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July 10th, 2018, 10:41 AM  #1 
Newbie Joined: Jul 2018 From: LA Posts: 2 Thanks: 0  Calculate probability of two polygons
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). I know that 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, but I am not sure how to get it. 
July 12th, 2018, 11:15 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,163 Thanks: 1135 
Sorry for the delay it took me this long to crack this one. 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 $3\cdot 1 + 6 \cdot 2 + 24 \cdot 3 + 696 \cdot 6 = 4263$ Last edited by romsek; July 12th, 2018 at 11:17 PM. 
July 14th, 2018, 11:24 PM  #3 
Newbie Joined: Jul 2018 From: LA Posts: 2 Thanks: 0 
Hi 1. What do you mean by match themselves? ie to match only 1 other, It will be 010101, 121212, 020202, 101010, 212121, 202020? To match only 2 other: is it first digit must match the 4th digit, 2nd digit must match the 5th digit? It will be 012012, 021021, 102102, 120120, i dont think can get 24 ways 2. How do you compute 696? 3. It looks like there is a pattern. So if I have n polygons with e edges with c colours. The answer is ? Its too tough!! 

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