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I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.
The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:
1 / [6! * 6!/2] = 1 / 259200
To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552
My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.
This might be the luckiest solve you could have on a 4x4x4!
My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.
This might be the luckiest solve you could have on a 4x4x4!
My friend (who is not very good at cubing) was solving his 4x4x4 and he got a 3x3x3 skip NO JOKE. He finished pairing up and after that, he had to do a D' and a U2 THAT'S IT.
This might be the luckiest solve you could have on a 4x4x4!
I want to know the chances of COLL cases and ZBLL cases. Cride5 has tried several times (maybe only once?) to explain this to me with complicated language. I might know what is going on now and I will try to explain.
So for OLL, the chance of Pi, Sune, Anti-Sune, T, U, and L is 1/54 while H is 1/108 and a skip occurs every 1/216 (these statistics are pretty well known). Just taking these cases for EO (setting 1/54 to be the standard):
Now the others should be divided evenly amongst themselves, if my reasoning is correct (no reason you should trust this). Which means every other case should have a nice 49/432 chance of occurrence. What? Hmmm. Weird number. I guess I'll post it with PLL chances:
I postulate that ZBLL has the same chance for every edge permutation. There are 12 possible edge permutations and COLL + EPLL generates each edge permutation equally often. Therefore the chance of each ZBLL case is that of the chance of the respective COLL case divided by 12. I don't want to have to post all of those probabilities.
This is pretty easy when you think about it for a moment. You need the first edge in place - that's 1/12. Second edge must now go in place - that's 1/11. Etc...so you have 1/12*1/11*1/10...*1/2=1/12!. Then they all have to be orientated correctly (1/2 for each) but the last edge orientation is defined by the other 11, so you get (1/2)^11 for orientation
Final answer: 1/(12!*2^11) = 9.81*10^11
EDIT: I hope I got this right...I shouldn't be doing maths at uni otherwise...
