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 we were all wrong
by Ady TZIDON)
What you're not taking into account (which is what Charlie seems to have been after), are the parts of the problem which read:
"Two dominoes are picked at random from a standard set of double-sixes...",
"You look at only one of the two numbers on each domino, choosing at random which end to look at."
It is not merely sufficient to identify the possible outcomes, but also the likelihood of those outcomes....
(This would be similar to incorrectly saying that with two normal dice there is a 1/11 chance of rolling each of the 11 totals from 2-12).