MoMath Mindbender: Six Dice, Four Distinct Numbers

Posted on July 16, 2026

Source

Monthly Mindbenders (July 2026). A counting problem using integer partitions and the inclusion-exclusion principle.

Problem

If you roll six dice, you could get the same number on every die, or (in contrast) all different numbers. What is the probability of getting exactly four distinct numbers?

Answer

\[\boxed{\dfrac{325}{648}} \approx 50.15\%\]

Solution

Step 1: Count all outcomes

Each die has 6 faces and rolls independently, so the total number of outcomes is

\[6^6 = 46{,}656.\]

Step 2: Classify the favorable outcomes by multiplicity pattern

For exactly 4 distinct values to appear across 6 dice, the counts of each value must be a partition of 6 into exactly 4 positive parts. The only such integer partitions are:

Pattern	Description
\((3,1,1,1)\)	One value appears three times; three values appear once each
\((2,2,1,1)\)	Two values appear twice; two values appear once each

Step 3: Count pattern \((3,1,1,1)\)

There are four choices to make:

Choose which 4 of the 6 die faces are present: \(\dbinom{6}{4} = 15\)
Choose which of those 4 faces is the triple: \(4\)
Choose which 3 of the 6 dice show the tripled face: \(\dbinom{6}{3} = 20\)
Assign the remaining 3 faces to the remaining 3 dice (one each): \(3! = 6\)

\[15 \times 4 \times 20 \times 6 = 7{,}200\]

Step 4: Count pattern \((2,2,1,1)\)

Similarly:

Choose which 4 faces appear: \(\dbinom{6}{4} = 15\)
Choose which 2 of those 4 faces are the doubled ones: \(\dbinom{4}{2} = 6\)
Choose which 2 dice show the first doubled face: \(\dbinom{6}{2} = 15\)
Choose which 2 of the remaining 4 dice show the second doubled face: \(\dbinom{4}{2} = 6\)
Assign the remaining 2 singleton faces to the last 2 dice: \(2! = 2\)

\[15 \times 6 \times 15 \times 6 \times 2 = 16{,}200\]

Step 5: Combine and simplify

Total favorable outcomes:

\[7{,}200 + 16{,}200 = 23{,}400\]

Probability:

\[\frac{23{,}400}{46{,}656} = \frac{325}{648} \approx 0.5015\]

Answer: \(\displaystyle\frac{325}{648} \approx \boxed{50.15\%}\)

Remarkably, getting exactly 4 distinct numbers is the single most likely outcome — it is more probable than any other value of distinct counts.

Aside: all distinct-count probabilities

For completeness, here are the probabilities for every possible number of distinct values when rolling 6 dice. The count for each \(k\) is \(\binom{6}{k}\) (choose which \(k\) faces appear) times \(k!\,S(6,k)\) (surjections from 6 dice onto those \(k\) faces), where \(S(6,k)\) is a Stirling number of the second kind.

Distinct values \(k\)	\(S(6,k)\)	Outcomes	Probability
1	1	6	\(\approx 0.013\%\)
2	31	930	\(\approx 1.99\%\)
3	90	10,800	\(\approx 23.15\%\)
4	65	23,400	\(\mathbf{325/648 \approx 50.15\%}\)
5	15	10,800	\(\approx 23.15\%\)
6	1	720	\(\approx 1.54\%\)

The totals sum to \(6+930+10800+23400+10800+720 = 46{,}656 = 6^6\) ✓

The symmetry between \(k=3\) and \(k=5\) (both 10,800 outcomes) is a pleasant coincidence of the numbers; it does not persist for other dice counts.

Interactive: Roll the Dice

Click Roll once to see a single roll of 6 dice. Use Roll 10,000× to run a simulation and compare to the theoretical probability.

—

More MoMath Mindbenders: Previous | Next

Back to blog