Probability All Dice Show the Same Value
You roll $k$ fair dice. Die $i$ has $n_i \geq 1$ sides showing the values
- Find the probability that all $k$ dice show the same value, in terms of $n_1, \ldots, n_k$.
- Compute the answer for $k = 4$ dice with side counts given by $n_i = 8 - 2|i - 2.5|$ for $i = 1, 2, 3, 4$.
Hints
- For all dice to show the same value $v$, every die must be able to show $v$. What is the maximum possible shared value? How many shared values are there?
- The number of favorable outcomes equals the number of shared values: $\min(n_1, \ldots, n_k)$. The total outcomes equal the product $\prod n_i$.
- For the specific case, compute $n_i = 8 - 2|i - 2.5|$ for $i = 1, 2, 3, 4$ to get $n_1 = n_4 = 5$ and $n_2 = n_3 = 7$.
Worked Solution
How to Think About It: For all dice to show the same face value, there must exist some value $v$ that every die can show -- meaning $v \leq \min(n_1, \ldots, n_k)$. So first figure out how many common values are possible (that's $\min(n_1, \ldots, n_k)$), then count the total outcomes. Each "all match" event corresponds to picking one of those common values and having every die land on it.
Quick Estimate: With heterogeneous dice, the most restrictive die dominates. If the smallest die has $m$ sides, there are at most $m$ values they can all share. The total outcomes is the product of all side counts. The probability is roughly $m / \prod n_i$, which gets tiny fast as $k$ grows.
Formal Solution:
Part 1 (General formula):
The total number of equally likely outcomes is $\prod_{i=1}^{k} n_i$ (by the multiplication rule, since dice are independent).
For all dice to show value $v$, we need
Each gives exactly one outcome (all dice show $v$), so: $P(\text{all match}) = \frac{\min(n_1, \ldots, n_k)}{\prod_{i=1}^{k} n_i}$
Part 2 (Specific values):
For $k = 4$ and $n_i = 8 - 2|i - 2.5|$: - $i = 1$: $n_1 = 8 - 2|{-1.5}| = 8 - 3 = 5$ - $i = 2$: $n_2 = 8 - 2|{-0.5}| = 8 - 1 = 7$ - $i = 3$: $n_3 = 8 - 2|{0.5}| = 8 - 1 = 7$ - $i = 4$: $n_4 = 8 - 2|{1.5}| = 8 - 3 = 5$
So $\min(n_1, n_2, n_3, n_4) = 5$ and $\prod n_i = 5 \cdot 7 \cdot 7 \cdot 5 = 1225$.
$P(\text{all match}) = \frac{5}{1225} = \frac{1}{245}$
Answer: $\dfrac{\min(n_1, \ldots, n_k)}{\prod_{i=1}^{k} n_i}$. For the specific case, $\dfrac{1}{245}$.
Intuition
The formula $\min(n_i) / \prod n_i$ is intuitive once you see why: the minimum dice count constrains the shared value range, while the product captures how many total configurations exist. As $k$ grows or the dice become more heterogeneous, the probability drops fast because the product in the denominator grows multiplicatively while the numerator stays bounded by the most restrictive die.
This is a clean example of how constraints ("all must match") naturally decompose into counting the feasible set (values all dice share) divided by the total space. In practice, similar logic appears when assessing joint event probabilities across instruments with different ranges or states -- the feasibility of the joint event is governed by the most constrained component.