Mixing
Probability of RGB cube with identically coloured opposite faces?
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I was given a really interesting problem at an interview. It seems like an elementary question but I just can’t think about it it properly. Apologies if the wording is a bit odd, I can’t think of better ways the phrase the question:
The six faces of a cube can be painted red, green or blue. What is the probability that each pair of opposite faces is coloured a distinct colour (i.e. one pair opposite faces is painted red, another pair painted green, the last pair painted blue) if:
a) the colour for each face is randomly chosen with probability $frac13$?
b) the cube has exactly 2 red, 2 green, and 2 blue faces?
I am not too sure how to begin because of the geometry/symmetry involved. For example, for a), is the probability of a fully red cube $frac1{3^6}$ or $frac6{3^6}$? In fact, is the number of ways to have three pairs of identically coloured opposite faces of distinct colours $3!$, or is there more geometry involved? I feel like I’m overthinking somewhere.
Tldr; how do I account for overcounting due to the symmetry in the problem? Am I even overcounting in the first place?
probability problem-solving
asked Jan 30 at 14:45
user107224user107224
475314
$endgroup$
add a comment |
$begingroup$
I was given a really interesting problem at an interview. It seems like an elementary question but I just can’t think about it it properly. Apologies if the wording is a bit odd, I can’t think of better ways the phrase the question:
The six faces of a cube can be painted red, green or blue. What is the probability that each pair of opposite faces is coloured a distinct colour (i.e. one pair opposite faces is painted red, another pair painted green, the last pair painted blue) if:
a) the colour for each face is randomly chosen with probability $frac13$?
b) the cube has exactly 2 red, 2 green, and 2 blue faces?
I am not too sure how to begin because of the geometry/symmetry involved. For example, for a), is the probability of a fully red cube $frac1{3^6}$ or $frac6{3^6}$? In fact, is the number of ways to have three pairs of identically coloured opposite faces of distinct colours $3!$, or is there more geometry involved? I feel like I’m overthinking somewhere.
Tldr; how do I account for overcounting due to the symmetry in the problem? Am I even overcounting in the first place?
probability problem-solving
asked Jan 30 at 14:45
user107224user107224
475314
$endgroup$
$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02
add a comment |
$begingroup$
I was given a really interesting problem at an interview. It seems like an elementary question but I just can’t think about it it properly. Apologies if the wording is a bit odd, I can’t think of better ways the phrase the question:
The six faces of a cube can be painted red, green or blue. What is the probability that each pair of opposite faces is coloured a distinct colour (i.e. one pair opposite faces is painted red, another pair painted green, the last pair painted blue) if:
a) the colour for each face is randomly chosen with probability $frac13$?
b) the cube has exactly 2 red, 2 green, and 2 blue faces?
I am not too sure how to begin because of the geometry/symmetry involved. For example, for a), is the probability of a fully red cube $frac1{3^6}$ or $frac6{3^6}$? In fact, is the number of ways to have three pairs of identically coloured opposite faces of distinct colours $3!$, or is there more geometry involved? I feel like I’m overthinking somewhere.
Tldr; how do I account for overcounting due to the symmetry in the problem? Am I even overcounting in the first place?
probability problem-solving
asked Jan 30 at 14:45
user107224user107224
475314
$endgroup$
I was given a really interesting problem at an interview. It seems like an elementary question but I just can’t think about it it properly. Apologies if the wording is a bit odd, I can’t think of better ways the phrase the question:
The six faces of a cube can be painted red, green or blue. What is the probability that each pair of opposite faces is coloured a distinct colour (i.e. one pair opposite faces is painted red, another pair painted green, the last pair painted blue) if:
a) the colour for each face is randomly chosen with probability $frac13$?
b) the cube has exactly 2 red, 2 green, and 2 blue faces?
