What is the probability of getting two of the same shape? We have 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).
I know that the total space or denominator is $3^{2\times6}$, therefore we have $3^{2\times6}$ at the bottom, but what is the numerator?
The answer is $4263$, but I am not sure how to get it.
 A: The tricky part of this question is in the statement that everything should be done up to rotation. For the purposes of this answer, I'm going to say that two hexagons are "the same" if they are coloured in exactly the same way, and "equivalent" if they're the same maybe after rotating one of them.
In other words: 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 ($3^3 = 27$, 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.
Putting these pieces together will give you the answer, and you should see where the $4263$ comes from too. Give it a go!
