New video! The hardest problem on the hardest test

Great video. I think you can explain to the solution with a even more "geometrical" intuition. (no coin toss) In the 2D case the farthest two points on a circle can be from each other is half the circle (more than that and they get closer). The chance for a third point to make the required triangle is the same as for it to be found between the first two points. The portion of the circle between the first to points ranges evenly from 0 to 0.5, so the chance for the third point to make the required triangle distributes evenly from 0 to 0.5, which makes it 0.25. For the 3D case the logic is the same (and animation would make it realy nice). The most the second point can be from the first is half a great circle, and the most the third can be is half a great circle from the second and the first (it will end on the same great circle but on the antipodal point). The area between three points covers at most 0.25 of the sphere's surface. The portion of this area from the sphere's surface area is the same as the fourth point chance of making the required tetrahedron - even distribution 0-0.25 goes to 0.125.

alon rothem

Can you...elaborate?

3blue1brown

Hi, Grant! I am despair 😢... I don’t see Logic en Number Theory 😕

Sure thing, it's in the video description: <a href="http://lsusmath.rickmabry.org/psisson/putnam/putnam-web.htm" rel="nofollow noopener" target="_blank">http://lsusmath.rickmabry.org/psisson/putnam/putnam-web.htm</a>

3blue1brown

Can you give us a link to the paper showing the solution in terms of Linear Algebra?

I had not heard of this, and I'll check it out thanks!

3blue1brown

grant, i've come up with a way that pascal's triangle is self- refrencing i would prefer that nobody takes credit for this, so if you comment on this, I will reply with what it is. I just think you might find it interesting.

would <a href="https://donorbox.org/" rel="nofollow noopener" target="_blank">https://donorbox.org/</a> work for you?

seerpea

This is great. Teaching you how to think and approach these problems. Similarly in computer science, we would always ask ourselves when faced with a tough problem: How can we solve it in the most naive way possible. Then worry about making it more efficient after.

This is a very cool video! And it fianally helped me distinguish what I take away from your videos vs formal maths education :)

ChalkyChalkson