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Probability Thread

What is the probability of getting a 3BLD Scramble with no cycle breaks? My Monty Carlo simulations give 736/100130 or roughly 0.74%.

That sounds right at least. I mean roughly it should be the odds that your buffer corner and buffer edge are exactly at one spot on the cube (namely, the end of the cycle), or (1/8)*(1/12)=1.04%. That's just an approximation though
 
What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)
What is the probability of having a 2x1 block (ce pair both with correct orientation) in the last layer solved if all of your edges are preoriented? As with the other question the pieces don't have to be solved relative to the whole cube.

Would also be interesting to see the probabilities without having edges preoriented.
 
What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)
What is the probability of having a 2x1 block (ce pair both with correct orientation) in the last layer solved if all of your edges are preoriented? As with the other question the pieces don't have to be solved relative to the whole cube.

Would also be interesting to see the probabilities without having edges preoriented.

first question= odds of having phased edges*2 (cause you can have the other two the wrong way around)*4(only 1 position for a corner*3(only one CO allowed/4(AUFS)
= (6*2*4*3)/4= 1/36 (I think)
2) 4*3/4= 1/3
 
What is the probability of having a 2x2 block in the last layer solved if all of your edges are preoriented? The block only needs to be solved relative to the other pieces in the last layer and not the whole cube. (It still counts as solved if it needs an AUF)

This is what I get:

Start with edges. We have 3 different possibilities for their permutation:
1) correctly permuted: probability 1/6
2) adjacent swap needed: probablility 2/3
3) swap across needed: probability 1/6 (but no possibility of a 2x2 block)

Edge case 1: edges correctly permuted
Corners must have even permutation because edge permutation is even. We have 3 cases:
1) Corners solved (with respect to edges): probability 1/12
2) Corners in a 3-cycle: probability 8/12 = 2/3
3) Corners in a double swap (adjacent or diagonal): probability 3/12 = 1/4

For corner case 3, no corners are in the correct place, so we have no possibility of a 2x2 block.
For corner case 1, all corners are in the correct place, so any oriented corner forms a 2x2 block. Of the 27 total orientation configuarations, 21 have at least one oriented corner. 21/27 = 7/9.
For corner case 2, we need the correctly positioned corner to be correctly oriented: 1 in 3 chance.

Edge case 2:
We can only form a 2x2 block if the corner between the 2 correct edges is correctly placed (1 in 4) and correctly oriented (1 in 3).

So the probability of at least one 2x2 block is (1/6)*(1/12)*(7/9) + (1/6)*(2/3)*(1/3) + (2/3)*(1/4)*(1/3) = 67/648 (or approximately 10.34%).

If we want the probablility of exactly one 2x2 block, then the 21/27 chance must be changed to 8/27. Thus, this probablility is (1/6)*(1/12)*(8/27) + (1/6)*(2/3)*(1/3) + (2/3)*(1/4)*(1/3) = 47/486 (or approximately 9.67%).
 
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