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:

- "How many positions require 2 moves to solve?"

OR
- "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.)