Answer
$\frac{1}{9}$
Work Step by Step
This is a two dice problem, the total outcomes are 36 ($6\times6$)
There are 1 chance (11) to roll a 2, 2 chances (12,21) to roll a 3,
and 1 chance (66) to roll a 12, so there are 4 chances to roll a 2,3, or 12,
the probability is $p=\frac{4}{36}=\frac{1}{9}$ of losing on the first roll.
