Probability Thread

[qqwref]

If x corners and y edges are unknown, and the rest is solved relative to the centers, then there are (x! * y!)/2 * 3^(x-1) * 2^(y-1) possibilities. So the probability it is solved is one over that, assuming all positions are equally likely.

So:
Probability of LL skip (no AUF): x=4, y=4, 1/62208.
Probability of a full cube skip after solving the cross: x=8, y=8, 1/227546313523200.

You can multiply by 4 (for 4 positions considered solved) if you want to also allow anything off by U, U2, U'.

 

[irontwig]

What's the probability of the remaining pieces being solvable with two three cycles:
5 corners?
6 corners?
5 edges?
6 edges?
LL?

 

[bluedasher]

What is the probability of a PLL skip when using 2 look OLL? I would assume it would be different.
what are the odds of after solving the F2L all edges are oriented on the Upper face?

 

[That70sShowDude]

Probability of 2 auf-less PLL skips in a row?

 

[kinch2002]

What is the probability of a PLL skip when using 2 look OLL? I would assume it would be different.
what are the odds of after solving the F2L all edges are oriented on the Upper face?

1. If you're blinding applying 2 look OLL then it would still be the usual 1/72. Of course you have more options to force a PLL skip as you're doing 2 algs, but this will clearly be difficult unless you learn specific cases, and then it's effectively 1 look OLL with 2 algs anyway.

2. 1/8. This is found by 1/2 chance of each edge being oriented with the last one having no choice. 1/2 x 1/2 x 1/2 = 1/8

What's the probability of the remaining pieces being solvable with two three cycles:
5 corners?
6 corners?
5 edges?
6 edges?
LL?

This one is a tricky one but I'll give it a go.
5 corners: I think this would have to be one 5-cycle. If there was a twisted corner in place then I think it wouldn't be solvable in 2 3-cycles. So the first piece has to go anywhere but to itself (i.e. not twisted), so that's 4/5 (maybe!). Or on second thoughts it might be 6/7 (12/14) if you think of it in terms of which sticker a specific sticker on it needs to go to. So after that, the second piece needs to go to another piece other than the first one, giving you 3/4, or the same 3/4 (9/12) if you think of stickers. The third piece now has to go to another piece other than the first one (it can't go to the second one clearly), giving 2/3 chance, or 2/3 (6/9) in terms of stickers. Now the fourth piece needs to go to the fifth piece, giving 1/2 chance, or 3/5 in terms of stickers (as it can't go back to the same sticker you started with on piece 1 otherwise the fifth piece would be solved rather than twisted in place.
So that gives me 4/5*3/4*2/3*1/2 = 1/5 the first way and 6/7*3/4*2/3*3/5 = 9/35 the second way. I have no idea whether either way is right so maybe someone else could help me out here! I'm not going to start on the others because this twisted pieces thing is doing my head in!

Probability of 2 auf-less PLL skips in a row?

Nice easy one PLL skip is 1/72, so AUF-less is 1/288. This gives 1/288 x 1/288 = 1/82944 for 2 AUF-less PLL skips in a row.

 

[That70sShowDude]

Nice easy one PLL skip is 1/72, so AUF-less is 1/288. This gives 1/288 x 1/288 = 1/82944 for 2 AUF-less PLL skips in a row.

O_O I got one of those.

 

[ben1996123]

2. separation skip on 3x3xn where n is 2 or 4+
4. Megaminx star skip

Anyone know any of these?

 

[qqwref]

Star skip is easy, assuming you're not color neutral. There are 30 edges with two orientations each, so you just need the probability that those five are solved, which should be (1/60)(1/58)(1/56)(1/54)(1/52) = 1/547223040, assuming a completely random scramble.


I'm not 100% sure what you mean by separation skip; I'll assume you mean you want all the pieces to be in the correct layers. Then, for a 3x3x2, you just need all four top pieces of each type in the four top slots, so the corners and edges each have a probability of 1/(8 choose 4), for a total separation skip probability of 1/4900.

For a 3x3x4 the problem is a little more complicated because in the inner layers you can't tell what layer the edges go to. The probability is the same for the corners, but for the edges we have to choose exactly one edge from each of the four pairs of identical edges. There are (8 choose 4) possible choices, of which 16 choices are acceptable (although I haven't factored in parity). So the inner layers have a separation skip probability of 4/1225. This probability will remain the same for inner layers of any higher 3x3xeven cuboid. The total separation skip probability for the 3x3x4 is 1/1500625.

 

[ben1996123]

I'm not 100% sure what you mean by separation skip; I'll assume you mean you want all the pieces to be in the correct layers. Then, for a 3x3x2, you just need all four top pieces of each type in the four top slots, so the corners and edges each have a probability of 1/(8 choose 4), for a total separation skip probability of 1/4900.

Yes that's what I meant. I have had 3 of those on 3x3x2 sim o.o

 

[bluedasher]

for the 3x3, what is the probability of, when using F2LL, you have a cross on top already made? It's more of a tri-cross, but O' well.

 

[hawkmp4]

for the 3x3, what is the probability of, when using F2LL, you have a cross on top already made? It's more of a tri-cross, but O' well.

Do you mean F2L? If so, (1/2)^3 = 1/8.

 

[Tim Major]

Do you mean F2L? If so, (1/2)^3 = 1/8.

F2LL = Winter Variation. I'm not good with probabilities, and even though this is simple, I don't want to be wrong.

 

[hawkmp4]

F2LL = Winter Variation. I'm not good with probabilities, and even though this is simple, I don't want to be wrong.

