Probability Thread

[Abram Grimsley]

What is the probability of after finishing first three cross edges and finishing the rest of centers on 4x4, that all edges or all but a pair of edges are solved and you skip right into 3x3 stage?

 

[OreKehStrah]

What is the probability of after finishing first three cross edges and finishing the rest of centers on 4x4, that all edges or all but a pair of edges are solved and you skip right into 3x3 stage?

If someone doesn’t beat me to it I’ll check later. Intuitively I would guess it’s 1/(18*16*14*12…*4*2*1)

 

[Isaiah Scott]

If someone doesn’t beat me to it I’ll check later. Intuitively I would guess it’s 1/(18*16*14*12…*4*2*1)

1 in 185,794,560

 

[Isaiah Scott]

Yeah its crazy. I didnt think itd be that high

 

[Abram Grimsley]

I know! Lol

 

[OreKehStrah]

Yeah its crazy. I didnt think itd be that high

Thinking about it more, I think it's actually lower at
1/(18*17*16*..*4*3)

The reason being after 3/4 cross is done, there are 18 positions a wing can be in. So if you randomly choose any wing to start, that removes 1 position so there are now 17 positions the other wing can be in but only 1 would form a paired dedge. Thus 1/18 * 1/17 for one extra dedge being paired which converts to 1/(18*17). Now there are 16 positions. The process repeats. It then stops, ending at 4*3 because once you get down to the last 3 wing locations, 1/3 will form the next to last dedge. It doesn't matter whether the last 2 wings need to swap since we are only concerned with the dedges being paired.

 

[NmeCuber]

Does anyone know the probability of getting an oll skip, then a normal solve, then another oll skip. This actually happened to me yesterday.

 

[cubenerd74]

Does anyone know the probability of getting an oll skip, then a normal solve, then another oll skip. This actually happened to me yesterday.

If you're asking about exactly that order, then it's 215/10077696 or 0.002%.

 

[EthanMCuber]

exactly that order

I don’t think it would change if the order of the three changed, right?

 

[cubenerd74]

I don’t think it would change if the order of the three changed, right?

No, there are three possible orders, so the probability will triple, at 0.006%.
But it is a pretty unrecognizable difference.

 

[NmeCuber]

If you're asking about exactly that order, then it's 215/10077696 or 0.002%.

No wayyyyyy. Rarer than LL Skip

 

[EthanMCuber]

No, there are three possible orders, so the probability will triple, at 0.006%.
But it is a pretty unrecognizable difference.

Ohhhhhhh. I get it now.

 

[Exmestify]

a couple of weeks ago, i as solving me mega, got an oll skip on that, then i solved my 3x3 and got an oll skip on that as well. they were like next to next solves. probability pls?

 

[Cuberstache]

If you're asking about exactly that order, then it's 215/10077696 or 0.002%.

@EthanMCuber this is a little misleading - this is the probability of being correct if you were to sit down and say "the next three solves will be OLL skip, not skip, and OLL skip". The probability of it happening anywhere in a session of 100 solves or any time in your entire cubing career or whatever is much much higher.

 

[xyzzy]

I don't know how I missed this ping by almost 3 years

IIRC I edited that in after the initial version of the post and I don't think mentions will show up in your notifs in that case.

 

[ruffleduck]

a couple of weeks ago, i as solving me mega, got an oll skip on that, then i solved my 3x3 and got an oll skip on that as well. they were like next to next solves. probability pls?

1/1296 megaminx OLL skip * 1/72 3x3 OLL skip = 1/93312

 

[OreKehStrah]

IIRC I edited that in after the initial version of the post and I don't think mentions will show up in your notifs in that case.

Ah I see. That makes sense.
Pretty cool to see I was right at the time. I had a lot of people tell me 3 adj colors was gonna be worse than op dual cross lol.

 

[EthanMCuber]

1/1296 megaminx OLL skip * 1/72 3x3 OLL skip = 1/93312

3x3 OLL skip is 1/216. 1/72 would be PLL skip

 

[xyzzy]

What is the probability of after finishing first three cross edges and finishing the rest of centers on 4x4, that all edges or all but a pair of edges are solved and you skip right into 3x3 stage?

See formula in my older post:

Let \( n \) be the number of edge pair locations in total. (12 for reduction, 9 for Yau after centres and 3/4 cross, 8 for Yau after centres and cross.) The number of ways of pairing up \( 2n \) items, labelled with two copies of \( 0 \) through \( n-1 \), such that no pair has items of the same label, is:
\( \displaystyle a_n:=\sum_{k=0}^n(-1)^k\binom nk(2n-1-2k)!! \)

(Note: !! means double factorial. The number of ways of pairing up \( 2n \) items, ignoring the order of the pairs and the order within the pairs, is \( \frac{(2n)!}{n!2^n}=(2n-1)!! \).)

If we want to count the number of ways to have \( m \) free pairs out of \( N \) total, we first choose which pairs are free (\( \binom Nm \)), then treat the remaining \( N-m \) pairs as an instance of the above problem (\( a_{N-m} \)). To wit:
\( \displaystyle(\text{# of ways to have }m\text{ free pairs out of }N)=\binom Nma_{N-m} \)

Plug in \( N=9 \), and we get:

# pairs	# cases	probability
0	20314880	0.58953044
1	10716048	0.31097582
2	2844288	0.08254020
3	507360	0.01472340
4	68544	0.00198912
5	7560	0.00021939
6	672	0.00001950
7	72	0.00000209
8	0	0.00000000
9	1	0.00000003

The probability of getting all paired edges is \( 1/17!! = 1/34459425 \approx 0.00000290\% \), while the probability of getting all-but-two paired edges is \( 72/17!! = 72/34459425 \approx 1/479000 \approx 0.000209\% \).

 

[Exmestify]

so, 1 in 279936 chance lol