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.
(In reply to re(2): we were all wrong
..".the answer is not as simple or intuitive as it at first seems, "
I never said it was. This is one of the best and certainly the most intriguing problems I found on fl.perp.
At present time I am on vacation in Austria and have a very limited access to Internet. I will organize my thoughts upon return and post a more detailed explanation- the way I see it. My humble opinion is STILL that it all boils down
to proper definition of how the randomness is achieved.
I hold in high esteem other people´´s considerations but STILL think that my solution is the right one- answering the problem the way it is defined.
The sequel to my memo is likely to appear about 48 hours from now.