Bertrand's paradox (with Numberphile)

nice... rather than thinking about translation in-variance, could it also be correct to think in terms of infinite sets of cords? Like should the solution not be the union of all infinite sets of possible cords and thus maximum of the set of probabilities of possible sets of cords? thus 1/2?

Bernard

This was very fun to watch and listen to. I also found the 1/2 answer to seem the least appropriate when originally looking at the problem. In a house-fire scenario, I would pick the 1/3 probability, as I define a chord as the line which connects two points on the perimeter of a circle. I'm now thinking about the very first question which is what is the mean length of that chord... How would you set that up for the third (1/2) definition of a chord? Could you just integrate the height of the circle over x and divide by x?

John Nichols

This feels like there should be a connection to that other numberphile video you did about the dart game with the area of the dart board shrinking based on the size of chord generated by a dart hit. Wouldn't the third method correspond to the dart throwing method where you are taking a random point drawing a line from the centre and then taking the chord? In that way the third method does feel somehow less biased than the other methods, in that it can be generated by uniformly blasting the circle with points but the first method still somehow feels right too.... I wonder what happens if instead of two random points you have one rando point on the circumference and then a distance traveled.

James Matheson

I thought of a variation on the 1/3 way, where the distribution is biased by the length of the lines (so corresponding with area). That one gives the area of the 60 degree sector (probability of around .609).

Kevin McCurdy

Hi Grant, great video as always. My take on the paradox is that the naive assumption that the number of point in the larger annulus (in method 2) is greater than the number of points in the smaller central circle. There are the same number of points in each and so the probability of choosing from one set is the same as from the other set. Likewise with method 1, although the length of the circle where the chords would be shorter is twice as long, the number of points in the longer and shorter sections are equal and so the probabilities are equal. The approximation methods, being inherently discrete, create distortions which throw off the estimates.

Other response. Here is a Desmos visualiser on the probabilities when you define the chord according the video's three ways: [ EDIT: https://www.desmos.com/calculator/hhebl403x8 ] Notice how a uniform probability over your starting choice is distorted by each definition. Here lies the secret. - Option 1 "choose randomly two points on the perimeter" essentially boils down to "choose randomly the subtended angle". One dice-throw equivalent over this angle exerts a cos^-1 distortion. - Option 2 "choose randomly a point in the circle" boils down to "choose some random radius but biased for larger r", whose dice-throw equivalent is tranformed by a Pythagorean relationship. - Option 3 is related to option 2, but no longer biased for larger r. What you've done is to shift the 2D dice-throw from (x,y) to (r,theta).

Poker Chen

Maybe an imgur link? Put it on the 3blue1brown subreddit?

3blue1brown

Within the context of Bayesian reasoning, your assumptions are free and you declare them together with the proposed answer. The usefulness is more about the process of deduction than the certitude of the final values. I.e., one would declare their beliefs on the possible definitions of a chord, resulting in a probability over the three answers 1/4, 1/3, and 1/2. It makes intuitive sense in the dice rolling analogy, where the more I know Grant (e.g. what table top games he plays), the better I can tell what kind of dice he has actual access to. He may have D100s, but he's unlikely to have stockpiled more than a few of them, whereas if he plays 40K Orcs he's might actually have 100 D6s, etc. So I can choose to believe that a proposed answer 1/100 is going to be a lot less probable than an alternative 1/6.

Poker Chen

I want to post my solution, but I can't paste an image... Sigh, Patreon....

Ah. I will just free hand them on the whiteboard like I have always been doing, then. But if you do find a way to mass produce them and sell them on your store, I'd be interested in buying one.

Kevin Iga

That was actually a custom-made piece from my dad, who enjoys woodworking on wanted to practice a gold leaf inlay on something.

3blue1brown

What's the wooden pi circle maker? Where can I get one?

Kevin Iga