Four fair dice are rolled. What is the probability of a 2-2 split? (For example, rolling 5, 3, 5, 3 is a 2-2 split, and a 1, 4, 1, 1 is a 3-1 split.)
+121 There are 15 splits if we do not count the order in which the 4 numbers are arranged, This comes from the observation that picking two distinct numbers out of the 6 possible, not counting order, is \({ 6\choose 2}\)=15. Each of the splits can be ordered in 6 different ways. For instance, two ones and two twos can be arranged as
(1, 1, 2, 2)
(1, 2, 1, 2)
(1, 2, 2, 1)
(2, 1, 2, 1)
(2, 2, 1, 1)
(2, 1, 1, 2)
so there are \({6 \times15}\)=90 ordered splits. There are a total of \(6^4\)outcomes when 4 six-sided dies are cast. So the probability of an ordered split is \(\frac {90} {6^4}=\frac{5}{72}\)
