When the Odds Are 1 in 241 Million

“Chat baby gronk just fanum taxed my nr"
- Jake Brown 2024

Chapters
0:00 Intro
0:45 3.91
1:39 The Solution
3:05 3.69
3:44 Breakdown
6:43 More Thoughts
7:20 Outro

Some footage from @AidanGrainger and @PieceByPiece1046

Scramble (both solves):
U L2 D2 B' R2 B' R2 D2 F2 U2 B L F2 L' U L' B' U' B2

*Credit to Stewy for the reconstruction*
Found at:

Use code “STU” to save on your order and help support the channel!

Music provided by:

LEMMiNO - Cipher
LEMMiNO - Triangular
LEMMiNO - Siberian
LEMMiNO - Aloft
CC BY-SA 4.0

#cubing #speedcubing

[JonathonGraves-qrfb]

While watching this video I got struck by an asteroid. I can guarantee I will not see another solve like this in my lifetime.

[speedcubingdotorg]

The 1 in 241 million odds are the odds of those two specific individuals both getting a last layer skip on one specific solve, the odds of any 2 individuals getting a last layer skip are far higher, especially when the solutions are somewhat similar, the thing is that there is a last layer skip possible on every scramble, it is just most of the time the F2L required to produce it would not be optimal and it wouldn't be predictable anyway. But two very impressive solves nonetheless.

[AidanGrainger]

Awesome video, I cant believe you thought I was skilled enough to be thinking about what solution was optimal for last 2 pairs in that pause though, ... in all honesty I was just looking for the pair to do next😅😅😅 Either way I cooked, absolutely ate, got the fries back out the bag and nommed them right up;

[ashwynwadhawan]

bro doesn't understand the difference between independent and dependent probability

[pepkin]

The odds in the title (1 in 241 Million) are incorrect, because those two events are not independent, as they came from the same scramble.

[duvi]

That's not really how probability works. The chances of 2 people getting a last layer skip in a competition in the same scramble is not only the chances of two last layer skips but, in fact, that times the number of participants and the number of scrambles done. I would estimate the probability is about 1/1000, still pretty rare, but not even close to This has probably happened before. What's actually crazy is that both competitors were able to take these opportunities and turn them into sub 4 solves. Unfortunately, that probability is impossible to calculate

[OvejaGD]

Its crazy that i havent solved a cube in a few months and literally the only speedcubing channel that i watch is yours. You make this topic so entertaining that even non-cubers will find this entertaining

[PieceByPiece]

thanks for this video!! i still can't believe all of this happened. can't speak to the probability of this but in a way, it doesn't matter, because however unlikely this was, it happened for real :D

[Da_Rizzley_Bear]

WOW the UK be popping of
this + FMC WR
Congrats to these guys

[curiousNic]

This is my favorite cubing story of the year. Getting a sub-4 out of the blue in the presence of two 3x3 World Champions has to be one of the best cubing dreams come true ever. PS: Not a fan of

[daniel-kidd]

Aidan Grainger and Jake Brown my goats

[RubiKubiK]

Bro your editing and storytelling is soo good I can’t believe the high quality sometimes! You deserve so many more subs!🎉

[MoreCubes]

If you were at that comp count yourself as one of the rarest people in the world,
To be a cuber, in the uk, at that comp, while 1 in chance solves happened

[TheCubicWizard]

man this is just crazy, also we cant leave the good work he makes, loved the video, and especially those solves where its *exactly* 2 seconds from their previous pb, thanks for the great video!

4:45 you can do a wide U exept for D and if you does the same solution he will have to do only one rotation and the solve will be the same.

[orngng]

Not sure if this works but if we use your assumptions on squaring last layer skip chances you also have to multiply the "number of competitors choose 2" which means 62 choose 2 = 1891 (So this is saying for any two people getting a last layer skip, this could be 30th and 59th person)

But this assumes that any competitor there can get a last layer skip which may depend on their solution

and 62 because thats how many people competed 3x3 2nd round
So we have to multiply the last layer skip calculation with that

We assume combinations without repetition too, i got 1 in 127903 as a result

[thebeastbros]

Lol I remember you mentioning FLOCCINAUCINIHILIPILIFICATION on Dylan Millers stream. Great video btw!

[JordanMerrill-mxcy]

I never would have thought I could get goosebumps while watching someone solve a Rubik’s cube haha 😂

[DVSS]

your solve breakdowns and the stats are the best.

[Ackermanmedia]

As a kid who was alive for the invention of the Rubik's cube I find this totally amazing. I would sit around and try my hardest to solve that thing all day and not get it. Watching these guys do it in 4 seconds is not easy to comprehend.