I am not too sure how to begin because of the geometry/symmetry involved. For example, for a), is the probability of a fully red cube $frac1{3^6}$ or $frac6{3^6}$? In fact, is the number of ways to have three pairs of identically coloured opposite faces of distinct colours $3!$, or is there more geometry involved? I feel like I’m overthinking somewhere.
Tldr; how do I account for overcounting due to the symmetry in the problem? Am I even overcounting in the first place?
probability problem-solving
probability problem-solving
asked Jan 30 at 14:45
user107224user107224
475314
asked Jan 30 at 14:45
user107224user107224
475314
edited Jan 30 at 15:04
user107224
asked Jan 30 at 14:45
user107224user107224
475314
asked Jan 30 at 14:45
user107224user107224
475314
asked Jan 30 at 14:45
user107224user107224
475314
475314
$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02
add a comment |
$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02
$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02
add a comment |
1 Answer
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The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $frac{6}{729} = frac{2}{243}$.
(b) Now there are only ${6 choose 2} cdot {4 choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $frac{6}{90} = frac{1}{15}$.
answered Jan 30 at 15:21
KlausKlaus
2,945214
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Hi, could you explain a bit more on why the geometry is irrelevant?
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– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
add a comment |
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$begingroup$
The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $frac{6}{729} = frac{2}{243}$.
(b) Now there are only ${6 choose 2} cdot {4 choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $frac{6}{90} = frac{1}{15}$.
answered Jan 30 at 15:21
KlausKlaus
2,945214
$endgroup$
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
add a comment |
$begingroup$
The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $frac{6}{729} = frac{2}{243}$.
(b) Now there are only ${6 choose 2} cdot {4 choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $frac{6}{90} = frac{1}{15}$.
answered Jan 30 at 15:21
KlausKlaus
2,945214
$endgroup$
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
add a comment |
$begingroup$
The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $frac{6}{729} = frac{2}{243}$.
(b) Now there are only ${6 choose 2} cdot {4 choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $frac{6}{90} = frac{1}{15}$.
answered Jan 30 at 15:21
KlausKlaus
2,945214
$endgroup$
The fact that these faces are on a cube is irrelevant for the problem. The only thing that matters is, you have 3 pairs of faces to paint.
(a) There are $3^6 = 729$ ways to paint the cube. There are $3!$ ways to assign the colors to the three pairs. Therefore the probability is $frac{6}{729} = frac{2}{243}$.
(b) Now there are only ${6 choose 2} cdot {4 choose 2} = 90$ ways to paint the cube. Therefore the answer to (b) is $frac{6}{90} = frac{1}{15}$.
answered Jan 30 at 15:21
KlausKlaus
2,945214
answered Jan 30 at 15:21
KlausKlaus
2,945214
answered Jan 30 at 15:21
KlausKlaus
2,945214
answered Jan 30 at 15:21
KlausKlaus
2,945214
2,945214
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
add a comment |
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
Hi, could you explain a bit more on why the geometry is irrelevant?
$endgroup$
– user107224
Jan 30 at 15:22
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
@user107224 Because you could just peel off the faces and put them on the table. The problem would remain the same, right?
$endgroup$
– Klaus
Jan 30 at 15:24
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
wow didn’t think of it that way! Spent so long dividing things by 6 and 2 that I lost track of everything! Thanks!
$endgroup$
– user107224
Jan 30 at 15:30
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
$begingroup$
It really depends on the probability space. If one understands the second question as “probability of every possible cube is the same”, then geometry plays role, and the answer is 1/6
$endgroup$
– Vasily Mitch
Jan 30 at 15:34
add a comment |
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$begingroup$
When you say "each pair of opposite faces is coloured a distinct colour" does that mean the two sides in each pair have the same colour - so there is a pair of red sides, a pair of green sides and a pair of blue side ? Or does it mean the two sides in each pair of opposite sides have different colours ?
$endgroup$
– gandalf61
Jan 30 at 14:58
$begingroup$
@gandalf61 the former!
$endgroup$
– user107224
Jan 30 at 15:02