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Domino Chain (Posted on 2004-02-11) Difficulty: 4 of 5
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.

See The Solution Submitted by Charlie    
Rating: 3.7143 (7 votes)

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Solution we were all wrong | Comment 13 of 19 |
(In reply to re(6): solution :-) by Charlie)

I am a long way from home & cannot write a long explanation- may be later. however the answer to the problem as it is worded is 19/49.
Shortly: the only pieces that count are 13 dominoes- 7 one-pip pieces and 7 two-pip pieces , 1-2 being counted twice.
Let us divide the population into 3 subsets:
A 6 one-pip pieces ,( all but 1-2)
B 6 two-pip pieces , ,( all but 1-2)
C the 1-2 domino.

The first domino is drawn from the union A U C
with probability of 6/7 belonging to A and
probability of 1/7 belonging to C.
If it is from A then it can be complemented by one matching piece from B or the one piece from C 6/7*2/7 =12/49
If the 1st domino is from C than any one of the B set qualifies: 1/7*6/6=1/7

Sum up 12/49+1/7= 19/49 - thats my answer.

btw - 1-1 or 2-2 do not have any special standing .
If you choose to look on the wrong end of the 1-pip pr 2-pip
it counts exactly as drawing a piece not in A U B U C-
conradicting the conditions stated in the problem.
Simulation taking into account only those 13 pieces will surely confirm my result.


Edited on February 13, 2004, 7:07 pm
  Posted by Ady TZIDON on 2004-02-13 19:05:35

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