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## 1 Answer

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.)

by Top Rated User (680k points)

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