My friend recently sent me a reconstruction of his new 6 seconds PB single of the 3 x 3 rubix cube. I was shocked as he is sub-40 on the 3x3. I myself am sub-15 and have a PB of 10.41 seconds. The reconstruction shows the solution is of 16 moves only. The reconstruction link is given below. The scramble is 20 moves long which turns out to seem legit (and I know he has used csTimer for this). Yet, I am not completely sure if the scramble is an official WCA scramble or not. Thus, is there any way by which I can check if the scramble is an official WCA generated scramble?


  • $\begingroup$ You really cannot find out whether it is an WCA "official" scramble, as WCA accounts basically any scramble that doesn't make the cube solved or less than 2 moves to solve as a official scramble. It took your friend 16 moves: seems legit to me. $\endgroup$
    – Stevo
    Commented Jan 8, 2023 at 10:33
  • $\begingroup$ @Stevo, I have added in the reconstruction link for his solve. Could you please check that once? $\endgroup$ Commented Jan 8, 2023 at 10:42
  • $\begingroup$ unfindable. The thing with WCA scrambles, is that you don't know whether it was "official" or not. As I have said, as long as it is 2 moves or more (the solution), its a good scramble. All possible scrambes are WCA scrambles that don't repeat moves. Just looks like your friend got lucky there, might be a good idea to check how lucky he was. $\endgroup$
    – Stevo
    Commented Jan 9, 2023 at 1:08
  • $\begingroup$ I'm sub-50, and there's no way I'd complete this in even 10 or 15 seconds on my first solve. If this was my friend, I'd assume he tried it multiple times before he got a 6. A free accidental double x-cross and PLL skip are great, but not enough for someone to pull off a PB that is 400% faster than their average. Call me a cynic, but I wouldn't believe it in a hundred years. But yeah, that scramble could show up in a WCA comp, theoretically. $\endgroup$
    – Stevish
    Commented Jul 6, 2023 at 20:28

1 Answer 1


WCA regulations 4b3: "An official scramble sequence must produce a random state from all states that require at least 2 moves to solve (equal probability for each state)."

To make sure the probability for each state is equal, the scramble program can first generate random state for each corner piece and edge piece, (Some parity laws have to be followed in order for the cube to be solvable. So the last corner piece and last edge piece are actually not random".), then use a solver to solve the cube within 20 moves, then reverse the solution steps to get the scramble that can be used in competition.

As long as equal probability is hold, for about 95% of chances, you will need at least 17 or 18 moves to solve the cube. The chance to get a scramble that can be solved within 16 moves is less than 3%. (The reconstruction mentioned is actually 20 moves in HTM since M and M' are counted as 2 moves. But there does exist a 16 move solution: U' F2 L D F2 D2 L F2 R2 U' L2 B2 R2 D F2 U2)

WCA doesn't care how easy you can find a XX cross or how possible you can skip OLL or PLL. It is purely luck. Regulation 4b1 says "Generated scrambles sequences ... must not be filtered or selected in any way by the WCA Delegate." So WCA delegate cannot say "Ah, I see white cross is already solved in this scramble, it is too easy, let's regenerate one."

To summarize, WCA official scramble follows the equal probability principle so that it is fair for everyone. In general the scramble is complex enough, but to a particular method/person, an easy case is possible.

  • 1
    $\begingroup$ Is there a link to an online source for the WCA regulations that you're quoting? $\endgroup$ Commented Jan 14 at 7:46
  • $\begingroup$ WCA regulations are available on its website: worldcubeassociation.org/regulations $\endgroup$
    – Guoyang
    Commented Jan 15 at 6:04

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