What is the probability that I roll a 2 before I roll two odd numbers?

Solution 1:

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 $1-\dfrac9{16}=\dfrac7{16}$.

Solution 2:

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 $p=\frac{7}{16}$.