The chances of a cross skip is 1/15840What are the chances of having a cross already done during a scramble?
The calculation is 1/24 x 1/22 x 1/20 x 1/18 x 6
The chances of a cross skip is 1/15840What are the chances of having a cross already done during a scramble?
(i) Please don't reply to nine-year-old posts unless it's absolutely relevant and necessary.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
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)
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
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 |
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 |
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 |
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% |
According to Robert Yau, skipping F2L has a 1 in 3.66 billion chance (1/3657830400).What are the chances of having F2L skip?
Dunno, but all you need to find is the chance of a pair skip and the chance of a cross skip, then multiply them.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.
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.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.
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.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.
Dude, Kit's a literal statistics professor. I don't think you should argue with him.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.
That's what I'm thinking. Solved and placed.I'm talking about scrambling the cube, and there being a solved cross and a solved(not just paired up) F2L pair
Ya I think there was a misunderstanding between us.Dude, Kit's a literal statistics professor. I don't think you should argue with him.
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.
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