Cube maths puzzle

[capo]

Hey there. This is one I made for my friends (which they struggled with somewhat), so when I saw the puzzles section in these forums I thought I might as well share it...

 

Q: Imagine a cube made of smaller cubes (like a Rubik's cube). Instead of measuring 3 by 3 by 3 however, this one measures 5 by 5 by 5. If you pick any two of these smaller cubes at random, what is the probability that their faces will be touching one another?

 

 

[kruzzgoldfish]

This solutions seems kind of simple so I'm not sure if it's correct, but since it's 5x5x5, that means there's 125 smaller cubes right? That means that the probability of choosing a specific cube is 1/125. However, since the second cube is really the only one that matters in this case, the probability of choosing a correct second cube will be 6/124.

 

Of course that doesn't include the edges so it's wrong. If it's a corner piece (4 pieces), it has 3 possible sides. The side pieces have 5 possible sides and there are 80 side pieces. That leaves 45 inside pieces that do have 6 possible sides. So the first cube chosen is gonna make a difference.

 

P(1st cube is a corner piece) = 4/125

P(1st cube is a side piece) = 80/125

P(1st cube is inside piece) = 45/125

 

P(2nd cube touches corner piece) = 3/124

P(2nd cube touches side piece) = 5/124

P(2nd cube touches inside piece) = 6/124

 

Therefore:

 

P(2nd cube touches first cube) = 4/125*3/124 + 80/125*5/124 + 45/125*6/124

= .044

Edited November 15, 2010 by kruzzgoldfish

[timo]

[math] \frac{1}{5^3} \left( 3^3 \frac{6}{5^3-1} + 6\cdot 3^2 \frac{5}{5^3-1} + 12\cdot 3^1 \frac{4}{5^3-1} + 4 \frac{3}{5^3-1} \right)[/math] would be my guess. But statistics has the nasty habit of fooling me pretty often.

[capo]

Unfortunately neither answers are correct, but they're very close and you're along the right lines.

 

 

[kruzzgoldfish]

Oops I just realized I made another mistake:

 

P(1st cube is a corner piece) = 4/125

P(1st cube is a side piece) = 80/125

P(1st cube is inside piece) = 41/125

 

P(2nd cube touches corner piece) = 3/124

P(2nd cube touches side piece) = 5/124

P(2nd cube touches inside piece) = 6/124

 

Therefore:

 

P(2nd cube touches first cube) = 4/125*3/124 + 80/125*5/124 + 41/125*6/124

= .04245

[capo]

That's not right either I'm afraid...but you know, when I said my friends struggled with it, what I really meant was that none of them could solve it. So you're already doing better than they did!

[Sisyphus]

Probability 1st one is corner piece = 8/125

3 other cubes touch corner, so probability of second touching first = 3/124

Therefore, the probability that it will be a corner piece, touching another, is (8/125)*(3/124) = 24/15500

 

Then, just do the same for edge pieces (24 of them, each can touch 4 others), face pieces, (54 of them, each can touch 5 others), and interior ones (27 of them, each can touch 6 others). Since any one of those situations satisfies the conditions, just add those probabilities together.

 

[(8/125)*(3/125)]+[(24/125)*(4/124)]+[(54/125)*(5/124)]+[(27/125)*(6/124)]

 

Unless I made a mistake somewhere, it should be 432/15500, or 108/3875.

[timo]

... oh right ... I should probably have assumed 8 corners above, not 4

[capo]

You're again, very very close but unfortunately it's not quite the right answer. It's so close however that I feel it'd almost be wrong not to hand out congratulations...

Edited November 16, 2010 by capo

[imatfaal]

Probability 1st one is corner piece = 8/125

3 other cubes touch corner, so probability of second touching first = 3/124

Therefore, the probability that it will be a corner piece, touching another, is (8/125)*(3/124) = 24/15500

 

Then, just do the same for edge pieces (24 of them, each can touch 4 others), face pieces, (54 of them, each can touch 5 others), and interior ones (27 of them, each can touch 6 others). Since any one of those situations satisfies the conditions, just add those probabilities together.

 

[(8/125)*(3/125)]+[(24/125)*(4/124)]+[(54/125)*(5/124)]+[(27/125)*(6/124)]

 

Unless I made a mistake somewhere, it should be 432/15500, or 108/3875.

 

Your slip up was here

 

8 corners ( 4 per face each shared by 3 faces 24/3=8) ,

36 side pieces (12 per face - each shared by two faces 72/2=36),

54 mid pieces (9 per face), and

27 interior pieces (3*3*3)

8+36+54+27 =125

 

However! You cannot touch the interior faces - unless this is some new form of touch that involves my finger going straight through a solid object. Any answer that deals with internal cubes seems to me a bit screwy - the question clearly stated

Like a Rubik's cube

- and Rubik's cubes do not have internal cubes, that's where the mechanism is. Therefore all sums should work on the 98 cubes that you can actually touch.

 

[(8/98)*(3/97)]+[(36/98)*(4/97)]+[(54/98)*(4/97)] = (8.3+36.4+54.4)/(97.98) = 384/9506 = 0.040396

[capo]

Sorry, that's my bad. I did actually mean a cube with internal cubes as everyone else figured, but nice bit of maths there imatfaal.

 

Okay, it's time to round it up. The answer I was looking for was 6/155. Sisyphus basically got it right with his/her working, but (s)he (as you pointed out imatfaal) got one of the values wrong. The other mistake (s)he made was again a minor one (125 instead of 124 in one place):

 

[(8/125)*(3/125)]+[(24/125)*(4/124)]+[(54/125)*(5/124)]+[(27/125)*(6/124)]

should have been

[(8/125)*(3/124)]+[(36/125)*(4/124)]+[(54/125)*(5/124)]+[(27/125)*(6/124)]

which evaluates to

6/155 = 0.0387096774193548... = Approx. 3.871%.

 

Congrats to everyone who contributed.

[timo]

Hey, that's exactly what my proposal gives if you assume 8 corners instead of 4. I want my name in the credits, too! B)

[capo]

Congrats to you too matey (Everyone who wrote something) B)