Three dice, rising order

Roll three dice one at a time. What's the probability the values come out in strictly increasing order, like 1-3-5 or 2-4-6?

What do you think?

What is the probability of rolling three dice in strictly increasing order?

Roll some dice

Before any formulas, let's get a feel for what "strictly increasing" looks like. Roll one die at a time and see whether the values come out in rising order:

Roll three dice

Die 1

Die 2

Die 3

Notice how often you get repeats or wrong orderings. The requirement is strict — each die must be higher than the previous one.

Running many trials

Now let's run a batch of trials and see how close the success rate is to the theoretical answer:

Dice Order Simulator

Rolls

All different

—

Increasing

—

Theory

9.26%

After enough rolls, the success rate converges to about 9.3%.

The symmetry insight

When the three dice show distinct values, there are exactly 6 ways those values could have appeared. Only one is strictly increasing:

All orderings of three values

If three dice show distinct values 2, 4, 5, there are exactly 3! = 6 orderings. Only one is strictly increasing:

2·4·5✓

2·5·4

4·2·5

4·5·2

5·2·4

5·4·2

1 out of 6 orderings is increasing → P(increasing | all different) = 1/6

This is the heart of the argument. If you know the three values are all different, each of the $3! = 6$ orderings is equally likely. The probability of the increasing one is $1/6$ .

The conditioning trick

The direct approach counts favorable outcomes among all $6^3 = 216$ . That works, but conditioning is cleaner and generalizes better.

Split the event into two parts using the multiplication rule:

$P(\text{increasing}) = P(\text{all different}) \times P(\text{increasing} \mid \text{all different})$

Part 1: All three values are different

The first die can be anything (6/6). The second must differ from the first (5/6). The third must differ from both (4/6):

$P(\text{all different}) = 1 \times \frac{5}{6} \times \frac{4}{6} = \frac{120}{216} = \frac{5}{9}$

Part 2: Given all different, what's the chance they're increasing?

This is the elegant part. If three values are all different, they have some specific set of values, say $\{a, b, c\}$ with $a < b < c$ . Those three values can appear in $3! = 6$ equally likely orderings (since the dice are rolled independently). Exactly one of those orderings is $a, b, c$ — the increasing one.

$P(\text{increasing} \mid \text{all different}) = \frac{1}{3!} = \frac{1}{6}$

Combining

$P(\text{increasing}) = \frac{5}{9} \times \frac{1}{6} = \frac{5}{54} \approx 9.26\%$

Counting directly

We can verify by listing all $\binom{6}{3} = 20$ ways to choose 3 distinct values from $\{1, 2, 3, 4, 5, 6\}$ . Each such choice gives exactly one increasing sequence:

All 216 Outcomes

Of 216 outcomes, exactly 20 are strictly increasing. That gives P = 20/216 = 5/54 ≈ 9.26%.

Each of the 20 triples maps to exactly one increasing order out of $6^3 = 216$ total outcomes, confirming $P = 20/216 = 5/54$ .

Formal derivation

\Omega = \{(a,b,c) : a,b,c \in \{1,...,6\}\}, \;|\Omega| = 6^3 = 216

The sample space: all ordered triples of die values.

Step 1 of 4

Generalizations

The conditioning approach scales beautifully:

Dice	P(all different)	P(increasing \| all different)	P(increasing)
2	5/6	1/2	5/12 ≈ 41.7%
3	5/9	1/6	5/54 ≈ 9.3%
4	5/18	1/24	5/432 ≈ 1.2%
5	5/54	1/120	1/1296 ≈ 0.08%
6	5/324	1/720	1/46656 ≈ 0.002%

What do you think?

What about non-decreasing order (allowing ties, like 2-2-5)?

What do you think?

What's the probability of rolling 3 dice in strictly DECREASING order?

decimal like 0.09

Competition appearances

AMC/AIME: "N dice rolled. Probability of strictly increasing sequence?" The $\binom{6}{N}/6^N$ formula appears frequently
Programming contests: Counting ordered subsequences in random sequences
Quant interviews: The conditioning trick is a standard test of probabilistic thinking — it generalizes to order statistics of continuous distributions
IMO training: Symmetry arguments for equally likely orderings

P(two dice in strictly increasing order)? (decimal like 0.42)

P(3 dice strictly increasing, using 8-sided dice)? (decimal like 0.08)

Given 3 dice are all different, P(they form the sequence 1,2,3 specifically)? (decimal like 0.05)

Takeaway

Conditioning transforms a counting problem into a symmetry argument. Instead of enumerating the 20 favorable triples directly, we split the problem: first, what's the chance all values differ? Then, given they differ, how likely is the specific ordering we want? The second factor is always $1/n!$ , making this technique effortless to generalize.