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.

Upon returning home I carefully reassessed the domino problem, read all the relevant comments and found my error. ...Took some time.

My answer now : 5/16

Let us divide the relevant population into 3 subsets:

A 7 one-pip pieces,(all but 1-2 +1-1 counted twice)

B 7 two-pip pieces , ,( all but 1-2 +2-2 counted twice)

C the "1-2" domino.

The first domino is drawn from the union A U C

with probability of 7/8 belonging to A and

probability of 1/8 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 except 1-1 :the one piece from C only.

If the 1st domino is from C than any one of the B set qualifies: 1/8*6/6=1/8

Sum up 5/8*2/8+2/8*1/8+1/8= 20/64=5/16 thats my answer.

The claim-" 1-1 or 2-2 do not have any special standing " was the source of my blunder..

ady

Added after seeing the official solution-.....

19/63 is right/

Actually I HAVE STARTED with the "grid counting" but was

not careful enough.

Beautiful problem...

ady

*Edited on ***February 17, 2004, 6:06 am**