Question about placing stickers on the Rubik's cube

[Robert-Y]

If I peeled off every sticker of the Rubik's cube and then I placed them all back on the Rubik's cube at random, what are the chances that I will attain a legal cube position?

 

[JohnnyA]

43 quintillion out of however many possible ways you can put the stickers on. So 54 stickers/places - (43quintillion)/(54!)

This is just off my head and I'm tired so sorry if it's wrong.

 

[Gparker]

i really dont know, but itd be impossible if you had 2 of the same stickers on the same peice XD

 

[speedcuber1355]

Can a piece be flipped around, or does it have to be able to be solved from that position?

 

[speedcuber1355]

If its a solvable position around 1/5^51

position with pieces flipped around 1/4^50

these are obviously rounded and I may be way off so dont hate me if im wrong

 

[Lucas Garron]

(43252003274489856000*(9!)^6)/(54!)
=278628139008/651266248090757461677878416625
≈4.27825*10^-19

:rolleyes:

(Reassemble-solvable: *12 it.)

 

[Herbert Kociemba]

(43252003274489856000*(9!)^6)/(54!)
=278628139008/651266248090757461677878416625
≈4.27825*10^-19

I do not understand the (9!)^6.
There are 6! possibilities to put 6 colours onto the 6 centers, so I would prefer 43252003274489856000*6!/54!

 

[Johannes91]

(43252003274489856000*(9!)^6)/(54!)
=278628139008/651266248090757461677878416625
≈4.27825*10^-19

I do not understand the (9!)^6.
There are 6! possibilities to put 6 colours onto the 6 centers, so I would prefer 43252003274489856000*6!/54!

Stickers of the same color can be in any order. For every cube position, there are (9!)^6 sticker permutations that look like it.

So maybe Lucas' number should be:

(43252003274489856000*(9!)^6*6!) / (54!)
= (43252003274489856000*6!) / (54!/(9!)^6)
= 40122452017152/130253249618151492335575683325
≈ 3.08034*10^-16

 

[Herbert Kociemba]

(43252003274489856000*(9!)^6)/(54!)
=278628139008/651266248090757461677878416625
≈4.27825*10^-19

I do not understand the (9!)^6.
There are 6! possibilities to put 6 colours onto the 6 centers, so I would prefer 43252003274489856000*6!/54!

Stickers of the same color can be in any order. For every cube position, there are (9!)^6 sticker permutations that look like it.

So maybe Lucas' number should be:

(43252003274489856000*(9!)^6*6!) / (54!)
= (43252003274489856000*6!) / (54!/(9!)^6)
= 40122452017152/130253249618151492335575683325
≈ 3.08034*10^-16

Oh yes, I hope we finally got the right number.

 

[Lucas Garron]

(43252003274489856000*(9!)^6)/(54!)
=278628139008/651266248090757461677878416625
≈4.27825*10^-19

I do not understand the (9!)^6.
There are 6! possibilities to put 6 colours onto the 6 centers, so I would prefer 43252003274489856000*6!/54!

Stickers of the same color can be in any order. For every cube position, there are (9!)^6 sticker permutations that look like it.

So maybe Lucas' number should be:

(43252003274489856000*(9!)^6*6!) / (54!)
= (43252003274489856000*6!) / (54!/(9!)^6)
= 40122452017152/130253249618151492335575683325
≈ 3.08034*10^-16

Oh yes, I hope we finally got the right number.

Ah, indeed. If we go by WCA-legal, we should probably rescheming so I'd probably agree with the extra factor of 6! (Else just *24)