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): solution :-) by SilverKnight)
BTW, Charlie....
Regarding your problem's statement:
"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..."
I read this to be equivalent to:
"One domino has at least one '1' on it, and the other domino has at least one '2' on it."
It seems (to me) that Ady interpreted it in the same manner. Are we understanding the problem correctly?
-SK
re(3): solution :-)
