Great Players of the Past: Richard Réti 

He could do a mediocre players of the present and include us directly

@@chowder6148 I don't think Ben considers me mediocre

@@qazzaqstan I think Ben considers anyone besides people with the names "Benjamin" and "Finegold" not mediocre.

@@chowder6148 Jokingly probably true. That said I kind of think of a mediocre players of the present lecture as being anyone between 2450 and 2600, and then we feel extra bad.

@@qazzaqstan One more loss (or tie with a lower rated player) and Ben is officially a FIDE 2300-level player.

Wait, is that Ben’s life story?

Cobra Kai 😭😭

Takes some mental math, but I calculated to the same answer in the exact same way in my head in a minute and a half, so I certainly think a minute is within the realms of possibility :)

Thinking about it, I think I figured out how to derive the formula in combinatorics. The answer would be ₆C₃ × ₇C₁ + ₄C₂ × ₇C₃ + ₂C₁ × ₇C₅ + ₀C₀ × ₇C₇. Plugging in the values gives 20×7 + 6×35 + 2×21 + 1×1 = 393. That still seems like a huge amount of work to do in your head. It's certainly too much for my feeble mind.

I came up the same combinatoric argument as Zanti but I wasn't convinced it was right. To verify I also got 393 by reducing a recursive function counting the number of legal paths of a given length and sideways drift until it gave the answer. I knew this method was guaranteed to work modulo arithmetic errors. I needed about a page of math to do it.

I solved it in around 30 seconds by a combination of reading your comment and permuting the answer of 393 into this comment

More fun for the math geeks, and I do include myself in this group: Imagine the chessboard is 3×n, i.e. the board has just an a-file, a b-file, and a c-file. Start the king on b1, and count the total number of paths possible to each square as the king moves up the board. You get the following table: 2nd rank: 1 1 1 3rd rank: 2 3 2 4th rank: 5 7 5 5th rank: 12 17 12 6th rank: 29 41 29 7th rank: 70 99 70 8th: 169 239 169 This sequence ties exactly with the Pythagorean triples where the legs of the triangle differ by exactly 1. In geometry you learn that Pythagorean triples can be generated by seeding integers m and n into the equations m² - n², 2mn, and m² + n². The m/n values that yield Pythagorean triples where the legs differ by 1 are: 2/1 generates 3-4-5 5/2 generates 20-21-29 12/5 generates 119-120-169 29/12 generates 696-697-985 70/29 generates 4059-4060-5741 etc.

The answer is fries

He died of scarlet fever so it checks out

@@kmarasin lol 'checked out'

It made me check the date of the lecture because I was wasn't expecting it to be new but I was curious if he was referencing Magnus not defending his title.

Which is really prophetic, considering this lecture was recorded in 2017, haha. He KNEW.