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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.

  Submitted by Charlie    
Rating: 3.7143 (7 votes)
Solution: (Hide)
First, it must be shown that in order for such a chain to be formed, the numbers at the ends must be the same (by the way, thereby precluding the visible numbers we saw--1 and 2--from being the actual end numbers). Each number appears 8 times in the set of dominoes: once each in combination with zero through six except for twice on its appearance on the double (i.e., in combination with itself). In a chain of dominoes, the internal (touching) ends must contain an even number of each number, as adjoining ends match. That means the two end points must account for an even number of any number shown, and therefore must be two of the same number.

That shows that a matching number on the two tiles is necessary; is it sufficient? Yes. Call the number that matches between the two dominoes a, and the other number or numbers b (and possibly c). Then assign the remaining pip numbers to the other letters up to g and use one of these two layouts:

aa ac cb bd dc ce ed df fe eg gf fa ag gb bb be ee ea ad dd dg gg gc cc cf ff fb ba


ab bb bc cc cd dd de ee ef ff fg gg ga aa ad db be ec cf fd dg ge ea af fb bg gc ca

The way the two dominoes in the question were chosen, any of the 56 numbers (domino ends) in the set was equally likely to be chosen for each of the two visible positions. That includes each of the eight 1's and each of the eight 2's. The determination of which 1 or 2 is showing determines what the other number is on that domino. Note that the double-1 for example has twice as many ways of appearing as the 1 in, say, 1-3 (but as many ways as 1 OR 3 in 1-3), because regardless of which way it's oriented, it is a possibility for what has happened in this instance.

So there are 8 possibilities for the first domino (including two that are double-1), and 8 for the second. Ordinarily this would be 64 possible combinations, but the 1-2 domino cannot appear in both places, so there are really only 63 equally likely ways of the pair appearing as it did.

Out of these equally likely ways, which ones have one number from the first domino matching one number from the second, so that they can be placed on the ends of a chain? They are:
1-0; 2-0
1-1; 2-1
1-1; 2-1 (the other 1 being the one showing)
1-3; 2-3
1-4; 2-4
1-5; 2-5
1-6; 2-6

1-2; 2-0
1-2; 2-2
1-2; 2-2 (the other 2 being the one showing)
1-2; 2-3
1-2; 2-4
1-2; 2-5
1-2; 2-6

1-0; 2-1
1-3; 2-1
1-4; 2-1
1-5; 2-1
1-6; 2-1

This is 19 out of 63 possibilities, or 19/63 = .3015873015873...
A simulation confirms this:
DIM pip(28, 2)
FOR low = 0 TO 6
 FOR high = low TO 6
   s = s + 1
   pip(s, 1) = low: pip(s, 2) = high
   PRINT s, low, high

  dom1 = INT(RND(1) * 28 + 1)
    dom2 = INT(RND(1) * 28 + 1)
  LOOP UNTIL dom2 <> dom1
  sid1 = INT(RND(1) * 2 + 1)
  sid2 = INT(RND(1) * 2 + 1)
  pip1 = pip(dom1, sid1): pip2 = pip(dom2, sid2)
  IF pip1 <> pip2 THEN
    tr = tr + 1
    pip3 = pip(dom1, 3 - sid1): pip4 = pip(dom2, 3 - sid2)
    IF pip1 = pip4 OR pip2 = pip3 OR pip3 = pip4 THEN
      suc = suc + 1
    END IF
    PRINT USING "####### #.######## #.########"; tr; suc / tr; SQR((tr * suc / tr) * (1 - suc / tr)) / tr

Note that the simulation considers a valid trial any one in which the two showing numbers are unequal, as the probability just depends on the inequality of the two showing numbers, not on their identities.

When stopped after over 3,000,000 trials, the last line showed up to that point:

3378243 0.30136198 0.00024965

indicating 3,378,243 trials produced an average success rate of .30136198, and that if the theoretic success rate is within .00025 of that, that the results fall within one standard deviation of the expected results. The calculated probability and the simulation result do agree within that amount.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
cool problem!Thalamus2004-02-24 13:06:29
Solutionmy last answerAdy TZIDON2004-02-17 05:46:11
Some Thoughtsre(3): we were all wrong AND MAY BE STILL AREAdy TZIDON2004-02-15 04:09:02
re(2): we were all wrongTomM2004-02-14 19:58:15
re: we were all wrongCharlie2004-02-14 19:54:01
re: we were all wrongSilverKnight2004-02-14 18:46:58
Solutionwe were all wrongAdy TZIDON2004-02-13 19:05:35
re(6): solution :-)Charlie2004-02-12 14:06:40
re(5): solution :-)SilverKnight2004-02-12 12:15:05
re(4): solution :-)Charlie2004-02-12 08:39:37
Questionre(3): solution :-)SilverKnight2004-02-12 05:31:23
Solutionsolution?Tristan2004-02-11 23:53:25
re(2): solution :-)SilverKnight2004-02-11 17:54:56
re: solution :-)Charlie2004-02-11 17:52:55
solution :-)SilverKnight2004-02-11 16:23:22
SOLUTION nowAdy TZIDON2004-02-11 15:20:05
Hints/TipsNot sure- just intuitionAdy TZIDON2004-02-11 14:54:11
Hints/TipsHere is a picture of double-sixes dominoesPenny2004-02-11 14:35:09
Hints/TipsFor those who are not familiar with dominoes...Penny2004-02-11 14:22:18
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