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Find Probability of two polygons

cheerful

New member
Jul 10, 2018
11
There are five hexagons.

The edges of each hexagon have been colored with one of three colors randomly.

If you pick two hexagons randomly without replacement, what is the probability that they are the same? (Rotation is okay).

The total space or denominator is 3^(2×6), therefore we have 3^(2×6) at the bottom, but what is the numerator?

The answer is 4263, any ideas?
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
There are five hexagons.

The edges of each hexagon have been colored with one of three colors randomly.

If you pick two hexagons randomly without replacement, what is the probability that they are the same? (Rotation is okay).

The total space or denominator is 3^(2×6), therefore we have 3^(2×6) at the bottom, but what is the numerator?

The answer is 4263, any ideas?
Is this different from this question?

p(2nd hexagon is the same color as 1st hexagon)? Perhaps I don't understand the coloring scheme. Are the edges colored separately?
 

cheerful

New member
Jul 10, 2018
11
Is this different from this question?

p(2nd hexagon is the same color as 1st hexagon)? Perhaps I don't understand the coloring scheme. Are the edges colored separately?
The vertices or edges are coloured by three different colours randomly. So we need to find the probability that two hexagons are the same if we picked two of them out of 5 hexagons. They are the same when the sequences of colors are the same with rotations.
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
The vertices or edges are coloured by three different colours randomly. So we need to find the probability that two hexagons are the same if we picked two of them out of 5 hexagons. They are the same when the sequences of colors are the same with rotations.
Okay, so can we start with the probability that there ARE two the same?
 

cheerful

New member
Jul 10, 2018
11
Okay, so can we start with the probability that there ARE two the same?
Sure. We can eventually work towards the answer. I am very curious how they arrive 4263 for the numerator.

Thanks
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
Sure. We can eventually work towards the answer. I am very curious how they arrive 4263 for the numerator.

Thanks
How many DIFFERENT ways to color one hexagon?

What does "3^(2\\)6)" mean?
 

cheerful

New member
Jul 10, 2018
11
How many DIFFERENT ways to color one hexagon?

What does "3^(2\\)6)" mean?
3 different ways to colour each edge.

3^(2*6) is denominator which I think is all the possible combination.
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
3 different ways to colour each edge.

3^(2*6) is denominator which I think is all the possible combination.
You didn't answer my question. You have one hexagon. How many ways to color it?
 

cheerful

New member
Jul 10, 2018
11

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
6^3 ways to color a hexagon. =)
??

How many colors for side #1: 3
How many colors for side #2: 3
...
How many colors for side #6: 3

Try again and then account for the radial symmetry that you are allowing? You need DISTINCT colorings.
 

cheerful

New member
Jul 10, 2018
11
??

How many colors for side #1: 3
How many colors for side #2: 3
...
How many colors for side #6: 3

Try again and then account for the radial symmetry that you are allowing? You need DISTINCT colorings.
Oops. Its 3^6 ways to color a hexagon. Since I picking two hexagon, I will have 3^6 * 3^6 = 3^(6*2) ways of colors combination. This happens to be denominator.

What u mean by radial symmetry?
 

cheerful

New member
Jul 10, 2018
11
Oops. Its 3^6 ways to color a hexagon. Since I picking two hexagon, I will have 3^6 * 3^6 = 3^(6*2) ways of colors combination. This happens to be denominator.

What u mean by radial symmetry?
I got to submit this homework over the weekend. Able to give me an example for the different ways to compute the symmetry?

Thanks
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I got to submit this homework over the weekend. Able to give me an example for the different ways to compute the symmetry?

Thanks
In my mind, we need to count the DISTINCT coloring, accounting for the rotational symmetry.

Since there are 6 sides, we might THINK there are (3^6)/6 distinct colorings. Trouble is, that's not an integer and therefore makes no sense. The problem is that there are three colorings that have NO rotational partners. Can you describe these three?
 

cheerful

New member
Jul 10, 2018
11
In my mind, we need to count the DISTINCT coloring, accounting for the rotational symmetry.

Since there are 6 sides, we might THINK there are (3^6)/6 distinct colorings. Trouble is, that's not an integer and therefore makes no sense. The problem is that there are three colorings that have NO rotational partners. Can you describe these three?
I found some similar answer but I dont understand how to derive the answer.

if I colour my hexagons r(ed), g(reen), b(lue), then obviously the six hexagons

(r, r, r, g, g, b)
(r, r, g, g, b, r)
(r, g, g, b, r, r)
etc.
are different, but all equivalent. Whereas:

(r, r, r, r, r, r)
is the only one of its kind. Nothing else is equivalent to it. So if I need a hexagon that's equivalent to (r, r, r, g, g, b), then I have six chances of getting one, whereas if I need a hexagon that's equivalent to (r, r, r, r, r, r), then I only have one chance.

This suggests that we're going to need to look at symmetries.

