- Nov 5, 2018
This made me think of something.I agree that this WR is far from unbeatable, but some of this math is pretty far off.
Let's eliminate the probability of nerves since that's so hard to estimate, just so we can get an idea of how many solves have WR potential. Given your other probabilities, we end up with a 0.000946 chance of a potential WR solve. I'm very skeptical of your 29.6% estimate for 2 three-move pairs in a row (and @Sajwo brings up some good points above), so let's be conservative and estimate a 0.0001 chance of a potential WR solve.
You can't add probabilities like you did in your post; the correct formula for the chances of a WR-potential solve is:
\[ 1-(1-P)^n \] where P is the probability of a given single solve having WR potential, and n is the amount of solves done. Let's say we want a 50/50 chance of a WR-potential solve. This would require: \[ (0.9999)^n=0.5 \]
Solving for n we get approximately 6931 solves.
This means, assuming each solve has a 0.0001 chance of having WR potential if done by a world class cuber (which is a very hard number to accurately estimate), 6931 official solves would need to be done by world class 3x3 solvers in order for there to be a 50% chance of one of these solves showing up. Additionally like you mentioned, they would have to control their nerves, which is definitely quite difficult to do.
I'd be surprised if this record is broken before 2020, but it's totally possible for it to be broken. These calculations make too many assumptions no matter what, since we can't account for improvements in hardware and solving methods. There's always a chance of someone coming up with a method that gets 30-move solves on average, and in that case, a 10TPS average solve would beat the record.
I forgot to mention that the solve was a Petrus solve. That’s really cool even though it was meant to be CFOP. If he had meant to do Petrus and didn’t get an EO skip, it still could have been a sub 5 time if he was good at EO. This just shows that CFOP is not always the best method.