Probability Thread

Ok, that's one of the two. Because you had only mentioned the top layer and because you had explicitly mentioned 8! (hinting at distinguishing the different white corners), I thought you might've meant 4 possibilities to pick 3 white corners to be in the top layer, then 4 possibilities to place them there (disregarding the placement in the bottom layer). Which btw also leads to 1 + 4*4 + 6*6 + 4*4 + 1.
 
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O_O Whats the probability of finding a completed F2L PAIR during the 15 second inspection? Two pairs?? Three? If you only solve on the white cross, and if you are color neutral??

And I mean just finding the pair inside the cube, not joined with a cross or anything.
 
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For white, there are 4 total correct F2L pairs for white. And a total of (12 Left edges in the scramble).

So only 1/12 of those edges would match with ONE f2l pair corner. And there are 4 possible pairs, so (1/12) * 4 -> So .333 is the probability that AN edge is correct with a corner.
(SOMEONE correct me if I'm wrong.. Remember no cross is made so its 12 possible edges(including yellow))

_____________________________________________________________________________________________________________________________________________________________________

Now if you are color neutral, there are 6 possible sides. So there are going to be that many more possible f2l pairs, 6*4-> 24 F2L pairs Possible NOW.
But there is another variable to be careful for, instead of 12 possible edges to match with each f2l pair, its 6*12 -> 72

24/72 = 33% again!
WRONG!! I don't think that is correct xD :P because each of the probabilities for a single color(white) are independent of each other.
We want to find the full probability TOGETHER!

.333^6 is the probability that EACH cross color will have 33% chance of having an f2l pair = .001, or highly improbable.

But what is the probability that the cross colors combined together will have of .A. free f2l pair given that they have a 33% chance for 1?

(AGain plz correct me if I'm wrong I'm just saying stuff)


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I used Wikipedia I think I found out how to find out O_O

(6 choose 1) * .333^1 * .666^5

WHAT?? Only 26% chance for 1 f2l pair?? ok I did something wrong plz fix it someone

WAIT! I think I forgot to take in to account the probability that if you have 1 CE pair found, that the probability for another CE pair will INCREASE!
 
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O_O I think I'm right O_O

Color Neutral:
Zero pairs found: 0.0872660613 -> 8.7%
One pair found: 0.261798184 -> 26%
Two pairs found: 0.32724773 -> 33%
Three pairs found: 0.218165153 -> 22%
Four pairs found: 0.0818119325 -> 8%
Five pairs found: 0.0163623865 -> 1.6%
Six pairs found: 0.00136353221 -> .1%

Not taking into account this "WAIT! I think I forgot to take in to account the probability that if you have 1 CE pair found, that the probability for another CE pair will INCREASE!"
because I don't really know how to and whether it is already happening because the probability looks like a curve thing.
 
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The probabilities should be A LOT higher for pairs found, but I didn't make it so finding an CE pair will lower the # of edges possible for the next pair to 11, 10, 9 ect.
 
Wow I forgot orientation.

White cross solver:
8 possible matches for a corner, only 1 is correct.
So 1/8 chance of matching the corner, but there are 4 white-corners, so
50% chance(ignoring orientation) of having 1 f2l CE pair for white.

There is a 33% chance of the Corner being oriented the same way as the edge.
So a 1/6 chance of having a true F2l Pair for Color Neutral.

___________________________________________________________________________________________________



White cross solver:
8 possible matches for a corner, only 1 is correct.
So 1/8 chance of matching the corner, but there are 4 white-corners, so
50% chance(ignoring orientation) of having 1 f2l CE pair for white.

There is a 33% chance of the Corner being oriented the same way as the edge.
So a 1/6 chance of having a true F2l Pair for Color Neutral.

One F2L Pair:
1/8(Es left) * 4(Cs left) -> 50% -> 17% (With orientation)

Two F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
0.214285714 -> 7% (With Orientation)

Three F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
1/6(Es left) * 2(Cs left)
0.0714285714 -> 2% (With Orientation)

Four F2L Pairs:
1/8(Es left) * 4(Cs left)
1/7(Es left) * 3(Cs left)
1/6(Es left) * 2(Cs left)
1/5(Es left) * 1(Cs left)
0.0142857143 -> .0047 (With Orientation)
 
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cmhardw said:
I had a really interesting one come up a few days ago. After solving the cross and the first two pairs, all the remaining 12 pieces were in their correct locations, only 6 of them were misoriented.

The probability of having the last 12 pieces of the cube in the correct location after solving the cross and first two pairs is:

1 / [6! * 6!/2] = 1 / 259200

To give a comparison, the chance of a LL skip with no partial edge control, and with the possibility of AUF, is
1 / 15552

I thought it was pretty neat.

Chris
Click to expand...

cubersmith said:
Chances of a LL Skip
Click to expand...

Mission accomplished.
 
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