4x4x4 LL odds

[Hadley4000]

I was just solving 4x4x4, and my LL was OLL parity. What are the odds of this happening?

And I wasn't timing it >_<

 

[Kian]

I was just solving 4x4x4, and my LL was OLL parity. What are the odds of this happening?

1/2 chance no pll parity
x
1/2 chance of oll parity
x
1/72 chance of pll skip
x
1/216 chance of oll skip


=1/62208? Maybe? Someone correct me if I overlooked something, though.

 

[blah]

Pure or wide OLL parity? (Actually I don't think it makes a difference...)

And more importantly, which alg do you use?

 

[Hadley4000]

Pure or wide OLL parity? (Actually I don't think it makes a difference...)

And more importantly, which alg do you use?

Wide. (Rr)2 B2 (Rr)' U2 (Rr)' U2 B2 (Rr)' B2 (Rr) B2 (Rr)' B2 (Rr2) B2

 

[blah]

Kian's right.

 

[Stefan]

Kian's right.

You're wrong.

 

[mazei]

Erm what do you mean by pure or wide? As in the algs?

 

[Kian]

Kian's right.

You're wrong.

Stefan, I was unsure, as I noted. Could you explain the error?

I guess I could also look at it was a 3/4 chance of having either PLL or OLL parity, and then the same 1/72 for a PLL skip and 1/216 for an OLL skip, thus resulting in 3/62208? I could see a few ways to look at this, but from your response I would figure you believe there is only one. Could you enlighten us?

 

[Stefan]

The way I see it, Hadley didn't provide quite enough information. But I assume he asked for the probability to solve the last layer as AUF1+OLLParity+AUF2. What you computed was for OLLParity+AUF1, missing a factor 4. It's not like if Hadley came across the OLLParity "on the right" he would blindly apply his alg "on the front". He'd do U or y and then apply it.

So my answer is 1/15552. Computed your way multiplied by four, or alternatively like this:

Raw number of possible LL cases: 4! * 4! * 3^3 * 2^4 = 248832.

All different AUF1+OLLParity+AUF2 (I mean differing by AUF1 and AUF2) create different results, meaning you have 4*1*4 = 16 good cases.

Probability = good/all = 16/248832 = 1/15552

 

[blah]

The way I see it, Hadley didn't provide quite enough information. But I assume he asked for the probability to solve the last layer as AUF1+OLLParity+AUF2. What you computed was for OLLParity+AUF1, missing a factor 4. It's not like if Hadley came across the OLLParity "on the right" he would blindly apply his alg "on the front". He'd do U or y and then apply it.

So my answer is 1/15552. Computed your way multiplied by four, or alternatively like this:

Raw number of possible LL cases: 4! * 4! * 3^3 * 2^4 = 248832.

All different AUF1+OLLParity+AUF2 (I mean differing by AUF1 and AUF2) create different results, meaning you have 4*1*4 = 16 good cases.

Probability = good/all = 16/248832 = 1/15552

Actually I assumed OLLParity+AUF1. I didn't use Kian's working though, I did my own and arrived at the same answer.

12*9*6*8*6*4*2 = 248832
4/248832 = 1/62208

If I used 2 AUFs, then it would've been 16/248832 = 1/15552.

So, my bad assumption?

 

[Hadley4000]

The way I see it, Hadley didn't provide quite enough information. But I assume he asked for the probability to solve the last layer as AUF1+OLLParity+AUF2. What you computed was for OLLParity+AUF1, missing a factor 4. It's not like if Hadley came across the OLLParity "on the right" he would blindly apply his alg "on the front". He'd do U or y and then apply it.

So my answer is 1/15552. Computed your way multiplied by four, or alternatively like this:

Raw number of possible LL cases: 4! * 4! * 3^3 * 2^4 = 248832.

All different AUF1+OLLParity+AUF2 (I mean differing by AUF1 and AUF2) create different results, meaning you have 4*1*4 = 16 good cases.

Probability = good/all = 16/248832 = 1/15552

I can't remember the AUFs.

So it is the same as a LL skip on 3x3, and parity doesn't effect the odds?

 

[Kian]

The way I see it, Hadley didn't provide quite enough information. But I assume he asked for the probability to solve the last layer as AUF1+OLLParity+AUF2. What you computed was for OLLParity+AUF1, missing a factor 4. It's not like if Hadley came across the OLLParity "on the right" he would blindly apply his alg "on the front". He'd do U or y and then apply it.

So my answer is 1/15552. Computed your way multiplied by four, or alternatively like this:

Raw number of possible LL cases: 4! * 4! * 3^3 * 2^4 = 248832.

All different AUF1+OLLParity+AUF2 (I mean differing by AUF1 and AUF2) create different results, meaning you have 4*1*4 = 16 good cases.

Probability = good/all = 16/248832 = 1/15552

That makes sense. I didn't consider that he taking AUF's into account in that instance, probably because they weren't mentioned explicitly, but given that he said the only thing he did on the LL was an OLL Parity alg, I should have.

 

[Stefan]

I can't remember the AUFs.

You don't need to. Certainly you weren't asking about the exact single case you had but about all cases of the form AUF1+OLLParity+AUF2, right?

So it is the same as a LL skip on 3x3...?

It happens to have the same probability.

...and parity doesn't effect the odds?

It does, that's what the above factor 4 discussion was about.

 

[Stefan]

I didn't consider that he taking AUF's into account in that instance, probably because they weren't mentioned explicitly, but given that he said the only thing he did on the LL was an OLL Parity alg, I should have.

Not necessarily. I'd say either take his statement literally or not. Here my earlier "Hadley didn't provide quite enough information" comes into play. If you take him literally and he really wanted to know about LL=OLLParity, no AUF or cube rotations at all, then the probability is 1/248832, same as for LL skip. If you don't take him literally, then I'd make the in my opinion reasonable assumption that he wanted to know about AUF1+OLLParity+AUF2, with 1/15552 probability. He just didn't give a reason to assume he's asking about AUF1+OLLParity or about OLLParity+AUF1.

 

[mrbiggs]

I can't remember the AUFs.

You don't need to. Certainly you weren't asking about the exact single case you had but about all cases of the form AUF1+OLLParity+AUF2, right?

So it is the same as a LL skip on 3x3...?

It happens to have the same probability.

...and parity doesn't effect the odds?

It does, that's what the above factor 4 discussion was about.

So the AUF odds cancel out the parity odds?

Neat.