Probability Thread

[cubenerd74]

18 possible 1st moves, 18 possible second moves.

Actually, there are 15 possible second moves because moves on the same side(e.g. R R2) cancel out.

 

[Christopher Mowla]

How many 2 move states exist:
18 possible 1st moves, 18 possible second moves. If the moves are parallel, the order doesn't matter. We eliminate all scrambles that start with a U, F or L move and are followed by any of the 3 parallel moves, so that's 3*3*3 scrambles states eliminated. Therefore, there are 18^2 - 27 2 move states.

Divide this by the total number of states and simplify and you get a probability of 1 in 14562964065484800

If you want the probability of getting it in competition, there are 19 fewer total possible states so the probability is 297 in 43252003274489855981, non-simplifyable and basically the same probability, but since I accounted for the 27 duplicate 2 move states I couldn't just ignore 19 illegal states I suppose (although they're waaaayyy less significant than the 27 duplicate states considering their respective proportion to the the total they belong to)

If you look at cube20.org, they calculated 243 (rather than 297). FYI.

Also,

probability of a 2 mover on 3x3? idk if someone asked this before "two" is too short to look up on the site and there's so many posts in this thread it would take forever to find this specific question.

I'm not sure if you were asking:

1. "How many positions require 2 moves to solve?"
   OR
2. "How many (non-redundant) algorithms that are 2 moves long are there?"

If you were asking question #1, then it's 243. If so, this is the probability. (Using the same idea as @Jorian Meeuse, just used a different numerator.)

If you were possibility asking question #2, well, Herbert Kociemba (the mathematician who was a part of the team who found god's number, who programmed Cube Explorer, etc.) found an incredibly simple "generating function" which quickly gives you such numbers.

For convenience, here is the generating function computed at Wolfram|Alpha for the 3x3x3 (n->3) for move sequences ranging from 0 moves to 20 moves ({x,0,20}). If you know basic algebra, then you may be familiar with the terminology "coefficients of a polynomial". (Basically it's the number to the left of the x's... ignore the O(x^21)!)

1 + 18 x + 243 x^2 + 3240 x^3 + 43254 x^4 + 577368 x^5 + 7706988 x^6 + 102876480 x^7 + 1373243544 x^8 + 18330699168 x^9 + 244686773808 x^10 + 3266193870720 x^11 + 43598688377184 x^12 + 581975750199168 x^13 + 7768485393179328 x^14 + 103697388221736960 x^15 + 1384201395738071424 x^16 + 18476969736848122368 x^17 + 246639261965462754048 x^18 + 3292256598848819251200 x^19 + 43946585901564160587264 x^20 + O(x^21)

What this is saying is that there is/are:

• 1 algorithm that is 0 moves long (kind of redundant, but... )
• 18 algorithms that are 1 move long
• 243 algorithms that are 2 moves long (... maybe you can see by now that the number after the ^ represents the number of moves!)
• 3240 algorithms that are 3 moves long, etc.

And you can manually "verify" that these numbers are correct by right-clicking on all stickers on a cube in Cube Explorer (to make they gray)

and then click on the Add and Solve button. Let it run some (maybe until it reaches 5 move long algorithms) and count the number of algorithms that are 1 move to 4 moves (by putting all "maneuvers" in the main window and then exporting the maneuvers into a text file).

(It's crazy, but the 2 coincide! Wow!)

If you add all of these bold numbers together, you get: 47505455028489778073776. The total number of positions for the 3x3x3 is less: 43252003274489856000.

So comparing the bold numbers above to the numbers at cube20.org, this tells us that number of positions which require n moves <= number of algorithms which are of length n. (The first 3 numbers coincide, but then the sequence for the number of algorithms that are n moves long increases faster.)

Finally, if we wanted to see what the probability would be for (non-redundant) move sequences that are 2 moves long out of the sum of all move sequences that are from 1 move long to 20 moves (assuming that the WCA would actually support such scrambles... but they don't... and the number of moves the algorithms is does not always mean that they solve positions which require the number of moves that they are, right?), we simply divide 243 by the big number: 47505455028489778073775. (Subtract 1 for algorithms/"maneuvers" that are 0 moves.)

Probability.

P.S.

Mr. Kociemba's generation function supposedly works for the OBTM move metric for the nxnxn. For example, if you look back to his post and see that he mentioned:

This means, on the 4x4x4 there are for example 567 maneuvers with length 2

Well, I just typed 4 in place of 3, and 7 in place of 20 to get the same series that he got. Here.

(You can play with that all you want... and for those interested, there are more posts about his generation function following that first one.)

 

[Jorian Meeuse]

If you look at cube20.org, they calculated 243 (rather than 297). FYI.

LOL I completely forgot that things like R R aren't 2-movers, removing another 3*18=54 states, resulting in 243

Not my smartest move indeed

 

[Christopher Mowla]

LOL I completely forgot that things like R R aren't 2-movers, removing another 3*18=54 states, resulting in 243

Not my smartest move indeed

Don't sweat it. I have had my moments too. (And I'm certainly no "authority" on this topic.)

(I still give you an A for effort!)

 

[Jorian Meeuse]

Don't sweat it. I have had my moments too. (And I'm certainly no "authority" on this topic.)

(I still give you an A for effort!)

Haha no I don't mind a little mistake every now and then, the worst thing that can happen is that I learn from them, which is even a good thing!

 

[Discomantis]

What are the odds of a no-bar scramble on 2x2

 

[SpeakCuba]

What are the odds of a no-bar scramble on 2x2

I’m not going to do the math, but just based on the scrambles I’ve gotten, I would say it’s kind of common.

