All Evens Before Any Odd on a Fair Die
You roll a fair six-sided die repeatedly. What is the probability that you see all three even numbers (2, 4, and 6) before you see any odd number (1, 3, or 5)?
In other words, when the first odd number finally appears, all three even numbers must have already shown up at least once.
Hints
- Don't track every roll -- think about the order in which the six distinct values first appear.
- By symmetry of a fair die, the first-appearance ordering of the six faces is a uniformly random permutation. You need $\{2, 4, 6\}$ to occupy the first three positions.
- Count step by step: the first new value is even with probability $3/6$, the second (given the first was even) is even with probability $2/5$, the third is even with probability $1/4$. Multiply.
Worked Solution
How to Think About It: Picture rolling the die over and over. Each roll is equally likely to be any of {1, 2, 3, 4, 5, 6}. The process "ends" the moment an odd number appears. For us to win, we need to have collected all three even values -- 2, 4, and 6 -- before that happens. This is a race between collecting all even faces and hitting any odd face.
The elegant way to think about it: forget about repeated values entirely. Focus only on the order in which the six distinct values first appear. By symmetry (each face has probability $1/6$), every ordering of first-appearance times is equally likely -- it is a uniformly random permutation of $\{1, 2, 3, 4, 5, 6\}$.
Quick Estimate: There are $\binom{6}{3} = 20$ ways to choose which 3 of the 6 values appear first. We need the first 3 to be exactly $\{2, 4, 6\}$. So we should expect something around $1/20 = 0.05$. Let's verify.
Approach: Use the fact that the sequence of distinct values encountered is a uniformly random permutation.
Formal Solution:
Label the six die faces. Define $T_i$ as the roll on which face $i$ first appears. By symmetry of the fair die, the relative ordering of $(T_1, T_2, T_3, T_4, T_5, T_6)$ is a uniformly random permutation of 6 elements.
We need: every even value appears before every odd value. Equivalently, in the random permutation of first-appearance order, positions 1, 2, 3 must be occupied by $\{2, 4, 6\}$ (in any order).
We can compute this step by step using the sequence of new (previously unseen) values:
- The 1st new value: probability it is even $= 3/6 = 1/2$ (3 even faces out of 6 total).
- The 2nd new value, given the 1st was even: probability it is even $= 2/5$ (2 remaining evens out of 5 remaining values).
- The 3rd new value, given the first two were even: probability it is even $= 1/4$ (1 remaining even out of 4 remaining values).
$$P = \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{6}{120} = \frac{1}{20}$$
Equivalently, count favorable permutations of the 6 first-appearance positions: the 3 evens can fill the first 3 slots in $3!$ ways, and the 3 odds fill the last 3 slots in $3!$ ways, out of $6!$ total permutations:
$$P = \frac{3! \cdot 3!}{6!} = \frac{6 \cdot 6}{720} = \frac{36}{720} = \frac{1}{20}$$
Both methods agree.
Answer: $P = \dfrac{1}{20} = 0.05$
Intuition
The key insight is to ignore repeated values and focus on the order of first appearances. Each face of a fair die has the same probability, so by symmetry, the six faces are equally likely to appear in any order as "firsts." The problem reduces to: in a random permutation of 6 items (3 labeled "even" and 3 labeled "odd"), what is the probability the evens come first? This is just $1 / \binom{6}{3} = 1/20$.
This "first-appearance permutation" trick is extremely powerful and shows up constantly in interview problems about collecting items, coupon-collector variants, and racing processes. Whenever you have a fair die (or any symmetric setup) and ask "does event A happen before event B," check whether the problem reduces to a question about the ordering of first appearances. It usually does, and the symmetry makes the answer clean. The common mistake is to try conditioning on the number of rolls and doing geometric series -- it works but is far more painful than necessary.