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.
Now finish the thought and score the whole matchup, not just your side of it. Your friend beats you exactly when they roll $6$ and you roll below $6$: $P(Y > X) = \tfrac12 \cdot \tfrac56 = \tfrac{5}{12}$ -- identical to your $\tfrac{5}{12}$. The remaining $P(X = Y) = \tfrac12 \cdot \tfrac16 + \tfrac12 \cdot \tfrac16 = \tfrac16$ is the tie. So the two dice are exactly evenly matched: $\tfrac{5}{12}$ / $\tfrac{5}{12}$ / $\tfrac16$. Concentrating all the mass on the extremes did not buy the loaded die any edge, and it did not cost it any either -- both dice share the mean $3.5$, and here the head-to-head is a dead heat.
That symmetry is the point worth carrying away: a fatter-tailed distribution with the same mean is not automatically harder or easier to beat, so resist reading an edge into variance alone. Whether higher variance helps depends entirely on the payoff you are graded against -- symmetric win/lose comparisons like this one are exactly the case where it washes out, while convex payoffs (options, tournament prizes, stop-loss rules) are where it starts to matter.