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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(5): solution :-) | Comment 11 of 19 |
(In reply to re(4): solution :-) by Charlie)

Different (correct?) solution, agreeing with Tristan:
19/63
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Okay... here's my qualified response:

It's not clear to me that our previous interpretation is 'wrong'. But if I understand what Charlie is getting at, then the previous answers are not what he is looking for, and here's his line of reasoning. (I hope Charlie will continue to comment, so we're not left in limbo, wondering...)
_______________________________________

Let us 'name' each domino by using its 2 sides, lower number first (doubles are ok too)... Then the 28 dominoes are:

(0,0) (0,1) (0,2) (0,3) (0,4) (0,5) (0,6)
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,2) (2,3) (2,4) (2,5) (2,6)
(3,3) (3,4) (3,5) (3,6)
(4,4) (4,5) (4,6)
(5,5) (5,6)
(6,6)

We can then draw a grid 28x28 showing the intersections of the first domino with the second domino.... (the diagonal line down the center is thrown away, because we can't pull the same domino twice).

Okay, now, for simplicity (longer, but easier, IMHO, to understand) let us make four of these grids:

  1. the first is where we take the side 1 of the domino 1 & side 1 of domino 2
  2. the second is where we take the side 1 of the domino 1 & side 2 of domino 2
  3. the third is where we take the side 2 of the domino 1 & side 1 of domino 2
  4. the fourth is where we take the side 2 of the domino 1 & side 2 of domino 2


Grid 1
The possible dominos to be seen:

(2,2)(1,1) (2,3)(1,1) (2,4)(1,1) (2,5)(1,1) (2,6)(1,1)
(2,2)(1,2) (2,3)(1,2) (2,4)(1,2) (2,5)(1,2) (2,6)(1,2)
(2,2)(1,3) (2,3)(1,3) (2,4)(1,3) (2,5)(1,3) (2,6)(1,3)
(2,2)(1,4) (2,3)(1,4) (2,4)(1,4) (2,5)(1,4) (2,6)(1,4)
(2,2)(1,5) (2,3)(1,5) (2,4)(1,5) (2,5)(1,5) (2,6)(1,5)
(2,2)(1,6) (2,3)(1,6) (2,4)(1,6) (2,5)(1,6) (2,6)(1,6)

(1,1)(2,2) (1,2)(2,2) (1,3)(2,2) (1,4)(2,2) (1,5)(2,2) (1,6)(2,2)
(1,1)(2,3) (1,2)(2,3) (1,3)(2,3) (1,4)(2,3) (1,5)(2,3) (1,6)(2,3)
(1,1)(2,4) (1,2)(2,4) (1,3)(2,4) (1,4)(2,4) (1,5)(2,4) (1,6)(2,4)
(1,1)(2,5) (1,2)(2,5) (1,3)(2,5) (1,4)(2,5) (1,5)(2,5) (1,6)(2,5)
(1,1)(2,6) (1,2)(2,6) (1,3)(2,6) (1,4)(2,6) (1,5)(2,6) (1,6)(2,6)

Of these 60, 18 (in bold) will work.


Grid 2
The possible dominos to be seen:

(1,1)(0,2) (1,2)(0,2) (1,3)(0,2) (1,4)(0,2) (1,5)(0,2) (1,6)(0,2)
(1,1)(1,2) (1,3)(0,2) (1,4)(0,2) (1,5)(0,2) (1,6)(0,2)
(1,1)(2,2) (1,2)(2,2) (1,3)(2,2) (1,4)(2,2) (1,5)(2,2) (1,6)(2,2)
(2,2)(0,1) (2,3)(0,1) (2,4)(0,1) (2,5)(0,1) (2,6)(0,1)
(2,2)(1,1) (2,3)(1,1) (2,4)(1,1) (2,5)(1,1) (2,6)(1,1)

Of these 27, 7 (in bold) will work.


Grid 3

(0,1)(2,2) (0,1)(2,3) (0,1)(2,4) (0,1)(2,5) (0,1)(2,6)
(0,2)(1,1) (0,2)(1,2) (0,2)(1,3) (0,2)(1,4) (0,2)(1,5) (0,2)(1,6)
(1,1)(2,2) (1,1)(2,3) (1,1)(2,4) (1,1)(2,5) (1,1)(2,6)
(1,2)(1,1) (1,2)(1,3) (1,2)(1,4) (1,2)(1,5) (1,2)(1,6)
(2,2)(1,1) (2,2)(1,2) (2,2)(1,3) (2,2)(1,4) (2,2)(1,5) (2,2)(1,6)

Of these 27, 7 (in bold) will work.


Grid 4

(0,1)(0,2) (0,1)(1,2) (0,1)(2,2)
(0,2)(0,1) (0,2)(1,1)
(1,1)(0,2) (1,1)(1,2) (1,1)(2,2)
(1,2)(0,1) (1,2)(1,1)
(2,2)(0,1) (2,2)(1,1)

Of these 12, 6 will work.


Now.... we total the possibilities up: 60 + 27 + 27 + 12 = 126
we total the working combinations: 18 + 7 + 7 + 6 = 38

38/126 = 19/63
Edited on February 12, 2004, 12:18 pm
  Posted by SilverKnight on 2004-02-12 12:15:05
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