A revealing accident
- Did I tell you about my new neighbour?
- Can’t say that you did.
- Well, he is blind, at least that is what he claims.
- Why do you doubt it?
- You remember the 27 cubes I put together to make a bigger cube?
- How can I forget. I was the one who painted it black on the outside.
- My neighbour accidentally knocked the cube over last night, but when he reassembled the small cubes the big cube was still black on the outside!
- You mean all 54 faces?
- Yes.
- That man can’t be blind!
Problem source: Nick’s Mathematical Puzzles.
February 4th, 2010 at 3:29 pm
All 6 faces of all 27 cubes are black.
Painted faces feel smoother than unpainted ones.
February 4th, 2010 at 5:09 pm
If the minimal number of faces have been painted black and painted faces do not feel smoother than unpainted ones, what is the probability for the event described?
February 4th, 2010 at 8:59 pm
Lay the blocks under a lamp. Pull them out and feel which are warmer (the black ones). Turn the cooler blocks, place them ubder the lamp and try again until all of the black sides have been identified.
February 5th, 2010 at 7:28 pm
The arrangement must meet these criteria:
1) the sole unpainted small cube must be in its proper place, in the middle
2) the 6 small cubes with 1 painted face each must be in the face centres (in any order)
3) each of them must have its painted face showing
4) the 12 small cubes with 2 painted faces each must be in the edge centres (in any order)
5) each of them must have its edge where its 2 painted faces meet in the centre of an edge of the big cube
6) each of the cubes with 3 painted faces must have its vertex where those 3 faces meet at a vertex of the big cube
1) Probability of meeting criterion 1: There are 27 equally likely places for that cube, and only 1 is right, so 1/27.
2) Probability of meeting criterion 2, given that criterion 1 is met: There are 26 cubes left of which 6 have 1 painted face each. There are 26 C 6 ways to choose 6 places for those cubes (without regard to which of the 6 cubes goes in which of the 6 chosen places), and only 1 is right, so 1/(26 C 6) = 6! 20!/26!
3) Probability of meeting criterion 3, given that criteria 1 and 2 are met: For each face-centre cube there is a choice from among 6 faces for the face that shows, and only 1 is right, so 1/6 for that cube. There are 6 such cubes, so 1/6^6.
4) Probability of meeting criterion 4, given that criteria 1-3 are met: 12! 8!/20! (see para 2 above)
5) Probability of meeting criterion 5, given that criteria 1-4 are met: For each edge-centre cube there is a choice from among 12 edges for the edge that is in the centre of an edge of the big cube, and only 1 is right, so 1/12 for that cube. There are 12 such cubes, so 1/12^12.
6) Probability of meeting criterion 6, given that criteria 1-5 are met: For each corner cube there is a choice from among 8 vertices for the vertex that is at a vertex of the big cube, and only 1 is right, so 1/8 for that cube. There are 8 such cubes, so 1/8^8.
Thus
[6! 20! 12! 8!]/[27 * 26! * 6^6 * 20! * 12^12 * 8^8]
= [6! 12! 8!]/[27! * 6^6 * 12^12 * 8^8]
~~ 1.83 * 10^-37.