Beating a Loaded Die
You and a friend each roll a single die. Your die is a standard fair 6-sided die with faces $\{1, 2, 3, 4, 5, 6\}$, each equally likely. Your friend's die is non-standard: its six faces show the values $\{1, 1, 1, 6, 6, 6\}$, each face equally likely.
Both of you roll once. What is the probability that your roll is strictly higher than your friend's?
Hints
- Your friend's die only has two distinct outcomes. What are they, and how likely is each?
- If your friend rolls a $6$, can you possibly beat them with a standard die?
- Condition on your friend's result: you can only win when they roll $1$, and then you need $X \geq 2$. Multiply the probabilities.
Worked Solution
How to Think About It: Before computing anything, think about what your friend's die can actually show. It only has two possible outcomes: $1$ (with probability $1/2$) or $6$ (with probability $1/2$). If your friend rolls a $6$, you cannot beat them -- the best you can do on a standard die is also $6$, which ties but does not win. So the only way you win is if your friend rolls a $1$, and then you roll something strictly greater than $1$.
Quick Estimate: Your friend rolls $1$ half the time. When they do, you just need to beat $1$ on a standard die, which means rolling $2, 3, 4, 5,$ or $6$ -- that is $5$ out of $6$ outcomes. So the answer should be roughly $(1/2)(5/6) \approx 0.417$. That is a bit below $1/2$, which makes sense: the loaded die concentrates mass on the extremes, so it is harder to beat than you might expect.
Approach: Condition on your friend's outcome and use the law of total probability.
Formal Solution: Let $X$ be your roll and $Y$ be your friend's roll. We want $P(X > Y)$.
Your friend's die has three faces showing $1$ and three showing $6$, so:
$$P(Y = 1) = \frac{3}{6} = \frac{1}{2}, \qquad P(Y = 6) = \frac{3}{6} = \frac{1}{2}$$
Conditioning on $Y$:
$$P(X > Y) = P(X > Y \mid Y = 1)\,P(Y = 1) + P(X > Y \mid Y = 6)\,P(Y = 6)$$
- If $Y = 6$: no face on a standard die exceeds $6$, so $P(X > 6) = 0$.
- If $Y = 1$: you need $X \geq 2$, so $P(X > 1) = 5/6$.
Putting it together:
$$P(X > Y) = \frac{5}{6} \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} = \frac{5}{12}$$
Answer: The probability that you roll strictly higher than your friend is $\dfrac{5}{12} \approx 0.4167$.
Intuition
This problem is a clean exercise in conditioning on the right variable. The loaded die looks intimidating at first, but once you realize it only takes two values -- $1$ and $6$, each with probability $1/2$ -- the problem collapses into a simple case split. The broader lesson is that non-standard dice (and more generally, non-standard distributions) often have fewer effective outcomes than their number of faces suggests. Identifying the support of a distribution before doing any calculation is a habit that saves time in interviews and in practice.
Note also that the loaded die is actually quite strong despite having the same expected value as the standard die ($3.5$). By concentrating all its mass on the extremes, it wins less often but ties or wins big when it does. This is a miniature version of a concept that shows up constantly in trading: a high-variance strategy can be hard to beat even if its mean is the same as a low-variance one, because you need favorable realizations to come out ahead.