Combinatorics - Yahtzee Problem
Solution 1:
That'd be $1-P \space (\text{all dice show a different number})$
And $P \space (\text{all dice show a different number}) = \dfrac{{6\choose 5}5!}{6^5}$
As the numerator is the number of lists of 5 distinct integers, each of which is b/w $1$ and $6$ inclusive. This is relevant as each such list corresponds to a configuration of the 5 dice, also there are $6\choose 5$ ways to choose which 5 distinct numbers appear on the dice and $5!$ ways to assign these to the $5$ dice.