Question

In four throws with a pair of dice, what is the chance of throwing doublets twice at least?

Answer 1

In a single throw the chance of doublets is $\frac{6}{36},$ or $\frac{1}{6}$; and the chance of failing to throw doublets is $\frac{5}{6}$. Now the required event follows if doublets are thrown four times, three times, or twice; therefore the required chance is the sum of the first three terms of the expansion of ${\left(\frac{1}{6}+\frac{5}{6}\right)}^4$.
Thus the chance $=\frac{1}{6^4}(1+4.5+{.5}^2)=\frac{19}{144}$.