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Probability Thread

Joined
Mar 30, 2013
Messages
9
Location
The Netherlands
WCA
2012VERS03
i asked this 6 years ago but i didnt get an answer, ive been looking for it and the answer seems to be 10.xx, but i cant trust myself
whats the average number of targets if you are using 2 fixed buffers? (dont mind orientations; passive pieces unoriented count as 0 targets, parity edge and corner count as 1 each and breaking into a new n cycle counts as n+1 targets)
also whats the average of passive pieces to twist? not counting the buffers i guess, or whatever is easier to calculate
Note that this is not a very accurate approximation as you do need to memorize misoriented pieces, and you do need to execute algorithms to solve them, hence they affect both memorization and execution time. Parity will only affect execution time (but as the chance of parity is 1/2, we can easily deal with it).
Anyway, I brute forced this problem by varying the number of pieces P from 3 to 10 (unfortunately my computer couldn't handle 12 due to there being many permutations) and got a very clear linear trend T = 1.0916P - 1.7731 for the average number of targets T.
For P = 8 (corners) the exact average is 6.968 while the model predicts 6.959 targets, so it is clearly very accurate.
For P = 12 (3x3 edges) this model gives an estimated 11.32 targets with an estimated error smaller than 0.02. So it is definitely not 10.xx.
For P = 24 (big cube wings) the estimate is 24.4 targets (altough I'm not sure how accurate that estimate is as it is extrapolated so far)
This formula does not apply for big cube centers, because there are identical pieces which this formula does not take into account. For those, I would use the approximation that it is possible to avoid cycle breaks altogether (which is practically always the case) and that we cannot choose our orientation (like on a 5x5). In that approximation, the average number of solved pieces can be computed exactly. Every piece has a chance of 1/6 to be solved, so an average of 23 * 1/6 = 3.83 pieces is solved (remember: not 24, because we never don't consider the buffer to be solved). Neglecting cycle breaks this gives us an average of 23 - 3.83 = 19.17 targets.
On a 4x4 we can choose the orientation based on centers, and I don't know a nice way of handling the problem then, as we cannot consider the rotations to be independent positions. From experience, I guess that the average number of targets is about 15 if we choose the optimal orientation.
 
L

lucarubik

Guest
Note that this is not a very accurate approximation as you do need to memorize misoriented pieces, and you do need to execute algorithms to solve them, hence they affect both memorization and execution time. Parity will only affect execution time (but as the chance of parity is 1/2, we can easily deal with it).
Anyway, I brute forced this problem by varying the number of pieces P from 3 to 10 (unfortunately my computer couldn't handle 12 due to there being many permutations) and got a very clear linear trend T = 1.0916P - 1.7731 for the average number of targets T.
For P = 8 (corners) the exact average is 6.968 while the model predicts 6.959 targets, so it is clearly very accurate.
For P = 12 (3x3 edges) this model gives an estimated 11.32 targets with an estimated error smaller than 0.02. So it is definitely not 10.xx.
For P = 24 (big cube wings) the estimate is 24.4 targets (altough I'm not sure how accurate that estimate is as it is extrapolated so far)
This formula does not apply for big cube centers, because there are identical pieces which this formula does not take into account. For those, I would use the approximation that it is possible to avoid cycle breaks altogether (which is practically always the case) and that we cannot choose our orientation (like on a 5x5). In that approximation, the average number of solved pieces can be computed exactly. Every piece has a chance of 1/6 to be solved, so an average of 23 * 1/6 = 3.83 pieces is solved (remember: not 24, because we never don't consider the buffer to be solved). Neglecting cycle breaks this gives us an average of 23 - 3.83 = 19.17 targets.
On a 4x4 we can choose the orientation based on centers, and I don't know a nice way of handling the problem then, as we cannot consider the rotations to be independent positions. From experience, I guess that the average number of targets is about 15 if we choose the optimal orientation.
so for non parity 3x3 scrambles the average is 9,144, thats a bit less than I expected, not too much tho
what about parity scrambles, i would asume you dont just add one (two targets) do you, i wouldnt know where to start to calculate it, ive never been taught probability, i would defenetly f*** up at 5 points becouse i dont have any experience
what about passive pieces to twist, every corner has 2/24 chances to be on its place but twisted, right? so you have 14/24 chances of having at least one, exclouding the buffer, but this doesnt take into consideration that if 7 corners are in the correct orientation the 8th has to be oriented too, god im so bad at this... does anyone know about a tool i can use to "calculate" this, just by simulation?
thanks for the answer, thats 1/3 of the problem solved
 
