PLL strangeness

[MiloD]

This happens to me all the time!!!!

It is VERY common for to get the same PLL 2 times in a row. I have debated whether or not to say anything but its just ridiculous now...there I said it... I don't even care if anyone responds.

I just had 2 R's
then 2 T's
then the same G twice

all with different oll's!!??:confused:

 

[andrewvo1324]

yes it happens to me to i was going to post a topic but did not want to be the only one -_-!

 

[badmephisto]

yea sometimes happens here as well. slightly more times than is probably statistically correct. I think it's just that if you don't get two same PLL's your brain discards it, but if you do, you tend to remember it strongly.

it's like Murphy's laws... they only really work because if something goes as expected, you tend not to notice

 

[Lucas Garron]

Let's not overinterpret; this is not too unlikely to happen to anyone after a while of cubing.

I've had two consecutive 4x4x4 solves in competition whre PLL was parity (Dan Dzoan had one where both OLL and PLL were only their respective parity cases).

A short while ago, I was timing up to ELS and got 9.91, 9.92, then 9.93 (consecutively).

For a long while, my 3x3x3 PB's were:

• 10.00 lucky
• 11.00 non-lucky
• 10.00 Heise (it was non-lucky)

Actually, I think they all still hold, but I'm not so sure about the second, and have little idea about the third.

 

[jeff081692]

A couple of days ago I did a Fridrich solve and a VH solve and they were both the exact same time.

 

[cmhardw]

If you're curious, the probability that if you do exactly 2 solves that you get the same PLL case twice in a row is calculated below (I consider 2 PLL skips in a row as 2 of the same PLL cases in a row):

If you consider there to be 13 PLL cases (and one more that's the solved case) than your odds are:
(2/9)^2+4*(1/9)^2+4*(1/18)^2+3*(1/36)^2+2*(1/72)^2 = 11.4% chance

If you consider there to be 21 PLL cases (and one more that's the solved case) then your odds are less:

16*(1/18)^2+2*(1/36)^2+4*(1/72)^2 = 5.2% chance

If you do an average of 12 (so 12 solves), the approximate probability that you get, at some point at least once, the same PLL 2 or more times in a row is (again I'm fairly certain this is only an approximation and not the exact chance):

13 PLLs plus one solved: 1-(1-0.114)^11 = 73.6%
21 PLLs plus one solved: 1-(1-0.052)^11 = 44.4%

Chris

 

[MiloD]

If this was happening only once in 12 solves I wouldn't have started this thread.

I am going to write down every pll I get during my next cubing session.

 

[cmhardw]

If this was happening only once in 12 solves I wouldn't have started this thread.

I am going to write down every pll I get during my next cubing session.

I maybe didn't write the statement describing that last calculation exactly as I meant to. What those very last two odds find is the chance that you get two or more of the same PLL in a row at least once during an average of twelve.

So you could get the R permutation 4 times in a row, followed by the T permutation 8 times in a row. Or you could get different PLLs for the first 10 solves, then the last two are the same PLL case. Or you could get 2 R's followed by 2 T's followed by 2 G's, followed by 2 A's, followed by 2 U's, followed by 2 V's.

See what I mean? It just turns out that getting this is not all that uncommon, so it's not particularly strange to see the same PLL twice in a row even a couple times during an average with different PLLs for each pair (or more in a group). In fact it's somewhat likely to happen in some form or another during an average.

Chris

 

[MiloD]

If this was happening only once in 12 solves I wouldn't have started this thread.

I am going to write down every pll I get during my next cubing session.

I maybe didn't write the statement describing that last calculation exactly as I meant to. What those very last two odds find is the chance that you get two or more of the same PLL in a row at least once during an average of twelve.

So you could get the R permutation 4 times in a row, followed by the T permutation 8 times in a row. Or you could get different PLLs for the first 10 solves, then the last two are the same PLL case. Or you could get 2 R's followed by 2 T's followed by 2 G's, followed by 2 A's, followed by 2 U's, followed by 2 V's.

See what I mean? It just turns out that getting this is not all that uncommon, so it's not particularly strange to see the same PLL twice in a row even a couple times during an average with different PLLs for each pair (or more in a group). In fact it's somewhat likely to happen in some form or another during an average.

Chris

I understood what you meant. My intuition is always telling me these things are statistical outliers. Very interesting.

 

[Mik]

I've gotten 3 PLL skips in a row off of JNET using the Fridrich Method.

 

[tegalogic]

This is kind of weird; I got 5 of the same G perm in a row once (solid block vertical lower right).
In the past 2 hours, I was solving the 4x4x4 10 times, and on every solve, there was no orientation parity and always a permutation parity. :\

 

[hait2]

selective memory

 

[Swordsman Kirby]

Ever gotten 8 G-perms in a row?