WLOG A=w, so it follows E[G]=2E[|A'-w|]. But E[|A'-w|]=MAD(2w), and we conclude E[X] = 2w - E[G] = 2w - 2 MAD (2w). 3/

The quantity we're interested in is E[X], where X=A+B. Let the number of ghost games be G, and note X+G=2w=A'+B' and G=(A'-A)+(B'-B). As the competitors are equally matched, E[A-B]=E[A'-B'] and E[A'-A]=E[B'-B]. 2/

President Trump says it perfectly. A patchwork of 50 different state systems creates a maze of conflicting regulations, resulting in chaos. Follow me to join the conversation on leading the AI revolution.

Here's one argument. We use the technique of "ghost games". We're racing to w wins. Now, pretend our competitors continued until 2w games were played, with A & B wins when someone first reached w wins, and A' & B' wins after our 2w games. 1/

Algebra?

One additional comment - this is a nice example of a problem that looks trivial, and it does indeed have a nice closed-form solution, but the closed-form solution is quite difficult to demonstrate (unless you do something like leverage De Moivre's result).

It's surprising and worrying that the University of Nebraska is shutting down its department of statistics Usually we think of this happening to marginal departments, but one would think that a good state university should have statistics? 1/

This type of formula also has a continuous version. For example, consider two competitors racing to w, where each scores points following the same Poisson process. What's the expected game length?

Major hint. This has a nice expression in terms of the mean absolute deviation (MAD), which has a nice closed form (found by De Moivre ~ 300 years ago). E[games] = 2w - 2*MAD(2w) = 2w ( 1 - C(2w,w)/2^{2w} ), where w = n+1 i.e. the competitors are racing to w = n+1 wins.

Man goes to doctor. Says he's depressed. Says life seems harsh and cruel. Doctor says, "Treatment is simple. Great clown Pagliacci is in town tonight. Go and see him. That should pick you up." Man bursts into tears. Says, "But doctor... I am Pagliacci!"

I'll post the solution in a short while, but both ChatGPT and Gemini really struggled to solve this (relatively simple) problem in the best/simplest way.

Happy Caturday!

Hint - this can be a very tricky problem unless you reduce it to one with a known solution.

Nice guest post on Buzzard’s blog by Boris Alexeev on his and others’ use of AI systems to attack Erdos-type problems (thanks to Bloom’s @thomasfbloom website collating them): https://t.co/XBgHLly41H In particular, with systems like Harmonic’s @HarmonicMath Aristotle,

An interesting phenomenon is recent AI engines seem to have become reluctant to attempt a problem if it's known to be an open conjecture/unsolved beyond a certain level of notoriety. This may depend on prompting, as might searching for a simpler solution to a solved problem.

This can also be written as 1/m! sum k_m, where k_m is the falling factorial. We can now derive a Rota-style umbral proof by performing the analogous integral integral_{x=0}^{n+1} x^m = (n+1)^{m+1}/(m+1), so our answer is (n+1)_{m+1}/((m+1)*m!), or C(n+1,m+1).

Say two competitors are equally matched and compete at a best of 2n+1 games format. What is the expected number of games played?

Related to Pell-type equations and Binet's formula for Fibonacci numbers. The 5n²±4 test is equivalent to x²-5y²=±4, whose integer solutions are (Lucas_k, Fib_k). The quadratic behaviour can also be seen from the Fibonacci companion matrix via Cayley–Hamilton / finite fields.

It may ultimately generalize better with mathematics (and physics?), as we have engines that will verify a proof with certainty, and the library of known results (and tactics?) will grow over time. Library creation (and tactics generation) is a moonshot project, however.

Combining n Schwarzschild black holes with masses m_i and considering the Bekenstein-Hawking entropy before and after also yields the Cauchy-Schwarz inequality.

Here's my writeup of the proof of Telgarsky's conjecture. not independent of the AI proof. I have already seen the 'informal proof' file in the github repo. But it confirms my suspicion that it's similar to my earlier answer to a different (related) question by @aryehazan