campos20
Member
Have you ever considered the probability of getting a random state cube solvable optimally with n moves? Well, I have. Since normal probability are not that readable when talking about cube, one can picture the scene with exciting coin flips.
The chance you get a cube that can be solved with 10 moves is approximately the same as getting 28 Heads in a row in a coin flip (or 5.37e-07%. See, coin flips are better).
Here's a table:
[TR1][TD1] 0 [/TD1][TD1] 66 [/TD1][/TR1]
[TR2][TD1] 1 [/TD1][TD1] 62 [/TD1][/TR2]
[TR1][TD1] 2 [/TD1][TD1] 58 [/TD1][/TR1]
[TR2][TD1] 3 [/TD1][TD1] 54 [/TD1][/TR2]
[TR1][TD1] 4 [/TD1][TD1] 50 [/TD1][/TR1]
[TR2][TD1] 5 [/TD1][TD1] 47 [/TD1][/TR2]
[TR1][TD1] 6 [/TD1][TD1] 43 [/TD1][/TR1]
[TR2][TD1] 7 [/TD1][TD1] 39 [/TD1][/TR2]
[TR1][TD1] 8 [/TD1][TD1] 35 [/TD1][/TR1]
[TR2][TD1] 9 [/TD1][TD1] 32 [/TD1][/TR2]
[TR1][TD1] 10 [/TD1][TD1] 28 [/TD1][/TR1]
[TR2][TD1] 11 [/TD1][TD1] 24 [/TD1][/TR2]
[TR1][TD1] 12 [/TD1][TD1] 21 [/TD1][/TR1]
[TR2][TD1] 13 [/TD1][TD1] 17 [/TD1][/TR2]
[TR1][TD1] 14 [/TD1][TD1] 13 [/TD1][/TR1]
[TR2][TD1] 15 [/TD1][TD1] 9 [/TD1][/TR2]
[TR1][TD1] 16 [/TD1][TD1] 6 [/TD1][/TR1]
[TR2][TD1] 17 [/TD1][TD1] 2 [/TD1][/TR2]
[TR1][TD1] 18 [/TD1][TD1] 1 [/TD1][/TR1]
[TR2][TD1] 19 [/TD1][TD1] 5 [/TD1][/TR2]
[TR1][TD1] 20 [/TD1][TD1] 37 [/TD1][/TR1]
Now, how was that calculated? Well, first we can go to http://cube20.org/. The number of cubes solvable with 10 moves is c10 = 232,248,063,316 and but the total number is ctotal = 43,252,003,274,489,856,000. So, ctotal is approximately 186,231,922.2686 times c10. Taking log2 (to calculate coin flips) we have 27.4725. Its ceiling is 28. Repeating this process, we can build this table.
The chance you get a cube that can be solved with 10 moves is approximately the same as getting 28 Heads in a row in a coin flip (or 5.37e-07%. See, coin flips are better).
Here's a table:
Moves | Flips |
---|
[TR1][TD1] 0 [/TD1][TD1] 66 [/TD1][/TR1]
[TR2][TD1] 1 [/TD1][TD1] 62 [/TD1][/TR2]
[TR1][TD1] 2 [/TD1][TD1] 58 [/TD1][/TR1]
[TR2][TD1] 3 [/TD1][TD1] 54 [/TD1][/TR2]
[TR1][TD1] 4 [/TD1][TD1] 50 [/TD1][/TR1]
[TR2][TD1] 5 [/TD1][TD1] 47 [/TD1][/TR2]
[TR1][TD1] 6 [/TD1][TD1] 43 [/TD1][/TR1]
[TR2][TD1] 7 [/TD1][TD1] 39 [/TD1][/TR2]
[TR1][TD1] 8 [/TD1][TD1] 35 [/TD1][/TR1]
[TR2][TD1] 9 [/TD1][TD1] 32 [/TD1][/TR2]
[TR1][TD1] 10 [/TD1][TD1] 28 [/TD1][/TR1]
[TR2][TD1] 11 [/TD1][TD1] 24 [/TD1][/TR2]
[TR1][TD1] 12 [/TD1][TD1] 21 [/TD1][/TR1]
[TR2][TD1] 13 [/TD1][TD1] 17 [/TD1][/TR2]
[TR1][TD1] 14 [/TD1][TD1] 13 [/TD1][/TR1]
[TR2][TD1] 15 [/TD1][TD1] 9 [/TD1][/TR2]
[TR1][TD1] 16 [/TD1][TD1] 6 [/TD1][/TR1]
[TR2][TD1] 17 [/TD1][TD1] 2 [/TD1][/TR2]
[TR1][TD1] 18 [/TD1][TD1] 1 [/TD1][/TR1]
[TR2][TD1] 19 [/TD1][TD1] 5 [/TD1][/TR2]
[TR1][TD1] 20 [/TD1][TD1] 37 [/TD1][/TR1]
Now, how was that calculated? Well, first we can go to http://cube20.org/. The number of cubes solvable with 10 moves is c10 = 232,248,063,316 and but the total number is ctotal = 43,252,003,274,489,856,000. So, ctotal is approximately 186,231,922.2686 times c10. Taking log2 (to calculate coin flips) we have 27.4725. Its ceiling is 28. Repeating this process, we can build this table.
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