What's the probability that the first hexagon you choose has 6-fold symmetry (i.e. all sides the same colour)? What's the probability the second hexagon you choose has 6-fold symmetry?
What's the probability that the first hexagon you choose has exactly 3-fold symmetry (e.g. (r, b, r, b, r, b))? (Be careful to exclude those that have 6-fold symmetry, which you've already dealt with in the previous case.) What's the probability the second hexagon you choose has 3-fold symmetry?
2-fold symmetry?
1-fold symmetry?
Now suppose you have two hexagons and you know they both have (e.g.) 2-fold symmetry. What's the probability that they're equivalent? Well, you need to count how many different 2-fold-symmetric hexagons there are (33=27, minus the 3 which are 6-fold-symmetric, so 24), and how many different ones of those are equivalent to any given one (3, because of the symmetry). So 3/24

Any ideas?
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I found some similar answer but I dont understand how to derive the answer.

if I colour my hexagons r(ed), g(reen), b(lue), then obviously the six hexagons

(r, r, r, g, g, b)
(r, r, g, g, b, r)
(r, g, g, b, r, r)
etc.
are different, but all equivalent. Whereas:

(r, r, r, r, r, r)
is the only one of its kind. Nothing else is equivalent to it. So if I need a hexagon that's equivalent to (r, r, r, g, g, b), then I have six chances of getting one, whereas if I need a hexagon that's equivalent to (r, r, r, r, r, r), then I only have one chance.

This suggests that we're going to need to look at symmetries.

What's the probability that the first hexagon you choose has 6-fold symmetry (i.e. all sides the same colour)? What's the probability the second hexagon you choose has 6-fold symmetry?
What's the probability that the first hexagon you choose has exactly 3-fold symmetry (e.g. (r, b, r, b, r, b))? (Be careful to exclude those that have 6-fold symmetry, which you've already dealt with in the previous case.) What's the probability the second hexagon you choose has 3-fold symmetry?
2-fold symmetry?
1-fold symmetry?
Now suppose you have two hexagons and you know they both have (e.g.) 2-fold symmetry. What's the probability that they're equivalent? Well, you need to count how many different 2-fold-symmetric hexagons there are (33=27, minus the 3 which are 6-fold-symmetric, so 24), and how many different ones of those are equivalent to any given one (3, because of the symmetry). So 3/24

Any ideas?
I'm a little surprised no one else has chimed in, here. I can be boring and pedantic.

Anyway...
1) There are 729 different colorings before considering rotational symmetry.
2) 3 are monochromatic
3) All others have five (5) matches if we allow rotation.
4) This gives us 3 + 726/6 = 3 + 121 = 124 different classes.

If you pick the first from any of the monochromatic classes, you are done. You cannot pick a match on the second draw.

I'm thinking that you picked 5 rings and THEN colored them. Thus, there is no component of getting anything in the initial draw of 5 hexagons.

Are we getting anywhere?

I have to disagree with this paragraph:

"So if I need a hexagon that's equivalent to (r, r, r, g, g, b), then I have six chances of getting one, whereas if I need a hexagon that's equivalent to (r, r, r, r, r, r), then I only have one chance."

If you need an equivalent of (r,r,r,g,g,b) you have only 5 chances of getting one, since the necessity was created by the first draw - thus eliminating one from contention.
 

cheerful

New member
Jul 10, 2018
11
I'm a little surprised no one else has chimed in, here. I can be boring and pedantic.

Anyway...
1) There are 729 different colorings before considering rotational symmetry.
2) 3 are monochromatic
3) All others have five (5) matches if we allow rotation.
4) This gives us 3 + 726/6 = 3 + 121 = 124 different classes.

If you pick the first from any of the monochromatic classes, you are done. You cannot pick a match on the second draw.

I'm thinking that you picked 5 rings and THEN colored them. Thus, there is no component of getting anything in the initial draw of 5 hexagons.

Are we getting anywhere?

I have to disagree with this paragraph:

"So if I need a hexagon that's equivalent to (r, r, r, g, g, b), then I have six chances of getting one, whereas if I need a hexagon that's equivalent to (r, r, r, r, r, r), then I only have one chance."

If you need an equivalent of (r,r,r,g,g,b) you have only 5 chances of getting one, since the necessity was created by the first draw - thus eliminating one from contention.
I think it is still very far.

The answer is P(two hexagons with same colors) = 4263/ 3^(6*2)

I guess the question is too challenging for the rest to answer. I think you are much better than me.
 

cheerful

New member
Jul 10, 2018
11
I think it is still very far.

The answer is P(two hexagons with same colors) = 4263/ 3^(6*2)

I guess the question is too challenging for the rest to answer. I think you are much better than me.
I got one reply: But how come I dont understand! Do you?

The fact that there are 5 hexagons is a red herring. Everything is equiprobable so the problem is reduced to finding the number of matches that are possible.

What is tricky is that there are hexagons that match 0, 1, 2, and 5 other hexagons.

We have to count the number of each. For example the hexagons that only match themselves are those with the same color for each edge.

