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Domino Chain (Posted on 2004-02-11) Difficulty: 4 of 5
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.

See The Solution Submitted by Charlie    
Rating: 3.7143 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(6): solution :-) | Comment 12 of 19 |
(In reply to re(5): solution :-) by SilverKnight)

Thank you for making the grids. I was going to but then got daunted, but that's about what I would have done. I would not leave you folks in limbo; at the very least I'll make the solution public before the problem disappears from the bottom of the home page (and of course journeymen will see it sooner).

I tried to make the explanation of the randomization as explicit as possible by randomizing both the domino chosen and its orientation (vis-a-vis its visibility). That made it more explicit than the problem of Simple Coins, where one (I at least) had to assume that the narrator chose at random which coin to report about. Here we explicitly say that the end of the domino to look at is determined at random.

I notice the scare quotes around 'wrong' in regard to the previously proposed solution. A simulation program would come out with a certain percentage found. I can't think of a simulation program that agrees with the randomization in the puzzle that would produce the original proposed answer, except by forcing it in choosing the "other" side of each domino (but, anyway, that's not agreeing with the randomization specified). Though I didn't ask for simulation programs in the puzzle, they certainly are welcome in probability problems, to verify analytic solutions.
  Posted by Charlie on 2004-02-12 14:06:40

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