Duncan Bannon
Member
Probability of doing OLL parity, it solving OLL and U2 Auf PLL skip?
I just got two PLL skips in a row, then four full-step solves, then another PLL skip aided by a COLL.
The chances of a PLL skip is 1/72, so I'll start with 1/(72^2). The chance of not getting a PLL skip is 71/72, so I'll change to expression to (1/(72^2))*((71/71)^4)). The chance of getting a COLL case that leads to a PLL skip is 173/7776, so the expression turns to (1/(72^2))*((71/71)^4))*(173/7776), which is equal to 4396220813/(1.0833062*10^15), which is equal to about 1 in 246,417 or about a 0.00041% chance of occurring!
Someone tell me if this math is wrong
That crossed my mind and you are correct, but it was more remarkable to me that I got 3 PLL skips in such a short span of time (one with some control).The math sounds right, but I've always found stats like these particularly unremarkable. For example:
I got an Ra perm, followed by a T perm, followed by a Gb, followed by an E-perm! The chances of this occuring are (1/18)^3 * (1/36) = 1/209,952. How rare! Isn't that crazy?
But it really isn't that remarkable in the sense that four random PLLs in a row isn't particularly exciting, even though it has roughly the same probability of occuring as your two PLL skips, four full step, then EPLL skip.
That crossed my mind and you are correct, but it was more remarkable to me that I got 3 PLL skips in such a short span of time (one with some control).
How about what happened to me a few months ago, I was doing PLL training on csTimer and I got 71 solves in a row with not a single E-perm, but then the 72nd, 73rd, and 74th solves were all E-perms!Well the chance of getting exactly 3 PLL skips in 7 solves (ignoring COLL, obviously that significantly increases the probability) is 21(71^4)/(72^7), which is about 1/11280, already more likely than an LL skip.
What's the probability of getting the same VLS case twice in a row (without worrying about angle)?
Wrong for the specific part of the question you quoted. (Hint: If you flip a coin twice, what're the chances of getting the same outcome twice in a row? For bonus points: what if you get to flip the coin more than twice?)There are 216 VLS cases and becsude they have the Slot in them I believe there is no symmetry - every case has a probability of 1/216.
That means your probability is 1/46,656
Wrong for the specific part of the question you quoted. (Hint: If you flip a coin twice, what're the chances of getting the same outcome twice in a row? For bonus points: what if you get to flip the coin more than twice?)
There are often a lot of assumptions in a probability question; some are "obvious", most aren't. The answer to an underspecified question can vary a lot depending on what you choose to assume.
Just to be pedantic z is 1 in 72 and h is 1 in 36. Maybe I should do some more intuitive forced OLL skip ( it's not viable for speedsolving but still fun) but it's pretty obvious that each of the PLLs will show up as normal.… ?
Unless you do some mysterious LL influencing in the last slot, the PLL cases you get after an OLL skip are distributed the way you'd expect. Skip, H perm and N perms would be 1/72 each; Z perm and E perm would be 1/36 each; the rest would be 1/18 each. If it feels like every time you get an OLL skip the resulting PLL case is different, that's merely because you haven't gotten enough OLL skips for a repeat PLL case to be statistically probable.
something something muphry's lawJust to be pedantic z is 1 in 72 and h is 1 in 36.