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Probability Thread

What's the probability of a solved 1x2x2 block forming on LL after completing F2L?

What's the probability of a solved 1x1x3 block forming on LL after completing F2L? What's the probability of a flipped 1x1x3 block forming?
 
Anything. Which permutation case would be most common for Sune, for example. With correctly oriented corner on ULF, would the most common case have adjacent switch on right, back, left, or front? Or if it was diagonal...
 
MennoniteCuber1 said:
What are the most probable CLL cases for each orientation?
Click to expand...
MennoniteCuber1 said:
Anything. Which permutation case would be most common for Sune, for example. With correctly oriented corner on ULF, would the most common case have adjacent switch on right, back, left, or front? Or if it was diagonal...
Click to expand...
There are 43 CLLs (including skip)
39 of them have 4/162 probability
Both double-bar H cases: 2/162
Y perm: 1/162
Skip: 1/162
 
cmhardw said:
The probability of a solved face (not necessarily a layer) (3x3 square) is approximately: 1.3*10^(-7)

Clicky
Also clicky
Click to expand...

I fixed a few little mistakes and used what I think are the correct multipliers for the inclusion/exclusion. If I didn't make a mistake, there are 5,778,848,497,349 states with a solved face (happens to be a prime number).

Here's my code, I'd appreciate some checking :)
http://ideone.com/ESdzhS
https://gist.github.com/pochmann/21408b91d0876c5e5828
 
Last edited:
What is the probability of a paired (solved) edge with the use of reduction on...

5x5?
6x6?
7x7?

And could someone give me the general idea in how to go about finding this? I was thinking something along the lines of this for a 5x5:

pick a random midge (say the white-red one)
chances that one wing will be attached to it: 1/24 for 24 possible positions for a wing to be in
chances that another wing will be attached: 1/24
(1/24)*(1/24)= 1/576 for some reason this probability seems rather small, this has happened to me a few times on 5x5 and I've probably done about 600 solves in my lifetime

Is this the way to do it? This is the first time I've really ever dabbled in cube probability, so excuse any mistakes I made.

edit: do any problems I have with this arise from the probability of a connected midge to a wing?
 
Last edited:
goodatthis said:
What is the probability of a paired (solved) edge with the use of reduction on...

5x5?
6x6?
7x7?

And could someone give me the general idea in how to go about finding this? I was thinking something along the lines of this for a 5x5:

pick a random midge (say the white-red one)
chances that one wing will be attached to it: 1/24 for 24 possible positions for a wing to be in
chances that another wing will be attached: 1/24
(1/24)*(1/24)= 1/576 for some reason this probability seems rather small, this has happened to me a few times on 5x5 and I've probably done about 600 solves in my lifetime

Is this the way to do it? This is the first time I've really ever dabbled in cube probability, so excuse any mistakes I made.

edit: do any problems I have with this arise from the probability of a connected midge to a wing?
Click to expand...
Shouldn't it be (1/24)*(1/23) = 1/552? Once you have one wing paired up there are only 23 positions the other wing can be in.
For 6x6, you could use one wing as a reference, and then it would be (1/24)*(1/23)*(1/23) = 1/12696
7x7 would be (1/24)*(1/23)*(1/24)*(1/23) = 1/304704
This is just the probability of having one specific edge solved, I'm not sure how you would find the chances of having any one edge solved.
 
Here's my calculation for the 5x5x5 for the probability that at least 1 tredge is paired after finishing centers. This assumes you do not try to influence tredge pairing at all during centers.

Clicky

This probability is approximately 2.1%
 
Last edited:
guysensei1 said:
There are 3 probabilities for PLLs, 1/18, with 15 PLLs, 1/36, with 2 PLLs and 1/72, with 4 PLLs.

Thus the probability is (1/18)^2*15+(1/36)^2*2+(1/72)^2*4 which is roughly 4.8%

Am I right?
Click to expand...

Better round to 4.9%. But there are 16 PLLs with 1/18 anyway.

cmhardw said:
Here's my calculation for the 5x5x5 for the probability that at least 1 tredge is paired after finishing centers. This assumes you do not try to influence tredge pairing at all during centers.

Clicky

This probability is approximately 2.1%
Click to expand...

Got the same, though I like to use "choose":
http://www.wolframalpha.com/input/?...choose+i)+*+(24-2i)!+/+24!+for+i+from+1+to+12
 
I doubt anyone is gonna believe this, but I just had a last 2 layers skip, after I finished the cross I only had to do a u' and the cube was solved. What are the chances of this happening?

Edit: some additional information: I was just casually solving, so it was just a hand scramble, it probably wasn't 20 moves, but it must've been more than 12. I don't optimally solve the cross and I don't remember how many moves I took this time.
I don't think the exact probability is calculable because of the unknown scramble length and movecount to solve.
 
Last edited:
Chrizz said:
I doubt anyone is gonna believe this, but I just had a last 2 layers skip, after I finished the cross I only had to do a u' and the cube was solved. What are the chances of this happening?
Click to expand...

There are {8! \times 3^7 \times (12!/2) \times 2^{11}} = 43,252,003,274,489,856,000 possibilites for the cube. Divide by 24 for the first edge, by 22 for the second edge, by 20 for the third edge, by 18 for the last edge of the cross
{43,252,003,274,489,856,000 /24 /22 /20 /18} = 2.275,463,1 e+14. So 1/(2.275,463,1e+14). (I'm not sure at all)
 
I had a go at trying to figure out the probabilities of 'EPLLs' for when DF and DB are swapped.

Can someone have a go at them so I can understand how the calculation works please? :)

My idea is (if O and W aren't so frequent) to start cross with 2 opposite edges swapped at the start of the solve for leaving these cases after COLL. Maybe get lucky for an M2 U2 M2 PLL :P
 
cuber8208 said:
I had a go at trying to figure out the probabilities of 'EPLLs' for when DF and DB are swapped.

Can someone have a go at them so I can understand how the calculation works please? :)

My idea is (if O and W aren't so frequent) to start cross with 2 opposite edges swapped at the start of the solve for leaving these cases after COLL. Maybe get lucky for an M2 U2 M2 PLL :P
Click to expand...
Opp: 2/12
Adj: 4/12
W: 4/12
O(cw): 1/12
O(ccw): 1/12

Yes, doing oppcross can often be a good idea imo. I do it maybe 1 in every 30 solves or so,.and try a bit harder to force edges/do COLL /OLLCP. I know how to do the above EPLLs and they're certainly no worse than many PLLs
 
cuber8208 said:
Thanks Dan, did you have a source for those stats?
If you did them yourself can you point me in the right direction of how to figure them out by myself?
Click to expand...
I did them myself. I do it like this by thinking though how to create these cases by picking a location and working out probabilities of them being in the right places.
For Opp:
Pick a random location.
Option 1 is that it belongs opposite (1/4 chance). If that is the case, look at the location opposite to it, which has to belong back in the original location (1/3 chance).
Option 2 is that it is in the correct place (1/4). If that is the case, look at the location opposite to it, which has to also be solved (1/3).
1/4 * 1/3 + 1/4 * 1/3 = 2/12
Note that because of permutation parity, I don't need to think about where the last 2 pieces are.

That was a super long way of explaining what is a simple logical process to me. :/
 
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