Joined
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Messages
9
Location
The Netherlands
WCA
2012VERS03
so for non parity 3x3 scrambles the average is 9,144, thats a bit less than I expected, not too much tho
what about parity scrambles, i would asume you dont just add one (two targets) do you, i wouldnt know where to start to calculate it, ive never been taught probability, i would defenetly f*** up at 5 points becouse i dont have any experience
what about passive pieces to twist, every corner has 2/24 chances to be on its place but twisted, right? so you have 14/24 chances of having at least one, exclouding the buffer, but this doesnt take into consideration that if 7 corners are in the correct orientation the 8th has to be oriented too, god im so bad at this... does anyone know about a tool i can use to "calculate" this, just by simulation?
thanks for the answer, thats 1/3 of the problem solved
On average, there is one piece in its correct place (that is a well known problem in probability theory) regardless of the number of pieces we consider. This can be seen as follows. Suppose we have P pieces. There is a chance of 1/P that piece 1 is in its correct place, so the expected contribution of this piece (averaged over all permutations) is 1/P. This argumentation also holds for all other pieces so the expected number of pieces in solved position is 1/P * P = 1. For corners, this means that the average number of twisted corners is 1 * 7/8 * 2/3 = 7/12 = 0.583 (here 1 is the number of solved pieces, 7/8 because we do not consider the buffer, and 2/3 the chance that is it oriented incorrectly). The fact that the orientation of the 8th corner depends on the orientation of the other 7 corners is irrelevant, as we can choose the buffer corner as 8th and its orientation is irrelevant. For edges, the same argumentation goes, giving an expected number of 1 * 11/12 * 1/2 = 11/24 = 0.458 flipped edges.
If you wish to brute force a question like this, you will need either a programming code (more difficult to learn) or a commando-based mathematical program (easier, i use Mathematica but you will need a license to use it).
The chance of at least one corner being twisted is NOT 14/24. By that argument, the chance of throwing at least one six with a dice in six throws would be 6 * 1/6 = 1, which is false as it is perfectly possible for there not to be any sixes. The expected value, however, is equal to 14/24 = 7/12. However, it is possible that two or more corners are twisted. Those cases are counted multiple times in this average. Therefore, the actual probability will be lower than 14/24 = 0.583
Computing this is a well-known problem in mathematics: a group of people have a hat, they take it off, throw it in a bin, and everybody picks a hat at random. What is the chance that nobody picks his own hat?
This chance turns out to be 1/e for a sufficiently large (>=5 people) group, where e is the exponential constant. The chance of at least somebody picking up the right one is therefore 1 - 1/e.
In this case, the corners are the hats and the positions where they belong are the men, and by scrambling the hats are distributed.
Therefore, the chance of at least one misoriented corner is approximately (1 - 1/e) * 2/3 = 0.421.
Similarly, for edges, the chance is (1 - 1/e) * 1/2 = 0.316.
In my calculation of the average number of targets of 11.32 I did not consider parity. The total number of targets averaged over both parity and non-parity scrambles is therefore equal to 6.97 + 11.32 = 18.29. I don't know if there is a difference between parity and non-parity scrambles here, but I could easily calculate that using my algorithm if you're curious.
 