 

[Arcanist]

What are the chances of a premade V on pyraminx with all tips solved?

 

[OtaMota]

What are the chances of a premade V on pyraminx with all tips solved?

All tips solved is 1/81. Idk about V. So premade v is 1/81 * 1/v

 

[cubenerd74]

What are the chances of a premade V on pyraminx with all tips solved?

The chance of a solved V is somewhat easy to calculate. First, you need to calculate the chance of a specific V being made, and multiply by 12(try to figure out why).
For a specific V to be premade, all 3 centers need to be oriented which is a 1/3^3 = 1/27 chance. Also, the 2 edges need to be solved, which is a 1/12*10 = 1/120 chance. (The second edge has a 1 in 10 chance of being solved because there's 1 fewer spot the edge could be in.)
As Otamota said, the chance of having all tips solved is 1/81, so the final number is 1/81 * 1/27 * 1/120 * 12=1/21870. Seems about right to me.

 

[ruffleduck]

The chance of a solved V is somewhat easy to calculate. First, you need to calculate the chance of a specific V being made, and multiply by 12(try to figure out why).

This does not account for overlap. There are scrambles where multiple V's are made.

 

[Jason Tzeng]

Get this clock scramble few minutes ago, what's the odds of this kind of scramble exist? (Only need to scramble one side)

UR3+ DR1+ DL2- UL2- U2- R1+ D6+ L2- ALL0+ y2 U0+ R0+ D0+ L0+ ALL0+ @2024-03-09 13:41:38

 

[finger go brr]

i tried to calculate the chance of an lsll skip and i got 1/2,332,800 for unoriented and 1/145,800 for pre-oriented

is this right?

 

[Celeritardum]

Get this clock scramble few minutes ago, what's the odds of this kind of scramble exist? (Only need to scramble one side)

UR3+ DR1+ DL2- UL2- U2- R1+ D6+ L2- ALL0+ y2 U0+ R0+ D0+ L0+ ALL0+ @2024-03-09 13:41:38

Assuming that each move has an equal chance of being generated (ex: U1- and U5+) then that would be 1/12^5, or approximately 0.0004% chance

 

[OtaMota]

Try this: R U R' U' R' F R F' x2 R U R' U' R' F R F'
Spot the 4-mover.

Found it.

Generated By csTimer on 2024-04-27
single: 1.22

Time List:
1. 1.22 R2 U2 F' R2 U' R2 F' U2 R'

z x2 U2 L' U2 F

 

[Jorian Meeuse]

I seriously just got the exact same 2x2 scramble twice in a row but from a different angle.
First I get R U F2 R' U F2 U R2 U2
Immediately afterwards, F R U2 F' R U2 R F2 R2

The solutions:
1. z2 U2 R2 U' R2 U' F R2 U' R'
2. x' z U2 R2 U' R2 U' F R2 U' R'

Also notice how the second scramble is literally exactly the RUF notation for x' z' (first scramble).
Now this (excluding the scramble sequences also matching) is actually slightly more likely than getting a LL skip on a given 3x3 solve, but it sure doesn't feel that way

 

[FineGeoduck2211]

Get this clock scramble few minutes ago, what's the odds of this kind of scramble exist? (Only need to scramble one side)

UR3+ DR1+ DL2- UL2- U2- R1+ D6+ L2- ALL0+ y2 U0+ R0+ D0+ L0+ ALL0+ @2024-03-09 13:41:38

not to mention it's a 7 mover too?!?!?! (I know this doesn't contribute to anything pls don't get mad at me)

what is the probability of getting 7 mover on clock (single moves not simul)?

 

[Mike Hughey]

I seriously just got the exact same 2x2 scramble twice in a row but from a different angle.
First I get R U F2 R' U F2 U R2 U2
Immediately afterwards, F R U2 F' R U2 R F2 R2

The solutions:
1. z2 U2 R2 U' R2 U' F R2 U' R'
2. x' z U2 R2 U' R2 U' F R2 U' R'

Also notice how the second scramble is literally exactly the RUF notation for x' z' (first scramble).
Now this (excluding the scramble sequences also matching) is actually slightly more likely than getting a LL skip on a given 3x3 solve, but it sure doesn't feel that way

Interesting. What app generated these scrambles?

I'm guessing that it might make sense for the scrambles to be identical if they were random state generated; I would assume the solver would likely find identical solutions minus the rotations for two essentially identical states like that.

 

[Jorian Meeuse]

not to mention it's a 7 mover too?!?!?! (I know this doesn't contribute to anything pls don't get mad at me)

what is the probability of getting 7 mover on clock (single moves not simul)?

That depends on how smart you are with choosing your solving orientation, but if we assume you choose your orientation randomly, the probability of skipping a given move is 1/12, so the probability of skipping precisely 7 out of 14 is (14!/(7!^2))*(1/12^7)*(11/12)^7 ≈ 0.00005
But yeah it is quite optimizable by not being wasteful of skips
And also it should be mentioned this is the probability of your solution being 7 moves, not the probability of there existing a 7 moves solution.

Interesting. What app generated these scrambles?

I'm guessing that it might make sense for the scrambles to be identical if they were random state generated; I would assume the solver would likely find identical solutions minus the rotations for two essentially identical states like that.

Cstimer. That is an interesting insight, I hadn't thought about it like that.

 

[FineGeoduck2211]

Maybe someone already asked this, IDK

but what is the probability of getting a 4 mover on 2x2?