Okay. Then, from what I know of WV, the idea is that you orient corners while inserting the last pair. And edges are already oriented using another method (says the wiki). In that case, of course, it's probability 1. However, if you don't do anything with edges before you do WV, the chances you'll have all 4 edges oriented is as above, (1/2)^3=1/8.

 

[JeffDelucia]

Joey I think you are on the right track with this. I think you could argue that a really good estimate to the probability would be 1/15552. You could say that the probability that you get a LL case on the first solve is 1. Then the probability that you get the matching case to that on the second solve is 1/15552. I don't know if the above sum equates to this, and if it does that would be really cool!

Chris

Is this including the AUF? He didn't say that both solves had the same AUF so I'm not sure if its correct.

 

[ariasamie]

on a 4x4x4 what is the probability of... duh... I can't describe it in words!
what is the probability of getting AT LEAST on of these:

Spoiler

so i'm only talking about white and yellow which are two opposite centers and they can show up at the same time.
and also if i didn't want to attach pictures, how could I ask it using sentences?

 

[Johannes91]

Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
Answer: 69%

I can post more statistics if anyone's interested.

 

[mrCage]

Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
Answer: 69%

I can post more statistics if anyone's interested.

How about a V4 with black stickers instead of white??

Per

 

[ariasamie]

Question: What's the probability of having at least one 1x2 block of white or yellow centers on a random 4x4x4 position?
Answer: 69%

I can post more statistics if anyone's interested.

thank you.
can you tell me how you found that out please?
I may want to explain it to some people.

 

[Johannes91]

thank you.
can you tell me how you found that out please?
I may want to explain it to some people.

I wrote a program that goes through all positions and counts the number of pairs in each one.

It can also be calculated by hand, but at least my way is a bit tedious.

Spoiler

First, the number P of different positions for the 8 center pieces is C(24,4) * C(20,4) = 51482970.

We'll count all positions with no pairs, case by case.

A) W,W,W,W,-,- : 3840 cases
AA) WYY,WYY,W,W,-,- : 6 cases
AB) WYY,WY,WY,W,-,- : 108 cases
AC) WYY,WY,W,W,Y,- : 288 cases
AD) WYY,W,W,W,YY,- : 16 cases
AE) WYY,W,W,W,Y,Y : 64 cases
AF) WY,WY,WY,WY,-,- : 81 cases
AG) WY,WY,WY,W,Y,- : 864 cases
AH) WY,WY,W,W,YY,- : 216 cases
AI) WY,WY,W,W,Y,Y : 864 cases
AJ) WY,W,W,W,YY,Y : 192 cases
AK) W,W,W,W,YY,YY : 4 cases

B) WW,W,W,-,-,- : 1920 cases
BA) WWYY,WYY,W,-,-,- : 2 cases
BB) WWYY,WY,WY,-,-,- : 9 cases
BC) WWYY,WY,W,Y,-,- : 72 cases
BD) WWYY,W,W,YY,-,- : 6 cases
BE) WWYY,W,W,Y,Y,- : 48 cases
BF) WWY,WYY,WY,-,-,- : 12 cases
BG) WWY,WYY,W,Y,-,- : 48 cases
BH) WWY,WY,WY,Y,-,- : 216 cases
BI) WWY,WY,W,YY,-,- : 72 cases
BJ) WWY,WY,W,Y,Y,- : 576 cases
BK) WWY,W,W,YY,Y,- : 96 cases
BL) WWY,W,W,Y,Y,Y : 128 cases
BM) WW,WYY,WYY,-,-,- : 1 case
BN) WW,WYY,WY,Y,-,- : 72 cases
BO) WW,WYY,W,YY,-,- : 12 cases
BP) WW,WYY,W,Y,Y,- : 96 cases
BQ) WW,WY,WY,YY,-,- : 54 cases
BR) WW,WY,WY,Y,Y,- : 432 cases
BS) WW,WY,W,YY,Y,- : 288 cases
BT) WW,WY,W,Y,Y,Y : 384 cases
BU) WW,W,W,YY,YY,- : 12 cases
BV) WW,W,W,YY,Y,Y : 96 cases

C) WW,WW,-,-,-,- : 60 cases
CA) WWYY,WWYY,-,-,-,- : 1 case
CB) WWYY,WWY,Y,-,-,- : 64 cases
CC) WWYY,WW,YY,-,-,- : 16 cases
CD) WWYY,WW,Y,Y,-,- : 192 cases
CE) WWY,WWY,YY,-,-,- : 32 cases
CF) WWY,WWY,Y,Y,-,- : 384 cases
CG) WWY,WW,YY,Y,-,- : 384 cases
CH) WWY,WW,Y,Y,Y,- : 1024 cases
CI) WW,WW,YY,YY,-,- : 24 cases
CJ) WW,WW,YY,Y,Y,- : 384 cases
CK) WW,WW,Y,Y,Y,Y : 256 cases

The comma-separated lists tell how many white/yellow center each face contains.

Total cases for A: 3840 * (6 + 108 + 288 + ...) = 10379520
Total cases for B: 1920 * (2 + 9 + 72 + ...) = 5245440
Total cases for C: 60 * (1 + 64 + 16 + ...) = 165660

Total cases with no 1x2 blocks: N = 10379520 + 5245440 + 165660 = 15790620

Probability that there is at least one 1x2 block: (P - N) / P ≈ 69%

 

[Alcuber]

How about a V4 with black stickers instead of white??

Per

Lanlan 4x4