# Chance of getting an "easy scramble" with cross already solved?

I read a blog post recently about the limit of speed-solving records getting to a point that rely on easy scrambles to beat. One of the most obvious head starts would be to get a scramble that already has the first phase done.

For this question, assume we are using the CFOP method and the solver is color-neutral. What are the chances of a random scramble having the cross already solved on one side?

Just an anecdote: I have done thousands of solves and have never had this happen, at least not on the white side. I'm curious what the odds are.

• From seeing some top speed solver interviews, it seems that for them, the cross is already as good as solved on any cube, as they can "see" at least a couple of moves past the cross and into the F2L in the investigation phase, before making any actual moves on the cube. (And thus, before starting the clock.)
– Bass
Aug 6, 2018 at 1:46
• Agreed, but this would probably allow them to plan out an extra F2L pair during inspection. Aug 6, 2018 at 1:52

Ignoring any intricacies of the Rubik's cube (particularly any edge parity considerations), and assuming a random scramble has a given edge piece at any of the 12 edge locations in either orientation with equal probability, the white edge pieces can be scattered around the cube in

$12 \times 11 \times 10 \times 9 \times 2^4 = 190080$ different ways. Exactly one of these has the white cross already solved, so the probability of a random scramble having a solved white cross is 1 in 190080.

That's a very small chance, so we can just ignore any "multiple simultaneous solved crosses" situations, and approximate getting one cross of any colour by just multiplying by six, for a total probability of

roughly one in 30 thousand.

The exact maths escape me, but this should be a good enough estimate for any "practical" purposes.

• The first part especially makes sense. With such a limited number of solves per competition, it is extremely unlikely. Aug 6, 2018 at 16:19

The chances of a cross being solved is quite slim . Even if you are colour neutral. The best that you can do , is to just plan out the cross/x-cross in inspection , and execute it as fluidly as you can.