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Probability Thread

Herbert Kociemba said:
My suggestion ist
(Binomial[4,1]*Binomial[8,3]*1*2^3+Binomial[4,2]*Binomial[8,2]*3*2^2+Binomial[4,3]*Binomial[8,1]*7*2^1+Binomial[4,4]*Binomial[8,0]*15*2^0)/(Binomial[12,4]*2^4)= 4271/7920 = 0.539268
Click to expand...
I don't understand the math you did there, but if you're correct then my experimental value had a percent error of (0.539268-0.531)/0.539268 = 1.5%, not too bad given the sample size. :)
 
PurpleBanana said:
I don't understand the math you did there, but if you're correct then my experimental value had a percent error of (0.539268-0.531)/0.539268 = 1.5%, not too bad given the sample size. :)
Click to expand...
Consider the 4 edges with one face being white as indistinguishable. We can do this because the second colour does not matter. Then there are Binomial[12,4] possibilities to set these 4 edges on the 12 possible places. Because there are 2 orientations for each edge, we have to multiply by 2^4. This explains the denominator.

Now take for example the term Binomial[4,3]*Binomial[8,1]*7*2^1.
Binomial[4,3]*Binomial[8,1] describes the number of possiblities to have exactly 3 edges in the correct layer. You have Binomial[4,3] possibilites to put 3 edges in the correct layer and Binomial[8,1] possiblities to put the remaining edge in the two other layers. There are 2^3 ways to orient the 3 edges in the correct layer. Only in 1 case all orientations are wrong, in 2^3-1=7 cases at least one orientation is correct. This explains the factor 7. The remaining edge in the "wrong" layers has 2^1 possiblities. This explains the factor 2^1.

The other terms in the nominator describe the number of possiblities, that exactly 1, 2 or 4 edges are in the correct layer with at least one edge correctly orientated.
 
Herbert Kociemba said:
Consider the 4 edges with one face being white as indistinguishable. We can do this because the second colour does not matter. Then there are Binomial[12,4] possibilities to set these 4 edges on the 12 possible places. Because there are 2 orientations for each edge, we have to multiply by 2^4. This explains the denominator.

Now take for example the term Binomial[4,3]*Binomial[8,1]*7*2^1.
Binomial[4,3]*Binomial[8,1] describes the number of possiblities to have exactly 3 edges in the correct layer. You have Binomial[4,3] possibilites to put 3 edges in the correct layer and Binomial[8,1] possiblities to put the remaining edge in the two other layers. There are 2^3 ways to orient the 3 edges in the correct layer. Only in 1 case all orientations are wrong, in 2^3-1=7 cases at least one orientation is correct. This explains the factor 7. The remaining edge in the "wrong" layers has 2^1 possiblities. This explains the factor 2^1.

The other terms in the nominator describe the number of possiblities, that exactly 1, 2 or 4 edges are in the correct layer with at least one edge correctly orientated.
Click to expand...
It seems you're right. Your reasoning all seems sound. The value you obtained also seems reasonable and is corroborated by my experimental evidence.

Thanks for working on this problem.
 
The answer will be the total number of 2x1x1 blocks possible, divided by the total number of arrangements.

The total arrangements is:

62b1e28e5b06515f607d4e6df9406a53.png


I don't know the total number of 2x1x1 blocks though.
 
What's the probability of this:

I was doing a solve on the iitimer app, and I came across this one scramble that gave me a cross skip and pair built but not placed no joke!!! I just had to do an AUF, or ADF since it was the cross, and do R U' R. Surprisingly it was not a PB (I was a bit slow on the other pairs)
 
teboecubes said:
What's the probability of this:

I was doing a solve on the iitimer app, and I came across this one scramble that gave me a cross skip and pair built but not placed no joke!!! I just had to do an AUF, or ADF since it was the cross, and do R U' R. Surprisingly it was not a PB (I was a bit slow on the other pairs)
Click to expand...
1/7920
 
This actually happened to me a few years back. I was still solving beginner's method, and I had one solve where I finished putting in the last corner of the first layer, and all I had left was AUF. Basically, I skipped having to solve the last twelve pieces.
If my math is right, the chance of this happening is 4*(1/8!)*2*(1/2)^7*(1/4!)*2*(1/3)^3 = 4*2*2*1/40320*1/128*1/24*1/27 = 1/209,108,880
(I think I broke the universe.)
 
How many combinations are there on a Rubik's cube if you can disassemble it and put it back together? (One edge or corner COULD be flipped)
 
Joel Banks said:
How many combinations are there on a Rubik's cube if you can disassemble it and put it back together? (One edge or corner COULD be flipped)
Click to expand...
There are eight corners with three possible orientations = 1/8! * (1/3)^8
There are twelve edges with two possible orientations = 1/12! * (1/2)^12
1/40320 * 1/6561 * 1/479001600 * 1/4096 = 5.19 x 10^20 = about 519000000000000000000. (The calculator I was using didn't want to use exact numbers because it was too big.)
 
Joel Banks said:
Same question as last... Just with 4x4
Click to expand...
Since you're talking about the number of ways it could be put back together, I'm going to count two scrambles that are technically the same but have swapped center pieces as different. (I hope that makes sense.)

There are 8 corner pieces with 3 possible orientations = 1/8! * (1/3)^8
There are 24 edge pieces with 1 possible orientation = 1/24!
There are 24 center pieces with 1 possible orientation = 1/24!

1/8! * (1/3)^8 * 1/24! * 1/24! = 101,836,133,451,709,433,602,625,095,475,784,887,762,411,323,392,000,000,000

If you don't want to count swapped centers as different:

Each center color can be swapped with other center pieces of the same color 4! = 24 different ways. There are six colors, so 24 * 6 = 144.

101,836,133,451,709,433,602,625,095,475,784,887,762,411,323,392,000,000,000 / 144 = 707,195,371,192,426,622,240,452,051,915,172,831,683,411,968,000,000,000
 
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