jfgw wrote:I draw three dominoes from a standard "double-six" set of 28. What is the probability that I can lay all three down (in the usual manner) in a single chain?
Nice problem, with a number of ways to tackle it. Here's the best I've come up with so far...
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The probability is the number of sets of three dominos that can be arranged in such a chain divided by the total number of sets of three dominoes. The latter is easy to calculate as 28 choose 3 = (28*27*26)/(3*2*1) = 3276. For those who don't know the argument, I can pick a set of three by picking one of the 28 dominoes, then picking one of the remaining 27 dominoes, then picking one of the remaining 26 dominoes, leading to 28*27*26 ways of picking a set of three. But for each set of three, I could have picked any one of the 3 first, then either of the remaining 2 second, then the remaining one third, leading to 3*2*1 ways of picking that particular set of three. So the 28*27*26 ways of picking a set of three split up into collections of 3*2*1 ways to pick the same set, leading to the answer.
So we need to determine the number of sets of three dominoes that can be arranged in a chain. This can be done in the same way, counting the number of ways of selecting such a set, but taking care to divide out by the number of ways that select the same set. This needs to be done with care, because that number of ways differs for different sets.
So what does such a chain look like? Obvious answer:
+---+---++---+---++---+---+
| a | b || b | c || c | d |
+---+---++---+---++---+---+
It's going to be important which of a, b, c and d are equal to each other. So first observe that a and c are not equal, nor are b and d, as those imply a repeated domino. That implies that there are the following categories of such sets of three dominoes:
CATEGORY 1) a, b, c and d are all different. I can pick a in 7 different ways, then b in 6 different ways, then c in 5 different ways, then d in 4 different ways. That implies there are 7*6*5*4 = 840 ways of choosing such a chain; however, each such chain can be selected by either choosing a or d first (the two numbers that only appear once on a domino), and then either following through on the only ways to complete the chain, which are picks of a, b, c, d or of d, c, b, a respectively. So I've double-counted the sets of three dominoes that lead to such a chain, so there are (7*6*5*4)/2 = 420 such sets of three dominoes. Note that such sets contain two numbers once each and two numbers twice each.
CATEGORY 2) a=b, but a, c, and d are all different. I can pick such a set in 7*6*5 = 210 ways, and the only way to pick a particular set is to pick the double first, then the domino that can connect to the double, then the remaining domino (which cannot connect to the double). So there are 7*6*5 = 210 such sets of three dominoes. Note that such sets contain one number 3 times, another twice and a third once, so they do not overlap with the 840 category 1 sets, and we have 840+210 = 1050 sets of 3 dominoes that can construct a chain so far.
CATEGORY 3) c=d, but a, b and c are all different. Again, there are 7*6*5 = 210 ways of picking such a set, but each such way has already been counted as a category 2 set, by selecting a=b, c and d for the category 2 set as this category's c, b and a respectively. So we don't add any further sets this time, and remain on 1050 sets of dominoes that can construct a chain.
CATEGORY 4) b=c, but a, b and d are all different (note a and d cannot be the same, as that would imply a repeated domino). We can select such a set in 7*6*5 = 210 different ways, but any such set can be selected in two different ways (pick either of a and d first, then b, then the other one of a and d). So there are (7*6*5)/2 = 105 such sets. Note that such sets contain one number 4 times and the other two once each, so cannot overlap with the category 1-3 sets above, and so we now have 1050+105= 1155 sets that can construct a chain.
CATEGORY 5) a=d, but a, b and c are all different (note b and c cannot be the same, as that would imply a repeated domino). Yet again, we can select such a set in 7*6*5 = 210 ways. But each such set can be selected by picking any of a, b and c first, then either of the remaining two, then the remainiing one, so there are (7*6*5)/(3*2*1) = 35 such sets. Noting that such sets contain three numbers twice each, they don't overlap with any of the sets found in categories 1-4 above, and so we now have 1155+35 = 1190 sets that can construct a chain.
CATEGORY 6) a=b and c=d, but a and c are different. Now we have 7*6 = 42 ways of selecting such a set, but each such set is selected twice, one by selecting a and then c, and once by selecting c and then a, so there are (7*6)/2 = 21 such sets. Noting the such sets contain two numbers, three times each, and thus do not overlap with sets in any of categories 1-5 above, that means we finish with 1190+21 = 1211 sets that can construct a chain.
So the probability is 1211/3276 = 173/468.
Gengulphus