L

lucarubik

Guest
if a corner has 2/24 chances to be in its place but twisted that doesnt mean if i scramble 24 times its gonna be on its place but twisted two times, so i was right till that point
if every corner has 1/12 chances then given 12 there will be 1 twisted on average so given 7 there will be 7/12, cool, it makes sense doesnt it! im horrible at this, i struggled to come up with this logic proccess, im horrible at intelligence, feels bad.
did your brute force experiment break cycles as we blders do? would it be such a hard ecuation to formulate that you, that seem to know about odds and stuff, rather brute forces it?
it really surprises me the average of cycles, if you optimize parity and count it as a cylce is 9,114, i defently wouldve said more
i got the 10.xx off a post i read a couple of pages ago in this thread, where instead of directly taking the average of targets it counted the odds of each number of taregts, till 10, and the most probable were 4 corners and 6 edges i believe
thanks again by the way its really cool for me to know this stuff, damn i cant beleive i almost forgot to say thanks
 
L

lucarubik

Guest
Hello,
I've digged out an old thread from polish forum where I estimated probablilities of getting an specific bld case(number of targets & number of twists for each edges and corners) by doing simulation in R. Since I had lazy weekend I've decided to do write this code again. Here is the table with results of this simulation:

k=c()
twists=c()
for(j in 1:300000){
a=sample(1:12)
h=0
i=1
obr=sample(c(0,1),11,replace=T)
if(sum(obr)%%2==1){obr[12]=1}else{obr[12]=0}
obr=obr+1
twists[j]=sum(as.numeric(a==c(1:12))==obr)
while(a!=i&a!=0){
h=h+1
ind=a
a=a[a]
a[ind]=0}
for(i in 2:12){
if(a!=i&a!=0){h=h+2}
while(a!=i&a!=0){
h=h+1
ind=a
a=a[a]
a[ind]=0
}
}
k[j]=h
}

results=paste(k,twists,sep=',')
wyniki=c(table(results)[27:length(table(results))],table(results)[1:26])
#^here i had to "re-sort" this table results because it was previously#
#sorted automaticly like this:14,4 15,1 15,2 <-(!)-> 4,3 4,4 4,5... etc#
barplot(height=wyniki/length(k),main="probability barplot of (targets,twist) cases for edges")
tablica=table(targets=k,twists)
dd <- addmargins(tablica, FUN = list(Total = sum), quiet = TRUE)
write.table(dd,file="dane.csv",sep=",",row.names=T)
}

In this table first column means number of targets, first row- numbers of twists. For instance, if i want calculate probability of getting 7-9 targets with maximum 2 twists, i read records from a table: (70+314+471+627+2067+2229+4131+8941+6042)/300000=8,297333%
"targets-twists","0","1","2","3","4","5","6","7","Total"
"3",0,0,0,0,1,0,0,0,1
"4",0,0,2,0,0,0,0,0,2
"5",1,7,11,12,11,5,1,0,48
"6",6,44,67,86,42,18,0,1,264
"7",70,314,471,345,110,7,0,0,1317
"8",627,2067,2229,999,136,10,0,0,6068
"9",4131,8941,6042,1160,97,5,0,0,20376
"10",18423,25064,7096,709,51,0,0,0,51343
"11",51388,30150,4174,365,17,0,0,0,86094
"12",63957,17032,1846,114,0,0,0,0,82949
"13",34568,5636,583,6,0,0,0,0,40793
"14",8278,1385,69,0,0,0,0,0,9732
"15",752,237,1,0,0,0,0,0,990
"16",13,10,0,0,0,0,0,0,23
"Total",182214,90887,22591,3796,465,45,1,1,300000
"targets-twists","0","1","2","3","4","5","Total"
"0",1,2,3,5,3,1,15
"1",6,18,16,13,6,0,59
"2",44,111,95,63,10,1,324
"3",320,658,543,177,34,4,1736
"4",2096,3217,1914,394,43,2,7666
"5",10207,11040,3736,323,18,0,25324
"6",35277,23422,3306,226,8,0,62239
"7",74702,21358,1665,125,0,0,97850
"8",67991,8612,759,0,0,0,77362
"9",22453,2578,82,0,0,0,25113
"10",1815,497,0,0,0,0,2312
"Total",214912,71513,12119,1326,122,8,300000

If You want to calculate probability of getting certain case(for instance 10 or 12 targets for edges with maximum 2 twist AND 6-8 targets for corners with maximum one corner twist, You calculate probablilites separate for edges and corners, than multiply those and multiply by 2. P(A)*P(B)*2
this is the post i was refering, it counts the odds of having x targets, being x all possible values not just till 10, and the most probable are 7 corners and 12 edges then i saw 8 corners was way more probable than 6 so i rounded to 10.xx average instead of 9.5 wich i guess was a mistake if both posts are correct (or if both are proportionally incorrect lol)
 

xyzzy

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Joined
Dec 24, 2015
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Probably out there somewhere but What is the probability there are 2 oriented sides on 2x2 with any combination of opposite colors, like in Guimmond

Fix the DBL corner. This leaves seven corners to be oriented relative to that fixed corner.

