Cube probability

**Krizalid** · July 19th 2007, 09:47 AM

On choosing three vertex of a cube at random, what's the probability that the three are in a same face?

**galactus** · July 19th 2007, 10:49 AM

I would think it would be $(\frac{1}{6})(\frac{3}{4})=\frac{1}{8}$

It may be more involved than that.

**topsquark** · July 19th 2007, 11:06 AM

Originally Posted by Krizalid

On choosing three vertex of a cube at random, what's the probability that the three are in a same face?

The counting principle, of course.

The first pick if free, we may pick any vertex.

There are 8 vertices on the cube, we've picked one leaving 7. Of these only 6 share a face with the one we've picked. So we have
$1 \cdot \frac{6}{7}$
so far.

Now, we've picked two vertices.
1) If the two vertices are on the same edge, there are 4 possible of 6 vertices left to pick, so the probability will be
$1 \cdot \frac{6}{7} \cdot \frac{4}{6} = \frac{4}{7}$

2) If the two vertices are not on the same edge, they are on a diagonal across one face. In this case there is only one possible face that these could be on, so there are only two possible vertices out of 6, so the probability will be
$1 \cdot \frac{6}{7} \cdot \frac{2}{6} = \frac{2}{7}$

Only one of these choices is possible, so we add the probabilities:
$P = \frac{4}{7} + \frac{2}{7} = \frac{6}{7}$

(Which seems entirely too large a chance for my intuition. Did I goof somewhere?)

-Dan

**CaptainBlack** · July 19th 2007, 11:19 AM

Originally Posted by Krizalid

On choosing three vertex of a cube at random, what's the probability that the three are in a same face?

Number the vertices, assume the first vetrex choosen is vertex 1.

Now draw up a contingency tree and work out the probabilities for the
favourable leaves.

Answer appears to be 20/42~=0.476

RonL

**ThePerfectHacker** · July 19th 2007, 11:37 AM

Am I missing something in this problem? Because it seems easy.

The first vertex can be anywhere. The second vertex needs to be on the same face. There are 7 points and 3 favorable points. So (3/7). The next one is 2 favorable points out of 6 so it is (2/6). So (3/7)(2/6)=1/7

**Plato** · July 19th 2007, 11:42 AM

Well, Well. I seems that we don’t agree, any of us.
There are $\left( {\begin{array}{c} 8 \\ 3 \\ \end{array}} \right) =56$ possible sets of three vertices on a cube.
There are $\left( {\begin{array}{c} 4 \\ 3 \\ \end{array}} \right) =4$ sets of three on any face.
Having six faces that means that there are 24 sets of three that share a face.
The probability of sharing a face ought to be: $\frac{{24}}{{56}} = \frac{3}{7}$

**galactus** · July 19th 2007, 11:48 AM

Dagnubbit, plato, that's what I came up with first, but thought it was wrong.

Here's what I done originally:

$\frac{C(6,1)C(4,3)}{C(8,3)}=\frac{3}{7}$

Oh well, live and learn. I always second guess myself with probability and counting.

**ThePerfectHacker** · July 19th 2007, 11:53 AM

Originally Posted by Plato

Well, Well. I seems that we don’t agree, any of us.
There are $\left( {\begin{array}{c} 8 \\ 3 \\ \end{array}} \right) =56$ possible sets of three vertices on a cube.
There are $\left( {\begin{array}{c} 4 \\ 3 \\ \end{array}} \right) =4$ sets of three on any face.
Having six faces that means that there are 24 sets of three that share a face.
The probability of sharing a face ought to be: $\frac{{24}}{{56}} = \frac{3}{7}$

I first got that result when I was doing this problem first. But then I switched over and did my other approach.

EDIT: I know why I made a mistake now. And Plato seems to have it right.

**CaptainBlack** · July 19th 2007, 12:10 PM

Originally Posted by CaptainBlack

Number the vertices, assume the first vetrex choosen is vertex 1.

Now draw up a contingency tree and work out the probabilities for the
favourable leaves.

Answer appears to be 20/42~=0.476

RonL

Miss-counted vertices, now I know the answer I do indeed get 3/7.

RonL