Matching Multiset of Three Dice Rolls
You roll three standard fair 6-sided dice and observe the values on the upfaces. Then you roll all three dice again. What is the probability that the second roll produces the same multiset of values as the first roll? (The values on the second roll can appear in any order -- only the collection of numbers matters, not which die shows which number.)
Hints
- Condition on the structure of the first roll -- how many distinct values appeared among the three dice?
- There are three cases: all same, exactly two distinct, all distinct. Count the permutations of each pattern to find how many second-roll outcomes match.
- Use the law of total probability: $P(M) = \sum_{i=1}^{3} P(M \mid V_i) P(V_i)$, where $P(V_1) = 1/36$, $P(V_2) = 5/12$, $P(V_3) = 5/9$, and $P(M \mid V_i) = (\text{permutations of pattern})/216$.
Worked Solution
How to Think About It: The second roll needs to produce the same multiset of values -- the same numbers with the same multiplicities, but in any order. The key insight is that the number of favorable outcomes on the second roll depends on how many distinct values appeared on the first roll. If all three dice showed the same value, there is only one way to reproduce it. If two values appeared, there are a few permutations. If all three were distinct, there are $3! = 6$ orderings. So we condition on the number of distinct values and use the law of total probability.
Quick Estimate: There are $6^3 = 216$ equally likely outcomes for each roll. The "best case" for matching is all distinct values, which gives 6 favorable outcomes out of 216, roughly a $1/36 \approx 2.8\%$ chance of matching conditional on that case. Since all-distinct is the most common pattern (probability $5/9$), we expect the overall probability to be somewhat above $1\%$ but well below $3\%$. A rough guess: around $2\%$.
Approach: Condition on the number of distinct values $V_i$ ($i = 1, 2, 3$) on the first roll, compute $P(V_i)$ and $P(M \mid V_i)$ where $M$ is the matching event, then apply the law of total probability.
Formal Solution:
Let $V_i$ denote the event that exactly $i$ distinct values appear on the first roll.
*Step 1: Compute $P(V_i)$.*
All three dice the same ($V_1$): there are 6 such outcomes (one per face value), so
$$P(V_1) = \frac{6}{216} = \frac{1}{36}$$
All three dice distinct ($V_3$): the first die has 6 options, the second has 5, the third has 4, giving
$$P(V_3) = \frac{6 \cdot 5 \cdot 4}{216} = \frac{120}{216} = \frac{5}{9}$$
Exactly two distinct values ($V_2$): by complement,
$$P(V_2) = 1 - \frac{1}{36} - \frac{5}{9} = \frac{36 - 1 - 20}{36} = \frac{15}{36} = \frac{5}{12}$$
*Step 2: Compute $P(M \mid V_i)$.*
Each of the 216 equally likely outcomes on the second roll is equally probable.
- If the first roll is $(a, a, a)$: the only matching outcome is $(a, a, a)$, so $P(M \mid V_1) = \frac{1}{216}$.
- If the first roll has pattern $(a, a, b)$ with $a \neq b$: the matching outcomes are the permutations of two $a$'s and one $b$ among three dice. There are $\binom{3}{1} = 3$ such permutations (choose which die shows $b$), so $P(M \mid V_2) = \frac{3}{216} = \frac{1}{72}$.
- If the first roll has pattern $(a, b, c)$ with all distinct: there are $3! = 6$ permutations of three distinct values, so $P(M \mid V_3) = \frac{6}{216} = \frac{1}{36}$.
*Step 3: Law of total probability.*
$$P(M) = P(M \mid V_1)P(V_1) + P(M \mid V_2)P(V_2) + P(M \mid V_3)P(V_3)$$
$$= \frac{1}{216} \cdot \frac{1}{36} + \frac{1}{72} \cdot \frac{5}{12} + \frac{1}{36} \cdot \frac{5}{9}$$
$$= \frac{1}{7776} + \frac{5}{864} + \frac{5}{324}$$
Converting to a common denominator of 7776:
$$= \frac{1}{7776} + \frac{45}{7776} + \frac{120}{7776} = \frac{166}{7776} = \frac{83}{3888}$$
Answer: The probability that the second roll matches the multiset of the first roll is $\dfrac{83}{3888} \approx 2.14\%$.
Intuition
The core idea is that matching a multiset is easier when there are more permutations of that multiset -- three distinct values can be rearranged in 6 ways, while three identical values can only appear one way. So even though all-distinct rolls are harder to match in spirit (more specific values to hit), they benefit from having more favorable orderings. Conditioning on the multiplicity structure is the natural decomposition because it cleanly separates "which values" from "how many ways to arrange them."
This pattern -- conditioning on the type of outcome to handle symmetry-breaking -- shows up constantly in combinatorial probability. Dice, cards, balls-in-bins problems all benefit from partitioning by the "shape" of the outcome (how many repeats, how many distinct elements). Whenever you see a problem where order does not matter but you are counting ordered outcomes, think about grouping by multiplicity pattern first.