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(i) Please don't reply to nine-year-old posts unless it's absolutely relevant and necessary.

(ii) This is almost correct, but it doesn't account for scrambles with multiple crosses already solved… or it would be, if you actually calculated your approximation correctly. Your calculation is off by a factor of two and you should've gotten 1/31680.

Just got 6 center skips in 7 consecutive Master pyraminx solves. Each happens with 1/12 probability, so one should expect to see 6 or more out of 7 consecutive solves 78 times out of 12^7 solves, or roughly once every 459382 solves. Not quite as extraordinary as my accidental XXcross on megaminx, but quite rare.

My "first-order" estimate for quad CN cross is that it should be around 5.01 moves average, interpolating between the dual CN and full CN values on Lars's site.

Python:

dualcn = [53759, 806253, 8484602, 74437062, 506855983, 2031420585, 2311536662, 175751822, 3672]
fullcn = [30942374, 462820266, 4839379314, 41131207644, 239671237081, 543580917185, 151019930400, 258842496, 40]
cdf = [sum(dualcn[:k+1])/sum(dualcn) for k in range(9)] # this is exact
cdf2 = [1-(1-x)**2 for x in cdf] # computing quad CN from dual CN, assuming approximate independence
cdf3 = [1-(1-x)**3 for x in cdf] # computing full CN from dual CN, assuming approximate independence
extrapolated_quadcn_mean = sum(1-x for x in cdf2) # ~ 4.984
extrapolated_fullcn_mean = sum(1-x for x in cdf3) # ~ 4.753
# since we know the ground truth for the full CN mean, we can use that to roughly
# quantify how much the crosses on different axes fail to be independent
correction = sum(i*x for i,x in enumerate(fullcn))/sum(fullcn) - extrapolated_fullcn_mean
corrected_quadcn_mean = extrapolated_quadcn_mean + correction/2
print(corrected_quadcn_mean) # prints 5.012148701428693

Probably won't be too hard to compute a much better approximation using random sampling, or with enough CPU time, the true exact value.

---

Wrote a bit of code, spent a bit of CPU time. Quad cross with 2^27 ~ 134 million total samples:

cross colours: white, yellow, red, orange

depth

samples

proportion

0

2826

0.000021

1

42502

0.000317

2

443883

0.003307

3

3827135

0.028514

4

24023111

0.178986

5

70399118

0.524514

6

35158823

0.261954

7

320330

0.002387

8

0

0.000000

Mean: 5.01942(7) moves

My estimate above was pretty close! Also, nobody asked for this, but here's the same table for triple CN (@OreKehStrah may be interested?):

cross colours: white, red, green (three mutually adjacent choices)

depth

samples

proportion

0

2162

0.000016

1

31567

0.000235

2

331994

0.002474

3

2887422

0.021513

4

18711235

0.139410

5

63047910

0.469744

6

47742095

0.355706

7

1463343

0.010903

8

0

0.000000

Mean: 5.18663(7) moves

cross colours: white, yellow, red (two opposite choices + a third one)

depth

samples

proportion

0

2068

0.000015

1

31820

0.000237

2

333504

0.002485

3

2897351

0.021587

4

18868616

0.140582

5

63934436

0.476349

6

46916485

0.349555

7

1233448

0.009190

8

0

0.000000

Mean: 5.17570(7) moves

NOTE: 8-move crosses are vanishingly rare when you can do at least three cross colours and my sampling just happened to miss them. They're obviously not impossible.

To put all the information together:

crosses

mean

%

1

5.8120

0%

2 (adjacent)

5.4081

40.3%

2 (opposite)

5.3872

42.4%

3 (adjacent)

5.1866

62.4%

3 (opp + another)

5.1757

63.5%

4 (all except adj)

5.0259

78.4%

4 (all except opp)

5.0194

79.1%

5

4.9054

90.4%

6

4.8095

100%

The last column is the "percentage benefit relative to going from fixed cross to full CN", so e.g. going from fixed cross to quad CN is about 79% as good as going to full CN.

According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400).
Skipping F2L and OLL, and corner CP has the odds of 1 in 37 trillion (1/37924385587200).

You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.

You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.

If you already have a cross, then there are fewer possibilities for the edge piece in the first pair, making it more likely to get that pair solved than if a cross already was there. If Y is the event of a solved pair, you're describing the event Y | X, and then independence is not necessary for multiplying.

If you already have a cross, then there are fewer possibilities for the edge piece in the first pair, making it more likely to get that pair solved than if a cross already was there. If Y is the event of a solved pair, you're describing the event Y | X, and then independence is not necessary for multiplying.

I think there was a misunderstanding. I'm talking about the chance of getting an F2L pair pre-solved with cross accounted for. I probably should've been clearer.

I think there was a misunderstanding. I'm talking about the chance of getting an F2L pair pre-solved with cross accounted for. I probably should've been clearer.

You have X chance of getting a cross skip and Y chance of getting F2L pair skip. That means that only Y of X's possibilities are xcrosses. They're independent; getting a F2L pair skip accounts for the fact that the cross is constructed.

First, to answer the X-cross probability question:

fixing 5 edges and 1 corner, permuting and orienting and permuting the remaining 7 edges and 7 corners can be done in this many ways: 7!*7!*2^6*3^6/2 = 592568524800.

That assumes a specific cross color and a specific pair though. To account for one cross color but any of the four pairs, multiply these states by 4: 2370274099200

Dividing by the full number of cube states results in the probability of an XCross on a specific color: 2370274099200/(12!*8!*2^11*3^7/2) = 5.48*10^-8

Multiply it by 6 for an x-cross on any color: 3.29*10^-7

And there you go, probabilities that are so small that I wonder what purpose you had for even asking the question in the first place.

First, to answer the X-cross probability question:

fixing 5 edges and 1 corner, permuting and orienting and permuting the remaining 7 edges and 7 corners can be done in this many ways: 7!*7!*2^6*3^6/2 = 592568524800.

That assumes a specific cross color and a specific pair though. To account for one cross color but any of the four pairs, multiply these states by 4: 2370274099200

Dividing by the full number of cube states results in the probability of an XCross on a specific color: 2370274099200/(12!*8!*2^11*3^7/2) = 5.48*10^-8

Multiply it by 6 for an x-cross on any color: 3.29*10^-7

And there you go, probabilities that are so small that I wonder what purpose you had for even asking the question in the first place.

15.70 B2 U' B2 R2 D F2 L2 U R2 F2 D2 R' U' B D2 U' F' U' B L2 U'

x2 // inspection
F D F2 D L2 U F2 // cross
y' U' R' U2 R y U' L' U L // 1st pair
U' R' U R2 U' R' // 2nd pair
y2 U L' U2 L U L' U' L // 3rd pair
R U R' U2 R U2 R' y' U R' U' R // 4th pair
M2 U' M2 U2 M2 U' M2 // PLL
U // AUF

what's chance of getting OLL skip and H perm ?
and is it higher than OLL skip and Z perm ?