thejerber44
Member
How many combinations are possible with a 2-gen scramble?
How many combinations are possible with a 2-gen scramble?
Thanks for the quick reply The website is awesome! A friend told me that he was doing some 2-gen solves (hand scrambles) and when he finished scrambling, the cube was a mere U2 from being solved. He's either really lucky, bad at scrambling, or lying
What would the prob. be of getting 3 LL skips in about 17k solves? (this is my situation)
I believe this should just be a binomial probability, right?
(17000 choose 3) * (1/15552)^3 * (15551/15552)^16997 = 7.3% chance of 3 LL skips in 17000 solves.
This is all assuming, of course that no influencing methods are used, which is probably not true. Oh well.
Does anyone know the probability of getting 4 edges skipped after centres on a 4x4x4?
Thanks for that. I haven't had a LL skip in a year and a half so I was due some luck.
Interesting post Chris!Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.
The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.
See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.
Please take another look at my post above as I noticed an error in my calculation. I knew something looked wrong I had done this calculation before, and after checking my answer to my post above I noticed that I had forgotten something.
The chance of having at least 4 edges paired up after centers is far more likely than I calculated before. The chance is about 1/481.
See my post above for the corrected calculation. Also see the spoiler in my post above for a more detailed breakdown of all the cases, plus some verification of the result.
Would it be possible to get each case (0 edges 1,2,3 edges etc) with a fractional probability? Those numbers are to big to comprehend at that level