P(U and D oriented) = P(L and R oriented) = P(F and B oriented) = \( 3^{-6}=1/729 \)
P(oriented on all axes) = P(U and D oriented) P(corners separated into tetrads) = \( 3^{-6}\times\binom84^{-1}=1/51030 \)
P(oriented on at least one axis) = P(U and D oriented) + P(L and R oriented) + P(F and B oriented) − 2 P(oriented on all axes) = \( 104/25515\approx0.41\% \)
 

Hazel

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How many L8E cases are there where the equator edges and U layer edges are scrambled but the E layer edges are all in the E layer and all the U layer edges are in the U layer, and all the U layer edges are properly oriented?
 

Teoidus

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Without counting AUFs: 4!/2 in U * 4! in E * 2^4/2 EO = 12 * 24 * 8
With AUFs: (4 + 4) in U * (2 + 5 + 5) in E = 8 * 12? I think?
 

AlphaSheep

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How many L8E cases are there where the equator edges and U layer edges are scrambled but the E layer edges are all in the E layer and all the U layer edges are in the U layer, and all the U layer edges are properly oriented?
In the top layer, you have 5 EPLLs (1 is solved) and 5 parity EPLLs.

For each of those you have 12 permutationsof the E layer edges for parity and 12 for no parity, but if they are oriented, you can reduce it to 5 by symmetry (for no parity its solved, two 3 cycles, adjacent double 2 swap, and diagonal double 2 swap) You can y rotate to get specific cases. I think it's also 5 for parity (adjacent swap, diagonal swap, clockwise and anti clockwise cycles, and the weird case that looks like a T).

That gives 5*5 + 5*5 - 1 (solved) = 49 cases if all E edges are oriented.

If you allow E edges to not be oriented then I think it increases to (5*5 + 5*5)*(2^3) - 1 = 399 cases. It's too early in the morning to try thinking of symmetry with bad edges.
 

LexCubing

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Aug 16, 2015
Messages
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There 72 permutations for any F2L case. 7 F2L cases: Solved, Corner in, Edge in, 4 cases when they're both at the U layer. This last slot by the way.

So 504 permutations?

Also if I want to know how many LSLL cases would a I do this:

504 x 27

There are 27 unique CO cases. Correct me if I am wrong.

Or:

504 x 108

108 CO cases since I took into account permutation?
 

shadowslice e

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If you want to solve the whole cube after F2L-1, you would have roughly (5!*5!*3^4*2^4)/2/4/4=583200 though likely slightly more as many cases would have multiple rotation symmetries (like the h perm is the same from multiple angles) (see the lemma that is not Burnside’s)

This is because you have
5! permutations of edges,
5! permutations of corners
3^4 orientations of corners (last corner orientation is dependant on previous 4)
2^4 orientations of edges (same reasoning as above)
1/2 parity
1/4 preauf
1/4 postauf
 

LexCubing

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When calculating let's say ZBLL.You don't 4!^2 x 3^3 because that doesn't give us the unique cases. We times it by 8 since there's just really the Solved, H, Pi, S, As, L, T, and U.

So for LSLL do we

5!^ x 108 or 5! x 27!

There are 27 unique CO for LSLL but when I saw TSLE there's 108 cases?
 

AlphaSheep

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When calculating let's say ZBLL.You don't 4!^2 x 3^3 because that doesn't give us the unique cases. We times it by 8 since there's just really the Solved, H, Pi, S, As, L, T, and U.

So for LSLL do we

5!^ x 108 or 5! x 27!

