Probability Thread

[bluedasher]

how many moves can a "star" on the megaminx always be solved in? Just curious because I ordered mine today and I want to know some statistics prior to solving it.

 

[bluecloe45]

Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')

 

[bluedasher]

Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')

I'm not really good with probabilities, but here it goes. According to Jessica Fridrich's page...
http://www.ws.binghamton.edu/fridrich/Mike/orient.html
the probability of that algorithm occurring is 1/54. then *3 = 1/162

 

[That70sShowDude]

1/157464 for that oll 3 times in a row

 

[mr. giggums]

I'm not really good with probabilities, but here it goes. According to Jessica Fridrich's page...
http://www.ws.binghamton.edu/fridrich/Mike/orient.html
the probability of that algorithm occurring is 1/54. then *3 = 1/162

Actually you would do 1/54 ^3 not *3 which would = 1/157464

Edit: Ninja'd twice in less than 5 minutes but mine is more detailed

 

[bluedasher]

1/157464 for that oll 3 times in a row

That sounds more like it.

 

[bluecloe45]

1/157464 for that oll 3 times in a row

That just happened LOLOLOOL

 

[Rinfiyks]

Whats the probability of having the same oll 3 times in a row? (R U R' U' R' F R F')

Well there are 57 OLLs, 58 including the solved case.
20 and 58 have 4 symmetries.
1, 21, 55, 56, 57 have 2 symmetries.
The remaining 51 have 1 symmetry.
2/(4x) + 5/(2x) + 51/x = 1
x = 54
Your OLL has 1 symmetry, so its probability of occurrence is 1/54 (2 symmetries would be 1/108, 4 would be 1/216). Oh bother, I could've just divided 216 by 4 to get 54 instead of adding them all up :fp

Anyway... probability of getting your OLL thrice in a row is 1/54^3 = 1/157464
But of course, it's not less likely that you will get other OLLs 3 times in a row.
So,

1/29 chance you get one of 2 OLLs with 1/216
5/58 chance you get one of 5 OLLs with 1/108
51/58 chance you get one of 51 OLLs with 1/54
Edit: This is all wrong /\
This is right \/
2/216 chance you get one of 2 OLLs with 1/216
5/108 chance you get one of 5 OLLs with 1/108
51/54 chance you get one of 51 OLLs with 1/54

Square it, not cube it, because you don't care which OLL is repeated 3 times, so it is a 1 in 1 chance that you get the first one first try.
2/216 * 1/216^2 + 5/108 * 1/108^2 + 51/54 * 1/54^2
= 551/1679616
= 1 in ~3048
Thanks, Lucas Garron, for pointing that out.

That just happened LOLOLOOL

Yeah but you'd be just as pleased to get any other OLL 3 times in a row, so it's not that exciting.
It's like saying
OMG I got D2 B2 D' U' F D2 B2 U F R D2 L' R U' R B2 R B D' scramble! Probability 1 in 43 quintillion!

 

[Lucas Garron]

Square it, not cube it, because you don't care which OLL is repeated 3 times, so it is a 1 in 1 chance that you get the first one first try

Sure?

It sort of appears right, but my calculations get 551/1679616:

c = Select[Tuples[Table[{0, 1, 2}, {4}]], Mod[Total[#], 3] == 0 &];
e = Select[Tuples[Table[{0, 1}, {4}]], Mod[Total[#], 2] == 0 &];
p = (First[Sort[Table[RotateRight[#, i] & /@ #, {i, 0, 3}]]] & /@ Tuples[{c, e}] // Tally)[[All, -1]]/216;
Total[p^3]

EDIT: Oh, duh. Your weights in the sum are off:

1/29 * 1/216^2 + 5/58 * 1/108^2 + 51/58 * 1/54^2!

1/29, eh?
In practice, since most of the OLLs are the 51, the numbers work out approximately the same, but it's not a good way to do the math.

