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Probability Thread

PingPongCuber

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whats also the probability of getting a 2 mover in a 3x3?

Very rare, i know, but its still allowed in WCA regulations and rules.
This ones pretty simple. There are approximately 45 quintillion possible combinations on the cube. 1 is solved. From there there are 6 sides, and on each side you can do 3 types of moves. This means that there are 18 states where it is 1 move from solved. Then on each of those 18 states there are 18 more moves you could do. But, 3 of those moves (the ones on the same face) would result in it being still 1 move away, so in reality 15 more moves. 18 x 15 is 270, so the odds of there being a 2 move scramble (assuming it is a random state scrambler) is 270/45000000000000000000
 

BenChristman1

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hmm ok but what about a probability of a cube having 2 or more edges flipped while the rest of the cube is solved?
whats also the probability of getting a 2 mover in a 3x3?

Very rare, i know, but its still allowed in WCA regulations and rules.
Both of these are dumb questions.
 

Nir1213

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This ones pretty simple. There are approximately 45 quintillion possible combinations on the cube. 1 is solved. From there there are 6 sides, and on each side you can do 3 types of moves. This means that there are 18 states where it is 1 move from solved. Then on each of those 18 states there are 18 more moves you could do. But, 3 of those moves (the ones on the same face) would result in it being still 1 move away, so in reality 15 more moves. 18 x 15 is 270, so the odds of there being a 2 move scramble (assuming it is a random state scrambler) is 270/45000000000000000000
thanks!

Both of these are dumb questions.
if its dumb you dont have to post and say about it.

also, based on bens profile, since his cube is 9x9, can any odd numbered cube be able to have a super flip, or its the same for even numbered?
 
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Athefre

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I mentioned an idea before on Discord. I think a well-developed probability page on the wiki would be very useful. We could have all of the ones that are commonly asked and any others. The page could be structured by categories like steps, methods, states, and others. Then the main post of this topic can be edited to include a link to the wiki page. If someone still asks a common question, then they can be linked to the page.

For anyone that enjoys the probability subject, this would be a fun project and you would be helping the community.
 

erdish

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Of the 43,252,003,274,489,856,000 permutations of the 3x3 cube, how many have the maximal cycle length 1260?

I have an estimate, just generating a random scramble/permutation I get about a 1 in 800 chance of getting cycle length 1260. So maybe around 5.4*10^16?

But is there a way to calculate the exact number?
 

Devagio

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Of the 43,252,003,274,489,856,000 permutations of the 3x3 cube, how many have the maximal cycle length 1260?

I have an estimate, just generating a random scramble/permutation I get about a 1 in 800 chance of getting cycle length 1260. So maybe around 5.4*10^16?

But is there a way to calculate the exact number?
I calculated by hand to get ~8.58*10^15; this is off by a factor of 6 from your estimate.

What I did basically was 1260=2x2x3x3x5x7
The only way to have a 1260 cycle on a cube is if:
You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation.
(To prove that this is the only way is pretty tedious but straightforward, so I’m not including that here)

So the answer is (1/2) * (8c5 * 4!) * (3c3 * 2!) * (12c7 * 6!) * (5c2) * (3c2) * (18 * 3^4) * (12 * 2^7)
 

erdish

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I calculated by hand to get ~8.58*10^15; this is off by a factor of 6 from your estimate.

What I did basically was 1260=2x2x3x3x5x7
The only way to have a 1260 cycle on a cube is if:
You have a 3cycle and a 5cycle of corners, and 7cycle and two 2cycles of edges; such that the corner 3cycle has a net corner twist and at least one of the two edge 2cycles has net edge misorientation.
(To prove that this is the only way is pretty tedious but straightforward, so I’m not including that here)

So the answer is (1/2) * (8c5 * 4!) * (3c3 * 2!) * (12c7 * 6!) * (5c2) * (3c2) * (18 * 3^4) * (12 * 2^7)
Nice, thanks! I like the compact notation. Still I wonder why it's pretty far off the estimate based on throwing the dice. There is another possibility re the possible set of cycle lengths, no?--the case where 1 edge remains fixed? FWIW, here are some stats on 1000 random permutations (hex, so that E=14 and F=15):

1000 random permutations of cycle length 1260
{edge cycle lengths} {corner cycle lengths} N
{1 2 2 4 4 E E E E E E E} {9 9 9 F F F F F} 343
{1 4 4 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 169
{2 2 2 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 326

{2 4 4 4 4 E E E E E E E} {9 9 9 F F F F F} 162
 

Devagio

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Nice, thanks! I like the compact notation. Still I wonder why it's pretty far off the estimate based on throwing the dice. There is another possibility re the possible set of cycle lengths, no?--the case where 1 edge remains fixed? FWIW, here are some stats on 1000 random permutations (hex, so that E=14 and F=15):

1000 random permutations of cycle length 1260
{edge cycle lengths} {corner cycle lengths} N
{1 2 2 4 4 E E E E E E E} {9 9 9 F F F F F} 343
{1 4 4 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 169
{2 2 2 4 4 7 7 7 7 7 7 7} {9 9 9 F F F F F} 326

{2 4 4 4 4 E E E E E E E} {9 9 9 F F F F F} 162
Yep, all of these permutations are of the kind that I mentioned. (At least One two-Cycle in edges of length 4, Exactly one three-cycle in corners of length 9; cycles of size 1,2,2,7 in edges, And of size 3,5 in corners); so that part agrees in both theory and simulation. The mutual ratios of these are supposed to be 1/3, 1/6, 1/3, 1/6; so that seems correct too.
Turns out I made an error in plugging in the expression into a calculator; I re-did it on my phone calculator just now to get 5.15e16, which is quite close to your estimate.
Just try calculating the numerical value of my expression; because the expression seems correct.
 

erdish

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Turns out I made an error in plugging in the expression into a calculator; I re-did it on my phone calculator just now to get 5.15e16, which is quite close to your estimate.
Just try calculating the numerical value of my expression; because the expression seems correct.
Terrific!

Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE.

So the answer is 51,490,480,088,678,400.

A question I would have is are the four basic types seen in the simulation the only ones possible? They are certainly the vast majority.
 
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