What is the probability that I roll a 2 before I roll two odd numbers? Assuming that I use a standard die, what is the probability that I roll a 2 before I roll  two odd numbers? The odd numbers do not have to be distinct. For example, 1,6,4,2 wins and 3,3 loses.
 A: It is reasonably clear that this probability exists. Call it $p$, the probability of winning.  We use the method of conditioning on the result of the first toss.  
If the first toss is a $2$, we have won..
If the first toss is a $4$ or a $6$, the probability we ultimately win remains at $p$.
If the first toss is odd, then we want the probability of tossing a $2$ before the second odd. This is $\frac{1}{4}$, since there are $3$ odd and only one $2$.
Thus
$$p=\frac{1}{6}+\frac{2}{6}p+\frac{3}{6}\cdot\frac{1}{4}.$$
Solve for $p$. We get $p=\frac{7}{16}$.
A: First observe that the numbers $4,6$ do not affect the outcome at all here, so it is equivalent to consider a $4$ sided die with sides $1,2,3,5$. The fact that it does not matter which odd number shows up means that we could consider a $4$ sided die with sides:
$2$, odd, odd, odd.
The probability that I roll $2$ odd numbers before rolling a $2$ is then $$\frac34 * \frac34=\frac9{16}$$
Hence your probability is $1-\dfrac9{16}=\dfrac7{16}$.
