Sum of Three Largest Dice

Probability · Medium · Free problem

You roll $5$ standard fair six-sided dice and sum the $3$ largest values showing. What is the probability that this sum equals

8$?

Hints

  1. What is the maximum possible value each die can show? What does that force the three largest dice to be?
  2. If all three largest must be $6$, reframe the problem as counting how many sixes appear among the $5$ rolls.
  3. The number of sixes follows a $\text{Bin}(5, 1/6)$ distribution. Compute $P(N \geq 3)$ using the binomial PMF.

Worked Solution

How to Think About It: Each die maxes out at $6$, so the only way the three largest sum to

8$ is if all three are $6$. That means at least $3$ of the $5$ dice must show a $6$. So the question reduces to: what is the probability that a $\text{Bin}(5, 1/6)$ random variable is at least $3$? This is a common trick in order-statistic problems -- translate a condition on extremes into a counting condition on individual outcomes.

Quick Estimate: Each die has a

/6$ chance of being a $6$, so the expected number of sixes in $5$ rolls is about $0.83$. Getting $3$ or more is well into the upper tail. Rough guess: it is a small probability, probably in the low single-digit percent range. The dominant term is $P(N = 3) = \binom{5}{3}(1/6)^3(5/6)^2 = 10 \cdot (25/7776) \approx 250/7776 \approx 0.032$. The $N = 4$ and $N = 5$ terms are much smaller, so the total should be just above $3\%$.

Approach: Compute $P(N \geq 3)$ exactly using the binomial PMF.

Formal Solution:

Let $N$ be the number of sixes among the $5$ dice. Since each die independently shows a $6$ with probability

/6$, we have $N \sim \text{Bin}(5, 1/6)$.

The sum of the three largest values equals

8$ if and only if $N \geq 3$. We compute:

$P(N \geq 3) = P(N = 3) + P(N = 4) + P(N = 5)$

$= \binom{5}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^2 + \binom{5}{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right) + \binom{5}{5}\left(\frac{1}{6}\right)^5$

$= 10 \cdot \frac{25}{7776} + 5 \cdot \frac{5}{7776} + 1 \cdot \frac{1}{7776}$

$= \frac{250 + 25 + 1}{7776} = \frac{276}{7776} = \frac{23}{648}$

Answer: The probability is $\dfrac{23}{648} \approx 0.0355$, or about $3.55\%$.

Intuition

The key insight is recognizing that a constraint on order statistics (the three largest values) can often be translated into a simpler counting condition on individual outcomes. Here, "the top three sum to 18" sounds like it needs order-statistic machinery, but since each die caps at 6, the only way to hit 18 is for all three largest to equal 6 -- which just means at least three dice landed on 6. Once you see that, the problem collapses to a standard binomial tail probability.

This pattern comes up constantly in quant interviews and in practice. Whenever you see a condition involving extremes or order statistics, ask yourself: can I rephrase this as a condition on counts? It almost always simplifies the math dramatically. In trading, the same logic applies when you think about joint tail events -- the probability that multiple positions all hit their max loss is easier to reason about by counting how many positions are in the danger zone rather than working with the joint distribution directly.

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