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?
Let's choose the three dominos successively.
If the first one drawn is a double, you can't complete the triangle, as that would require the next two to be identical. So this 1/4 of cases is a failure from the start.
In the 3/4 probability that the first domino drawn is not a double, let's say 1 and 2, there are 5 usable remaining dominos that could be used to match the 1 (the ones that are not 1,1 and not the 1,2 that's already there) and 5 that could be used to match the 2. They are mutually exclusive sets, as the 1,2 is not included in either. So there are 10 dominos that allow continuation, on one side or the other, out of the remaining 27.
After the first two dominos are placed, there is only 1 which will match both ends. Since, in the example, the 1,2 is not present as a second domino also, and we ruled out the use of a double, there will indeed be 1 usable third domino out of the remaining 26.
So the probability of completing the triangle is (3/4)*(10/27)*(1/26) = 5/468 = 0.0106837606837606837... = 1/93.6.
Posted by Charlie
on 2011-03-14 16:48:10