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Probability Thread

Has anyone made a chart showing the probabilities of solve times? So you would for example combine the TPS of the best cubers, the number of official solves they do in a year and the probability of certain skips. From this you could calculate the rough probabilities of specific times occurring during the next year.
 
Tony Fisher said:
Has anyone made a chart showing the probabilities of solve times? So you would for example combine the TPS of the best cubers, the number of official solves they do in a year and the probability of certain skips. From this you could calculate the rough probabilities of specific times occurring during the next year.
Click to expand...

Wow, this is actually a really cool idea.
 
How many last slot cases are there when all edges and corners are permuted?

How do you account for the (several) rotational symmetries?
 
4Chan said:
How many last slot cases are there when all edges and corners are permuted?

How do you account for the (several) rotational symmetries?
Click to expand...

I would guess (3^4)*(2^4)/4(Aufs)/2(reflections)= 162. But how do you have LS when you have correctly permuted all pieces (unless you mean they are misoriented in their slot which is what I assumed)
 
What are the probabilities to get 5 PLL skips in a row, doesn't matter the AUF?

Is it 1/440? 22 PLL cases including solved, 4 angles, thus 88, 88*5 is 440. Then why didn't I ever get one? I'm cubing ever since 2006-7.

Where is my fail?

PS: I did get 5 G perms in a row, though, but I know that there are 4 types of G perms and it's more likely...
 
G2013 said:
What are the probabilities to get 5 PLL skips in a row, doesn't matter the AUF?

Is it 1/440? 22 PLL cases including solved, 4 angles, thus 88, 88*5 is 440. Then why don't I get it so often? Actually I never got that.

Where is my fail?
Click to expand...

1/72 chance to get a PLL skip (not all cases have the same probability).
(1/72)^5 = 1/1934917632 which is about 0.0000000517%
 
So, let's assume I have solved all of F2L, except for a single edge, which I place in its position, but flipped. What are the chances that all of the remaining pieces are, save for one edge, correctly oriented?
 
Given a random state scramble, what is the probability that at least one edge of the CFOP white cross is already solved? By "solved" I mean it has to be next to the white center and oriented correctly, but not necessarily matched up with its corresponding center.

For any individual edge, the probability would be 1/6. 1/3 = probability the edge is in the cross layer, 1/2 = probability the edge is oriented correctly, 1/3 x 1/2 = 1/6. So the probability that at least one of the edges is solved would be 1/6 + 1/6 + 1/6 + 1/6 = 2/3. That seems a bit high. Am I missing something?

Sorry if this question has already been answered but I did a search and couldn't find anything.
 
PurpleBanana said:
Given a random state scramble, what is the probability that at least one edge of the CFOP white cross is already solved? By "solved" I mean it has to be next to the white center and oriented correctly, but not necessarily matched up with its corresponding center.

For any individual edge, the probability would be 1/6. 1/3 = probability the edge is in the cross layer, 1/2 = probability the edge is oriented correctly, 1/3 x 1/2 = 1/6. So the probability that at least one of the edges is solved would be 1/6 + 1/6 + 1/6 + 1/6 = 2/3. That seems a bit high. Am I missing something?

Sorry if this question has already been answered but I did a search and couldn't find anything.
Click to expand...

What you're looking for is the odds of the first edge being 'solved' plus the odds of the second cross edge being 'solved', etc etc. However, you run up against the problem that the second, third, and fourth cross edges actually have to be really 'solved' relative to the first cross edge. So while odds for correct permutation for the first cross edge are indeed 4/12=1/3, the odds for the second cross edge's permutation being correct go down to 1/11, being that there is only one spot relative to the first 'solved' cross edge that it can go. and then down to 1/10 for the third cross edge, etc. It ends ups being 1/6+1/22+1/20+1/18=31.77%
 
Last edited:
obelisk477 said:
What you're looking for is the odds of the first edge being 'solved' plus the odds of the second cross edge being 'solved', etc etc. However, you run up against the problem that the second, third, and fourth cross edges actually have to be really 'solved' relative to the first cross edge.
Click to expand...
It wouldn't matter where the other cross edges were, all that matters is that there is at least one correctly oriented edge in that layer. There could be more than one and they don't have to be correctly permuted relative to each other.

Edit: For what it's worth, I just did an experimental test of 200 scrambles and 110 of them had a solved edge = 55%. This isn't a very large sample size so the value isn't going to be accurate but should be in the ballpark of the actual figure.

Edit2: I expanded the experiment sample size to 500 scrambles. Number with the solved edge = 262/500 = 52.4%.

Edit3: Expanded the sample size to 1000. 531/1000 = 53.1% with solved edge. This value should be pretty close to the actual probability. Right now I'm working on a different way of calculating it.
 
Last edited:
My suggestion ist
(Binomial[4,1]*Binomial[8,3]*1*2^3+Binomial[4,2]*Binomial[8,2]*3*2^2+Binomial[4,3]*Binomial[8,1]*7*2^1+Binomial[4,4]*Binomial[8,0]*15*2^0)/(Binomial[12,4]*2^4)= 4271/7920 = 0.539268
 
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