Using trinary numbers these would be 000000, 111111, 222222. There are 3 of these

Similarly there are hexagons that match only 1 other.
An example is 101010.
There are 6 of these.

There are 24 that match only 2 others, and 696 that match 5 others.

So the total number of matches is

3⋅1+6⋅2+24⋅3+696⋅6=4263
 

cheerful

New member
Jul 10, 2018
11
I got one reply: But how come I dont understand! Do you?

The fact that there are 5 hexagons is a red herring. Everything is equiprobable so the problem is reduced to finding the number of matches that are possible.

What is tricky is that there are hexagons that match 0, 1, 2, and 5 other hexagons.

We have to count the number of each. For example the hexagons that only match themselves are those with the same color for each edge.

Using trinary numbers these would be 000000, 111111, 222222. There are 3 of these

Similarly there are hexagons that match only 1 other.
An example is 101010.
There are 6 of these.

There are 24 that match only 2 others, and 696 that match 5 others.

So the total number of matches is

3⋅1+6⋅2+24⋅3+696⋅6=4263
When I add 3 + 6 + 24 + 696 = 729. This is the total colouring you have suggested. But I dont know how 24 and 696 is computed.

ANy ideas?
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
When I add 3 + 6 + 24 + 696 = 729. This is the total colouring you have suggested. But I dont know how 24 and 696 is computed.

ANy ideas?
I don't understand the need to consider matching 2 or 3 or 4 or 5 at a time.

Why isn't it just:

$\dfrac{3}{124}\cdot 0 + \dfrac{121}{124}\cdot\dfrac{5}{123}$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,686
When I add 3 + 6 + 24 + 696 = 729. This is the total colouring you have suggested. But I dont know how 24 and 696 is computed.

Any ideas?
Let me give this a shot.

The first class of colorings is if a rotation by one sixth gives us the same hexagon.
To achieve that, all colors must be the same.
That means there is 3 of them: 000000, 111111, 222222.
If the first hexagon is one of these, there is exactly 1 possibility for the second hexagon to be a match.

Second class is if a rotation by two sixths gives us the same hexagon.
To achieve that, the colors must alternate, such as in the pattern 010101.
We have 3 choices for the first color, and 2 choices for the second color, meaning there are 6 of them.
If the first hexagon is one of these, there are 2 possibilities for the second hexagon to get a match (010101 and 101010).

Third class if is a rotation by three sixths gives us the same hexagon.
Pattern is either 001001 (18 possibilities), or 012012 (6 possibilities).
To find all of them, it's safest to enumerate them.
If the first hexagon is one of these, there are 3 possibilities for the second hexagon to be a match.

Fourth class is all others, where no rotation gives us the same hexagon.
The number of possibilities is $3^6 - 3 - 6 - (18+6) = 696$.
If the first hexagon is one of these, there are 6 possibilities for the second hexagon to be a match.

The probability to get a match between 2 random hexagons is then:
$$\frac{1}{3^6\cdot 3^6}(3\cdot 1 + 6 \cdot 2 + (18+6)\cdot 3 + 696\cdot 6) = \frac{4263}{3^6\cdot 3^6}$$
 

hanli2018

New member
Aug 9, 2018
1
I'm a little surprised no one else has chimed in, here. I can be boring and pedantic.

Anyway...
1) There are 729 different colorings before considering rotational symmetry.
2) 3 are monochromatic
3) All others have five (5) matches if we allow rotation.
4) This gives us 3 + 726/6 = 3 + 121 = 124 different classes.

If you pick the first from any of the monochromatic classes, you are done. You cannot pick a match on the second draw.

I'm thinking that you picked 5 rings and THEN colored them. Thus, there is no component of getting anything in the initial draw of 5 hexagons.

Are we getting anywhere?

I have to disagree with this paragraph:

"So if I need a hexagon that's equivalent to (r, r, r, g, g, b), then I have six chances of getting one, whereas if I need a hexagon that's equivalent to (r, r, r, r, r, r), then I only have one chance."

If you need an equivalent of (r,r,r,g,g,b) you have only 5 chances of getting one, since the necessity was created by the first draw - thus eliminating one from contention.
Counting all the possibilities of coloring the hexagon is also interesting, but I don't agree with the method you provided here. You can disprove it by a square. With your method, you'd get $3+(3^4-3)/4$ ways of coloring, which is not even an integer.

Actually, we can use Burnside's Lemma, https://en.wikipedia.org/wiki/Burnside's_lemma
The symmetry group (just rotation) is $G=\{e,r,r^2,r^3,r^4,r^5\}$, where $r$ is rotation by one sixth. So, the number of possibilities is
$$N = \frac{1}{|G|}\sum_{g\in G} |Fix(g)|.$$
$|Fix(e)| = 3^6, |Fix(r)|=3, |Fix(r^2)|=3^2, |Fix(r^3)| = 3^3, |Fix(r^4)|=|Fix(r^5)|=3$. So, $N=\frac 1 6 (3^6+3+3^2+3^3+3+3) = 129$.