Two dominoes are picked at random from a standard set of double-sixes. Such a set contains all the possible combinations of two numbers of pips that are possible from zero to six. That includes all 7x6/2=21 combinations of two different numbers plus all seven doubles from double zero to double six.
You look at only one of the two numbers on each domino, choosing at random which end to look at. You see that the number you look at on the first domino is 1. The number you see on the second domino is 2 (of course represented as pips).
What is the probability that you will be able to use these two dominoes as the ends of a chain of dominoes using all 28 in the set, linked in the usual fashion of requiring a match between the two adjoining numbers of two touching dominoes?
Remember, the numbers you looked at need not be the end numbers--one or the other of the still-hidden numbers might be positioned at the actual end(s) of the chain.
17/48 - I agree with Ady...
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There are 7 number (0, 1, 2, 3, 4, 5, and 6), and each of them is displayed 8 times (for a total of 56). Since there are 2 on each domino, we should find 56/2 = 28 dominoes. Verified.
Now, to have a chain, clearly every number must match up with a match. Therefore, we must have an even number of them along the chain. Therefore, the two "ends" of the chain MUST match.
Equivalently,
If the ends of the chain don't match, then we can't make a chain using all the dominoes.
(Corollary: if the ends of the chain *do* match, then we CAN make a chain using all the dominoes... left to prove as an exercise to the reader.)
So, if, for example, the two dominoes are (
1,3) and (
2,4), then we will be unable to make a chain. (This is because they share no number, so we can't have the ends equal.)
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Not including the the 1,2 domino, there are:
6 x 6 = 36 different possibilities ( (0,1) & (0,2), (0,1) & (3,2), etc. )
Including the 1,2 domino, there are:
2 x 6 = 12 different possibilities ( (1,2) & (x,1), (1,2) (x,2))
Total: 48 equally likely possibilities
Of them, only the following have a *match* at the ends:
with (1,2) - 12 combinations
(1,2) (0,1)
(1,2) (1,1)
(1,2) (3,1)
(1,2) (4,1)
(1,2) (5,1)
(1,2) (6,1)
(2,1) (0,2)
(2,1) (2,2)
(2,1) (3,2)
(2,1) (4,2)
(2,1) (5,2)
(2,1) (6,2)
without (1,2) - 5 combinations
(0,1) (2,0)
(3,1) (2,3)
(4,1) (2,4)
(5,1) (2,5)
(6,1) (2,6)
Total of 17 successful combinations over 48 equally likely possibilities.
solution :-)
