AnastastZhivago
New member
- Joined
- Mar 22, 2021
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- 1
The Question:
A cube with edge length six is composed of unit cubes and has all six faces painted. One move consists of removing an exposed unit cube (i.e. a visible unit cube) and painting all exposed faces on the main cube. The removed unit cube is not painted. The game ends when all cubes have been separated and all moves are made. Assume moves which create 2 or more main objects are invalid, and so are moves which split off more than one unit cube. (aka it is not allowed to remove cubes if said cubes are holding together the main structure. cubes connected by an edge or a vertex are separate, only cubes connected by faces are one object.) If a monkey who moves randomly plays, what is the probability there will be exactly 29 cubes with exactly 5 sides painted at the end? What about 47 cubes with 3 sides painted?
The work I have done:
The cube has dimensions 6*6*6, so there is 216 cubes. The last cube and only the last cube will have all faces colored, working backwards we can also see that the second to last cube must have 5 faces colored, and likewise for the 3rd to last. I conjecture since 6 cubes surround the last cube there is at minimum 6 cubes that will have 5 sides painted, however I can not prove it so. Any face at any point in time on the main cube will either be exposed and painted, or touching a nonpainted face. When a unit cube is removed, all unpainted faces on said removed cube represent faces to be painted on the main cube, and thus the number of faces painted is equal to the original painted faces combined with half of the unexposed faces. 7 Great Faces each with 36 faces in the 3 Dimensions makes 756 painted faces in total, and the painted face to cube ratio is 3.5:1. The first cube randomly chosen is either could be one of the 8 corners, 48 edges, or 96 face cubes, with a total of 6^3-4^3 options, or 152. Suppose a central face cube is chosen. Then, for the second cube, the possibilities are 8 corners, 52 edges, and 92 face cubes (a face cube is created on the bottom of the "hole"), for a total of 152 options again. If a corner or edge piece is removed instead for the first cube, there are 151 options.
I can't seem to progress any farther than that, and everything above seems either obvious or irrelevant. Any ideas?
A cube with edge length six is composed of unit cubes and has all six faces painted. One move consists of removing an exposed unit cube (i.e. a visible unit cube) and painting all exposed faces on the main cube. The removed unit cube is not painted. The game ends when all cubes have been separated and all moves are made. Assume moves which create 2 or more main objects are invalid, and so are moves which split off more than one unit cube. (aka it is not allowed to remove cubes if said cubes are holding together the main structure. cubes connected by an edge or a vertex are separate, only cubes connected by faces are one object.) If a monkey who moves randomly plays, what is the probability there will be exactly 29 cubes with exactly 5 sides painted at the end? What about 47 cubes with 3 sides painted?
The work I have done:
The cube has dimensions 6*6*6, so there is 216 cubes. The last cube and only the last cube will have all faces colored, working backwards we can also see that the second to last cube must have 5 faces colored, and likewise for the 3rd to last. I conjecture since 6 cubes surround the last cube there is at minimum 6 cubes that will have 5 sides painted, however I can not prove it so. Any face at any point in time on the main cube will either be exposed and painted, or touching a nonpainted face. When a unit cube is removed, all unpainted faces on said removed cube represent faces to be painted on the main cube, and thus the number of faces painted is equal to the original painted faces combined with half of the unexposed faces. 7 Great Faces each with 36 faces in the 3 Dimensions makes 756 painted faces in total, and the painted face to cube ratio is 3.5:1. The first cube randomly chosen is either could be one of the 8 corners, 48 edges, or 96 face cubes, with a total of 6^3-4^3 options, or 152. Suppose a central face cube is chosen. Then, for the second cube, the possibilities are 8 corners, 52 edges, and 92 face cubes (a face cube is created on the bottom of the "hole"), for a total of 152 options again. If a corner or edge piece is removed instead for the first cube, there are 151 options.
I can't seem to progress any farther than that, and everything above seems either obvious or irrelevant. Any ideas?