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Domino Polygons (Posted on 2011-03-14) Difficulty: 4 of 5
Randomly draw three from a standard set of 28 Double Six Dominoes.

What is the probability they can be arranged in a triangle with each end touching another end with an equal number of spots?
For example: [1|3][3|0][0|1] can form such a triangle.
(Note: this triangle is not a legal configuration in an actual game of dominoes.)

Randomly draw four dominoes instead.
What is the probability of being able to form a square in the same fashion?

How about 5? 6? 27? 26?

There doesn't seem to be a general formula. Is there?

No Solution Yet Submitted by Jer    
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Some Thoughts 3! 4! (More eureka!) Comment 5 of 5 |
Let's just count combinations:

n = 3
------

Total combinations of 3 dominoes = C(28,3) = 28*27*26/(3*2*1) = 3276

There is exactly one combination that forms a triangle for every set of three different numbers out of 7 = C(7,3) = 7*6*5/(3*2*1) = 35

So, the required probability = 35/3276 = 5/468, which is a little under 1%.


n = 4
-----
Total combinations of 3 dominoes = C(28,3) = 28*27*26*25/(4*3*2*1) = 20475

A Square of dominos can be formed in one of two ways:

a) A triangle of dominoes with a double inserted at one of the three vertices = 35 * 3 = 105

b) Using 4 non-doubles with a different value at each corner.
   There are C(7,4) ways to draw 4 numbers out of 7 = 7*6*5*4/(4*3*2*1) = 35
   But, for each of these, the 4 values can be arranged in 3 ways to form a square,  using 3 different sets of dominoes.
     (If 1234 are the numbers, either 2 or 3 or 4 can be diagonally opposite to 1)
     So, total combinations = 35 * 3 = 105!
     
(Wow!  Again I am wondering about coincidence vs. deep patterns!)
   
The two are mutually exclusive, so the total combinations = 210

Probability = 210/20475 = 2/195 is a little over 1%.

  Posted by Steve Herman on 2011-03-16 23:52:39
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