Correct would be (2 * 1/216) * 1/216^2 + (5 * 1/108) * 1/108^2 + (51 * 1/54) * 1/54^2.
I did 2 * 1/216^3 + 5 * 1/108^3 + 51 * 1/54^3, which is the same, but I think clearer: What's the chance of getting three in a row, for each OLL?

 

[Rinfiyks]

EDIT: Oh, duh. Your weights in the sum are off:

Ohhhhhhh yeaaaaaaaah! Stupid mistake. I'll think twice before I post any maths in future.

 

[dada222]

Suppose you solve the cross on the bottom and top layer, how many algorithms would you need to make up to solve both faces simultaneously?

Forget it, just realized the number is huge.

 

[ben1996123]

Bump.

Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.

Would this just be 1/72^6, or would it be 4/72^6 since there are 2 different pll's (H/skip)?

EDIT: next solve forced skip 0.o

 

[Johan444]

Bump.

Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.

Would this just be 1/72^6, or would it be 4/72^6 since there are 2 different pll's (H/skip)?

EDIT: next solve forced skip 0.o

In that fixed order it'd be 1/72^5 * P(Forced skip)

 

[cuBerBruce]

Just did some 3x3 solves, and for the PLL's, I got skip, h perm, h perm, skip, skip, forced skip.

Would this just be 1/72^6, or would it be 4/72^6 since there are 2 different pll's (H/skip)?

The standard PLL probabilities assume the permutation of the LL pieces are ignored prior to PLL. If you're sometimes forcing a PLL skip, then clearly you're not always ignoring the permutation of the LL pieces until the PLL step. So without more information on how the PLL is being influenced, accurate probabilities can not be computed.

It's also not clear what exactly you probability you're trying to compute. Do you mean exactly that sequence skip, H, H, skip, skip, skip for a given set of six consecutive solves? (and not skip, H, skip, H, skip, skip; nor skip, skip, H, skip, H, H; nor any other combination of H's and skips)

 

[Rpotts]

What's the probability of getting No AUF LL skip followed by a solve where the last slot + LL is just J perm + AUF? I basically had two LL skips in a row, I inserted the last pair with j perm, giving me a U2 LL.

 

[cuBerBruce]

What's the probability of getting No AUF LL skip followed by a solve where the last slot + LL is just J perm + AUF? I basically had two LL skips in a row, I inserted the last pair with j perm, giving me a U2 LL.

Assuming method used was not trying to influence the last layer case...

Probability of no AUF PLL skip: 1/((1/2)*(4!)^2)*(2^3)*(3^3)) = 1/62208

Probability of (Last slot J-Perm with no AUF before) + AUF (after J-Perm, or no AUF):
8/((1/2)*((5!)^2)*(2^4)*(3^4)) = 1/1166400

If possibly an AUF before the J-Perm, then multiply by 4 (i.e., 4 times more likely).

Probability of precisely these two cases in that order on two consecutive solves: 1/(62208*1166400) = 1/72,559,411,200

 

[Andrew Ricci]

How many unique cross cases are there?

 

[nccube]

What is the probability of getting a cubeshape square-1. One scramble from Spanish Championships had it.

 

[Mike Hughey]

What is the probability of getting a cubeshape square-1. One scramble from Spanish Championships had it.

4 in 3678 (1 in 919.5). I have the odds of all of the shapes for square-1 in my blindfold solving page that's listed in my signature.

 

[cuBerBruce]

How many unique cross cases are there?

The 4 cross pieces can be arranged 12*11*10*9 ways with 2^4 orientation possibilities for a total of 190080 arrangements.

If you consider rotational symmetry, I calculate this reduces the number of cases to 47652. If you also consider mirror symmetry, I calculate 24098 cases.

If you consider allowing AUF & ADF, I calculate 5582 cases, not counting mirrors as the same; and 2909 cases if you count mirrors as the same.

Subtract 1 from the above numbers if you don't want to count the solved case.