All Odds Before Even

Question

You are rolling a fair 6-sided die repeatedly. What is the probability that all odd values $(1,3,5)$ appear before any even value $(2,4,6)$?

QuantSoc · Accepted Answer

There are $6! = 720$ possible ways to arrange the numbers 1 through 6. To ensure all odd numbers $(1,3,5)$ appear before any even numbers $(2,4,6)$, the 3 odd numbers must occupy the first 3 positions, and the 3 even numbers must occupy the last 3 positions.

The number of favorable arrangements is:

$$
3! 	imes 3! = 6 	imes 6 = 36
$$

Therefore, the required probability is:

$$
\frac{36}{720} = \frac{1}{20}
$$

Final Answer:

$$
\boxed{\frac{1}{20}}
$$