will be longer than s is 0.


Any chord will be between 0 and 2r in length. s=r*sqrt(3). s/2r=sqrt(3)/2. Therefore the probability that a randomly selected chord will be shorter than s is sqrt(3)/2.

[jrcarlyon]

This collab is a better Christmas gift than anything I have ever got from my family

[bennmurhaaya]

This reminds me a little of one of my math exams questions back at uni, where it was about hitting a dart in the inner 2/3 of the target. I simplified it to 1D (2/3 of a line vs 1/3) not taking into account the fact that I cannot just get rid of polar coordinates like that. Here the different distribution of cords in the circle remind me of that. That some simplification that lead to a wrong distribution took place...

[TheNameOfJesus]

Most people don't even know about probability density functions, so this sort of topic will really challenge them. As someone once said, "The generation of random data is too important to be left to chance."

[rgfs]

Great example of why it’s important to state your assumptions! The problem isn’t in dealing with infinite spaces, it is in how you state what you know about one aspect of the space vs another.

In each example the probability of the first point being chosen is assumed to be uniform in a particular space. The first student, for example, assumes points are equally likely to lie on the perimeter, but in the second example points are equally likely to lie within the circle, not the perimeter. These don’t describe the same space and therefore lead to different results.

[Ganerrr]

"It's implicit when you're told to choose a random number between 0 and 1 you would use some kind of uniform distribution"

...wait there's multiple kinds?

[diobrando]

Couldn't expect less form a collab from my two favorite math channels. Awesome stuff!

[Awsom-]

I didn't know where it was going exactly, but immediately when he said that you pick a "random chord" I instantly thought "Define random. What's the distribution?"
And lo and behold, that turned out to be where the paradox comes from.
- If you pick a random chord by taking two random points uniformly distributed on the edge of the circle you get student #1's answer: 1/3
- If you pick a random chord by taking a random point uniformly distributed within the circle and make that the midpoint of the chord you get student #2's answer: 1/4
- If you pick a random chord by generating a random radial line and then choosing a random point uniformly distributed on that line to be the center of the chord you get student #3's answer: 1/2

[kubaissen]

The answer is simple. First, we simplify to a lower dimension. Now we randomly choose a value from the length of the circle (2 * pi * r), and then divide it by the leg length of the triangle sqr(3). So the answer is (2 * pi * 1) / sqr(3)

[nigeldepledge]

It seems to me that the biggest clue here was that the second method looked sparser in the centre than the first.

Clearly, the three different methods of "choosing a random chord" bias the outcome towards the three different results. The first and third methods are about choosing one or two random points along a line, whereas the second method is about choosing random points within an area. With the first method, there is twice as much circumference that will give you a short chord as there is circumference that will give you a long chord. With the third method, the randomly-chosen radius is irrelevant : what matters is the point on that radius. This method is giving equal probability to those points that will give either a long chord or a short chord.

The second method, randomly selecting a point within the area of the circle, has three times as much area that will give a short chord as there is area that will give a long chord. (Points that will give a long chord must occur within circle that has half the radius of the initial circle, and this smaller circle will possess an area that is a quarter of that of the main circle.)

And, making a welcome change, I worked this out for myself before the end of the video!

[henrym]

Choose a random point (uniformly) outside the circle.
The two tangents from that point to the circle intersects the circle at 2 points, which defines a chord.
Chord length is greater than s <=> distance of point to circle is greater than some distance d (I think it’s 2 but doesn’t really matter)

The probability is given by area outside circle of radius d / total area with the limit tending to 1