Blind Man with Cubes.

Probability Level 5

A large white cube is painted red, and then cut into $27$ identical smaller cubes. These smaller cubes are shuffled randomly.

A blind man (who also cannot feel the paint) reassembles the small cubes into a large one. Let $P$ denote the probability that the outside of this large cube is completely red?

What is the value of $10^{37} \times P$ ? (Answer to two decimal places)

The answer is 1.83.

2 solutions

Aditya Kumar
Mar 5, 2015

Complete answer:

First, number the small cubes to make them distinguishable.

Without taking into account the orientation of the smaller cubes, there are 27! possible ways to reassemble them into a big cube.

There is 1 way to place the center cube correctly
There are 6! ways to place the face cubes correctly
There are 8! ways to place the corner cubes correctly
There are 12! ways to place the edge cubes correctly

So there are 6!8!12! correct cubes (without taking into account the orientation).

Now take such a cube.

The center cube has probability 1 of being in the correct orientation.
Each face cube has probability 1/6 of being in the correct orientation.
Each corner cube has probability 1/8 of being in the correct orientation.
Each edge cube has probability 1/12 of being in the correct orientation.

This means that the probability of getting a red cube must be

$(6!8!12!)÷27! * ((1÷6)^6 ) *((1÷8)^8) * ((1÷12)^12)$

@Calvin Lin sir I actually saw your report today. Sorry for that. But the solution provided seems to be correct.

Aditya Kumar - 6 years, 2 months ago

There was a typo in your original question which has since been fixed. You asked for $10 ^ { - 37 } \times P$ , as opposed to $10 ^ {37} \times P$ .

Calvin Lin Staff - 6 years, 2 months ago

Thank you sir:)

Aditya Kumar - 6 years, 2 months ago

Nice solution!!!! @Aditya Kumar .. Can u suggest me some good links to study combinatorics.. I am lagging behind in that damn topic and I need to improve.. Help required =D

Hrishik Mukherjee - 6 years, 2 months ago

sorry @Hrishik Mukherjee , I myself is searching help for combinatorics. :(

Aditya Kumar - 6 years, 2 months ago

@Vaibhav Prasad how did you solve it?

Adarsh Kumar - 6 years, 2 months ago

I saw this

Vaibhav Prasad - 6 years, 2 months ago

So you first understood that and then answered this?

Adarsh Kumar - 6 years, 2 months ago

@Karthik Venkata how did you do it?

Adarsh Kumar - 6 years, 2 months ago

Refer to Miki Monigkai's solution in the same problem posted by Andrei Golovanov. I did it in his method.

Venkata Karthik Bandaru - 6 years, 2 months ago

have you seen this problem before?Karthik Venkata

Adarsh Kumar - 6 years, 2 months ago

@Adarsh Kumar – Nope, the problem was nice ! A really nice way of using Rule of Product.

Venkata Karthik Bandaru - 6 years, 2 months ago

Nicola Mignoni
Jul 8, 2019

It's possible to generalize this problem for $n^3$ cutted cubes. There are $4$ type of small cube:

Corner cubes : there are always $8$ of them and the probability of placing them with the right orientation is $\displaystyle \frac{1}{8}$ ( $3$ faces painted)
Side cubes : there are always $12(n-2)$ of them and the probability of placing them with the right orientation is $\displaystyle \frac{1}{12}$ ( $2$ faces painted)
Mid cubes : there are always $6(n-2)^2$ of them and the probability of placing them with the right orientation is $\displaystyle \frac{1}{6}$ ( $1$ face painted)
Internal cubes : there are always $(n-2)^3$ of them and their placement is irrelevant, since they are internal.

You can check that $(n-2)^3+6(n-2)^2+12(n-2)+8=n^3$ .

The probability $P(n)$ of correctly reassembling the large cube with $n^3$ smaller cubes is

$\displaystyle P(n)=\frac{[6(n-2)^2]![12(n-2)]!8!}{(n^3)!} \bigg(\frac{1}{8}\bigg)^8 \bigg(\frac{1}{6}\bigg)^{6(n-2)^2} \bigg(\frac{1}{12}\bigg)^{12(n-2)}$

In our particular case,

$\displaystyle 10^{37} \cdot P(3)=\boxed{1.83}$

Blind Man with Cubes.

The answer is 1.83.

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