Two dominoes are picked at random from a standard set of doublesixes. 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 numbersone or the other of the stillhidden numbers might be positioned at the actual end(s) of the chain.
(In reply to
we were all wrong by Ady TZIDON)
"btw  11 or 22 do not have any special standing "
The special standing of the doubles is that when initially chosen they will always fall into the observed set of events. When, say the 14 domino is initially chosen, half the time it can't fall into the observed situation because the 4 side will be visible. So when the observed situation comes about half of the 14 dominos are filtered out, but all of the 11 choices remain.

Posted by Charlie
on 20040214 19:54:01 