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?
(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.
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Posted by Jer
on 2011-03-15 16:05:20 |
re: 25! (Eureka!)