There are 27 unique CO for LSLL but when I saw TSLE there's 108 cases?
That's not quite right. With ZBLL, you do
4! (corner permutation)
* 4! (edge permutation)
* 3^3 (corner orientation)
/ 2 (parity)
/ 4 (AUF afterwards doesn't matter)
= 1944 states.
You can't just times by 8 because solved and H appear in different proportions to the others. This doesn't account for symmetry, and as has been pointed out a few times, the easiest way to deal with symmetry is to list the symmetric cases. Listing these and removing the symmetry cases gives 494 cases (493 excluding solved).

For LSLL, you've also got it wrong. 5! x 27! is around 30 billion times larger than the number of cube states (43 quintillion).
The number of LSLL states (with EO done) is
5! (corner permutation)
* 5! (edge permutation)
* 3^4 (corner orientation)
/ 2 (parity)
/ 4 (AUF afterwards doesn't matter)
= 145800 states.
Counting out the symmetric cases is actually easier than you'd think. The trick is that a case can only have rotational symmetry if both the LS corner and edge are correctly placed. If either one of them is in the last layer, then rotational symmetry is impossible.

The same trick can be used in TSLE. So for TSLE, we are concerned with the orientation of 4 corners (the 5th one is determined by the other 4), and the position of 1 edge. So the number of states is
5 (number of positions for the edge)
* 3^4 (corner orientations)
= 405 states.
Note in this case there is no parity (because we're not completely solving permutation) and there is no correction for final AUFs because AUFs afterwards do not change the goal state.

Now we need to count symmetries. Remember what I said about states with a LS edge in LL not having any symmetries? We can use that here. Of the 405 states, there are 81 with the LS edge in place and 324 with it in LL, so those 324 states can't have any symmetry. Therefore, we can do an AUF before these cases to transform these into another case, so just by adding AUFs in front, these 324 states represent (324 / 4) = 81 unique cases.

That leaves the 81 remaining cases with the LS edge placed. Of these, there are three possible twists for the LS corner. If it's solved, we list out all the possible arrangements of the top layer - there are 27 of them. Of these, we note that H has a symmetry in U2 (U2 doesn't change the H case) and the solved state has symmetry in U. So the solved case can't be reduced, and only 2 states can be reduced to H. So, of the 27, 1 is solved, 2 are H, leaving (24 / 4) = 6 other cases, for 8 in total (or 7 excluding the solved case).

If the corner is twisted, then there are similarly 27 states for each twist direction, but one of them has symmetry in U (the one where all for corners in LL are twisted the same direction), and one state has symmetry in U2 (the one where two diagonal corners are twisted in the same direction and the other two are solved). That gives again (1+2/2+24/4) = 8 cases. for each twist direction.

Therefore the total number of unique TSLE cases taking symmetry into account is 81 + 7 + 8 + 8 = 104 cases (or 105 if you include the solved case).

If you want to work out the unique cases for LSLL, then you have the same symmetries to deal with. It's not nearly as difficult as it seems. Why don't you give it a try?
 
M

Malkom

Guest
What's the probability 5/10 solves end up in Vperms? Is it as simple as (4/74)^5 x (70/74)^5?
 

xyzzy

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Messages
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What's the probability 5/10 solves end up in Vperms? Is it as simple as (4/74)^5 x (70/74)^5?

For a fixed set of 10 solves, assuming you don't do OLLCP in any form (this includes not using COLL), a V perm would show up with probability 1/18. Getting five V perms in these ten solves would then have a probability \( \binom{10}5(1/18)^5(17/18)^5\approx0.01\% \). (The probability doesn't change substantially if you ask about having at least 5 V perms, rather than exactly 5.)

But of course you're not doing only 10 solves, so if you're looking for the probability of 10 consecutive solves in a large session having 5 V perms, this probability tends to 1 (exponentially) as the session gets larger.
 
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LexCubing

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Aug 16, 2015
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LL skip is 1/72 and H PLL is 1/108. Why? How do you do this?

How do you know a step is N moves efficient on average? How do you say 1LLLL is 15 moves on avg and Cross is 4.9 moves efficient?

How do you calc that?
 
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