Probability Thread

[Miklopy]

What are the chances of having a cross already done during a scramble?

The chances of a cross skip is 1/15840
The calculation is 1/24 x 1/22 x 1/20 x 1/18 x 6

 

[xyzzy]

The chances of a cross skip is 1/15840
The calculation is 1/24 x 1/22 x 1/20 x 1/18 x 6

(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.

(iii) The exact probability is around 1/31704.

 

[whatshisbucket]

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.

 

[ottozing]

http://www.cubezone.be/crossstudy.html

I'd like to know what the numbers are for "Quad Cross" aka Dual on two sets of colours (Half turn only because who cares about Quarter turn)

I suspect that the numbers would be basically the same as CN

I'm also interested because I know Max Park was Quad Cross for a very long time (I believe he's CN now)

 

[Tranman64]

What are the chances of having F2L skip?

 

[Kit Clement]

http://www.cubezone.be/crossstudy.html

I'd like to know what the numbers are for "Quad Cross" aka Dual on two sets of colours (Half turn only because who cares about Quarter turn)

I suspect that the numbers would be basically the same as CN

I'm also interested because I know Max Park was Quad Cross for a very long time (I believe he's CN now)

Quoting this request, as it's a fairly interesting question and has implications for the relative gains of levels of color neutrality.

 

[xyzzy]

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.

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.

 

[mencarikebenaran]

What are the chances of having F2L skip?

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).

 

[ProStar]

Chances for an Xcross to already be finished?

 

[brododragon]

Chances for an Xcross to already be finished?

Dunno, but all you need to find is the chance of a pair skip and the chance of a cross skip, then multiply them.

 

[Kit Clement]

Dunno, but all you need to find is the chance of a pair skip and the chance of a cross skip, then multiply them.

That assumes Independence, which is not true in this case.

 

[brododragon]

That assumes Independence, which is not true in this case.

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.

 

[Kit Clement]

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.

 

[brododragon]

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.

 

[ProStar]

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'm talking about scrambling the cube, and there being a solved cross and a solved(not just paired up) F2L pair

 

[Cuberstache]

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.

Dude, Kit's a literal statistics professor. I don't think you should argue with him.

 

[brododragon]

I'm talking about scrambling the cube, and there being a solved cross and a solved(not just paired up) F2L pair

That's what I'm thinking. Solved and placed.

Dude, Kit's a literal statistics professor. I don't think you should argue with him.

Ya I think there was a misunderstanding between us.

 

[Kit Clement]

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.

 

[ProStar]

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.

I think I'd better just hope for the 2 mover

 

[mencarikebenaran]

got second OLL skip and H perm

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 ?