Ickathu
Member
What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
Does that make sense?
Does that make sense?
So far: 1/LifetimeLast 2 Center skip on 5x5?
So far: 1/Lifetime
I was just casually solving while my kids played on the playground, so I wasn't paying 100% attention to the solving, so when I had solved the first 4 centers I was turning the cube around to find the next unsolved centers. After a few x-y-z's I couldn't find any
What's the probability of each of the LL cases on pyraminx? I'm referring to the last layer that you get with Oka, not the LBL LL. This LL involves 3 edges and 3 centers.
Does that make sense?
Thanks! I have some more questions!
What is the probability of a last layer skip on a 3x3 given that the cube is in a state that can be solved 2 generator <RU> and has the F2L completed?
How does this compare with a standard LL skip after standard F2L?
What is the chance of two N's in a row then a Z, then a H, then an E, in that order. It's all the unlikely PLL's. (This just happened to me!)
1/483729408, but it's not really that special since there are 4084101 different combinations of PLLs you can get in 5 solves, so its only really ~118 times rarer than most other combinations.
Yeah, this seems right. Super rare!I think you made a mistake.
P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36
P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352
The number of "different combinations" depends on your definition of what a combination is. Is Ua the same as Ub? Since the OP says only "N" instead of Na and Nb, I assume he groups them together. Therefore, there are only 13 "different" permutations (U, H, Z, A, E, J, T, R, F, G, V, N, and Y) so the number of combinations is 13^5 = 371,293. (I'm ignoring PLL skip as a case because I'm lazy.)
The original sequence of permutations is rarer than the expected combinations, but not 1/118. The PLLs that the OP got were N, Z, H, and E, which makes up 7/72 of all PLL cases. To only get one of these PLLs each time and not any of the others will occur on average (7/72)^5, which is about 1/115125. The most common combinations will involve one of the other PLLs, which account for 8/9 of PLL cases. "Most other combinations" will have only PLLs from that group, and the probability of getting 5 PLLs from that group is (8/9)^5 ~ 5/9, which makes the OP's combination about 63,886 times rarer than "most other combinations."
....I think.
I think you made a mistake.
P(N) = 1/36
P(Z) = 1/36
P(H) = 1/72
P(E) = 1/36
P(N-N-Z-H-E) = (1/36)^4 * 1/72 = 1/120932352
This is the chance of gettting these permutations, but not getting them in a specific order
I just get 6 R perm in a row (all PLLs was on right hand). What's the chance fot something like that?
For it to happen in those specific six solves:
Chance of R(b)-perm = 1/18
(1/18)^6 = 1/34012224
But for it to happen in any random set of six consecutive solves, it's a lot higher.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
Cube probability and coin flips | Puzzle Theory | 2 | ||
Probability of having a pre-made pair in a ZZ solve? | Puzzle Theory | 11 | ||
bloc 1x1x2 on corner probability | Puzzle Theory | 7 | ||
T | Joseph Bertrand Math Problem in Probability | Off-Topic Discussion | 4 | |
H | Probability Problem | Off-Topic Discussion | 12 |