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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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Comments: ( Back to comment list | You must be logged in to post comments.)
re: 25! (Eureka!) | Comment 4 of 5 |
(In reply to 25! (Eureka!) by Steve Herman)

Nice reasoning!

That p(three doubles) = p(triangle) is not just a coincidence.

It looks like a coincidence at first because the calculations for the numerator are 21*10*1 and 7*6*5 but
if d=the size of the set (d=6 for a double six set)
the formula for each simplifies to the same rational function
p(d) = [8(d-1)] / [(d+2)(d+3)(d^2+3d-2)]

I am not sure how deep a pattern it is because the probability for four doubles is not the same as that they form a square.  I do not have general formulas for either.   

It should be clear now that if you can solve five dominoes you can easily get 24. 


  Posted by Jer on 2011-03-15 16:05:20

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