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Probability Thread

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Then it was a misscramble and you should contact someone to get that changed to a DNF (assuming it's too late to get an extra attempt).
Wouldn't they take that time out of my average?
I'm pretty sure they were doing average of 3, and taking out the best time and the worst.
That's how most competitions do it I think.
Anyway, I'm probably gonna contact them.
 
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(ps sorry for the double post i couldn't figure out how to edit in quotes)
The way I would do it would be to edit it. Once you click edit, type left bracket, then say the word quote then right bracket then copy in the message they say then type the same thing as the quote thing but add a forward slash between the [ and the Q

It’s hard to explain in words without it making it a quote :) So feel free to ask questions.
 

Aerma

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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 :p
 
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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 :p
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.
 

Aerma

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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).
 
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this is a little more remarkable: A few weeks ago I had an accidental XXstar on megaminx.
That's (1/60)(1/57) for the corners and (1/50)(1/48) for the edges, with 10 possible pairs of F2L pairs, for 1/820,800.
This overcounts a little bit (because of XXXstars) but who cares. For reference, a LL skip comes 1 in every 933120 solves on megaminx.
 
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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).
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.
 

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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.
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!
 
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What's the probability of getting the same VLS case twice in a row (without worrying about angle)?
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
 
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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.
 
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If you pick up a 3x3 and turn it randomly after 2.99800 × 10^19 turns it will have a 50% chance of having been solved. I assumed that each time you turn the cube it goes into a random state but this should still be close to the answer. Turning at 10 TPS, this would take 9.5*10^10 years or 6.9 times the age of the universe.
 
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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.
I think you need to explain something to me. I quoted all the information. The specific case is not important as it's a different case from all angles.
Gonna embarras myself but I try.
You have a 50% chance because the first outcome doesn't matter. What matters is the 50/50 chance on the second flip.

Ok, I get that hint. Needed some days for it. The real chance is 1/206.
 
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… ?

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.
 
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… ?

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.
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.
 
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