Probability Thread

[VincentCuber]

odds of a centers skip on 5x5?

 

[OtaMota]

Does anybody know the odds of a PBL skip on square-1?

If you do parity before, I think 1/10368.

 

[Deleted member 74176]

If you do parity before, I think 1/10368.

Wow. I don't trace CSP or anything. I beat my PB though with a PBL skip.

 

[OtaMota]

odds of a centers skip on 5x5?

If you mean all centers, I got a number that had like 47 digits.

If someone wants to check my math, I did 24! and squared it.

 

[Jorian Meeuse]

If you mean all centers, I got a number that had like 47 digits.

If someone wants to check my math, I did 24! and squared it.

You need to account for the fact that there are many center states that are considered solved. Also that a 2-swap of centers is not possible. The total number of center states is (24!)^2 / 2
The number of solved states can be computed like this:
6 (faces) x (4!)^2 (permutations per face that are considered solved) / 2 (eliminate parity)
So the probability is
((24!)^2/2) / (6x(4!)^2/2) = (24!)^2/(6x(4!)^2)
Lmk if something is wrong

 

[THE_V01DL0RD]

You need to account for the fact that there are many center states that are considered solved. Also that a 2-swap of centers is not possible. The total number of center states is (24!)^2 / 2
The number of solved states can be computed like this:
6 (faces) x (4!)^2 (permutations per face that are considered solved) / 2 (eliminate parity)
So the probability is
((24!)^2/2) / (6x(4!)^2/2) = (24!)^2/(6x(4!)^2)
Lmk if something is wrong

111387794911264837077470197382224281600000000

 

[THE_V01DL0RD]

odds of a centers skip on 5x5?

1 in 1 if you cheat
Edit: why does the devil emoji look so cringy?

 

[Discomantis]

Does anybody know the odds of a PBL skip on square-1?

Odds of PBL skip without CSP is 1/20736

Odds of PBL skip with CSP is 1/10368

 

[Discomantis]

And how about a V skip in pyraminx

1 in 4320

 

[Discomantis]

You need to account for the fact that there are many center states that are considered solved. Also that a 2-swap of centers is not possible. The total number of center states is (24!)^2 / 2
The number of solved states can be computed like this:
6 (faces) x (4!)^2 (permutations per face that are considered solved) / 2 (eliminate parity)
So the probability is
((24!)^2/2) / (6x(4!)^2/2) = (24!)^2/(6x(4!)^2)
Lmk if something is wrong

Can't you just "solve" parity by just doing a swap of one green center and another green center while doing another swap?

Since green swapping with green isn't visible, I think it would do the equivalent of a one swap, while technically being a two swap alg.

 

[Jorian Meeuse]

Can't you just "solve" parity by just doing a swap of one green center and another green center while doing another swap?

Since green swapping with green isn't visible, I think it would do the equivalent of a one swap, while technically being a two swap alg.

Yeah now that I think about it, you can swap two centers and two wings, amounting to any center state being possible. This doesn't matter for the probability though, because I divided both the numerator and the denominator by 2 and cancelled them.
Thanks for the correction!

 

[xyzzy]

Also that a 2-swap of centers is not possible.

Yeah now that I think about it, you can swap two centers and two wings, amounting to any center state being possible.

The t-centres and wings have the same parity, so you can do 2-cycle t-centres + 2-cycle wings.

The x-centres and corners and midges have the same parity, so you can do 2-cycle x-centres + 2-cycle corners + 2-cycle midges. (edit: forgot about midges, oops)

 

[Jorian Meeuse]

The t-centres and wings have the same parity, so you can do 2-cycle t-centres + 2-cycle wings.

The x-centres and corners have the same parity, so you can do 2-cycle x-centres + 2-cycle corners.

Thanks, I didn't know that.

 

[Jake - Speedcuber]

I just got here LOL, and don't know if this exists. BUIT it would be very cool to make a Google Sheet/Doc of all the probablities!

 

[Caden Fisher]

I don‘t know how easy this would be to calculate, but what is the probability of having the same EP on both sides of a Sq1

 

[Jorian Meeuse]

I don‘t know how easy this would be to calculate, but what is the probability of having the same EP on both sides of a Sq1

No CSP:
10 EP cases so 1 in 10
With CSP:
5 EP cases so 1 in 5

This is not considering post or pre AUFs

 

[Caden Fisher]

No CSP:
10 EP cases so 1 in 10
With CSP:
5 EP cases so 1 in 5

This is not considering post or pre AUFs

Thanks!

 

[joshsailscga]

No CSP:
10 EP cases so 1 in 10
With CSP:
5 EP cases so 1 in 5

This is not considering post or pre AUFs

Are all EP cases equal probability? I know that's not the case for 3x3 PLL but don't know the math behind it so not sure if that translates over to SQ1.

 

[Discomantis]

Does anyone know how many 4 Gen scrambles there are?

I'm wondering the same for 2 Gen and 3 Gen.

 

[Discomantis]

Does anyone know the odds of, after solving your third pair, you get a fourth pair skip into an OLL skip?