Let the letters a, b, c, d, e, f represent the numbers 1-6, but not necessarily a=1, b=2, etc. The three dice are thrown. The first shows value a. There are 10 ways to get another a (and no more):

aab, aac, aad, aae, aaf, aba, aca, ada, aea, afa.

And there are another 5 ways where the other two dice duplicate their values:

abb, acc, add, aee, aff.

So thatâ€™s 15 duplications out of a possible 36 combinations of the second and third dice, where a is the value of the first die.

What applies to a symmetrically applies to the other values b-f. So the probability is 15/36=5/12. (Or we could simply state this as 90/216 for all possible outcomes. That is, out of the 216 possible outcomes for throwing three dice, 90 of them contain exactly one pair of duplicated values.)