Nine fair dice are rolled simultaneously. What is the probability of getting three pairs?

My attempt:

$$P(A) = \frac{\binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times3\times2\times1}{6^{9}}$$

We first choose which three numbers will be pairs, and which dice will be pairs, then we are left with three numbers and three positions for them. Is this correct?

  • 1
    $\begingroup$ So... nine dice, three pairs, that only accounts for six of the dice. How about the remaining three? Are you looking to only count those scenarios where you have three numbers appearing twice each, each of which different, and the remaining three dice all being different than eachother and different than all that came before? For example 112233456? Or are you asking for the probability that there are at least three numbers each of which appeared at least two times each? For example 111222334? How about three pairs and a triple like 112233444? $\endgroup$ – JMoravitz Feb 12 at 1:17
  • $\begingroup$ Do you mean exactly three pairs? Because if so, I counted them all on computer and number is different. $\endgroup$ – enedil Feb 12 at 1:18
  • $\begingroup$ Your analysis and methodology are correct. $\endgroup$ – Robert Shore Feb 12 at 1:19
  • $\begingroup$ Assuming you mean the first interpretation in my comment... your answer is close but incorrect. Take note... Why did you choose the number $10$ in the second binomial coefficient in the numerator? You then used $8$ which is two less than $10$... but again, why $10$? The error continues to propagate itself in the next binomial coefficient too. $\endgroup$ – JMoravitz Feb 12 at 1:19
  • $\begingroup$ @JMoravitz He means exactly three pairs, with the other three numbers appearing once each on the remaining three dice. The last three elements of the product in his numerator assign the last three dice to the three missing numbers. $\endgroup$ – Robert Shore Feb 12 at 1:21

Here's another approach to counting the numerator:

Take the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9.$ Arrange them in any sequence. Then pick three of the numbers $1, 2, 3, 4, 5, 6$ and replace $7, 8, 9$ with those numbers in increasing order (that is, $7$ is replaced by the smallest number, $9$ by the largest).

This produces every possible sequence of nine numbers selected from $\{1, 2, 3, 4, 5, 6\}$ with three pairs and three singletons, that is, the number of rolls that have exactly three pairs and no other matches, but it produces each such sequence more than once. In particular, each of the numbers $7, 8,$ and $9$ could originally have appeared either before or after the numbers they eventually are made to duplicate, so each roll of three pairs is produced exactly eight times.

After correcting for this overcounting, the number of rolls is

$$ \frac{9! \binom63}{8} = \frac52(9!).$$

This turns out to be equal to

$$ \binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times 6 = \binom{6}{3} \frac{9!7!5!\times6}{(2!7!)(2!5!)(2!3!)} = \binom{6}{3} \frac{9!}{(2!)^3},$$

confirming your answer for the probability of exactly three pairs.


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