# What probability means: counting equally likely outcomes

> Is a total of 7 with two dice three times as likely as a 12, or six times? Settling it leads to sample spaces, events and the rules every probability obeys.

Series: Probability (https://www.monocurl.com/monocurl-essays/probability/)
Canonical: https://www.monocurl.com/monocurl-essays/probability/sample-spaces/
Keywords: what is probability, sample space, event, equally likely outcomes, two dice probability, probability axioms, complement rule, addition rule

[Figure 1, animation: The 36 ways two dice can land, one cell for each pair of faces, with each cell showing its total. Sweeping through the totals from 2 to 12 lights up one diagonal at a time, and a total of 7 has the longest diagonal: 6 cells of the 36.]

Roll two dice and add them up. Everyone who has played a board game knows that a total of 7 comes up more often than a total of 12. The harder question is how much more often, and two reasonable people can count their way to two different answers.

The first person says: a 12 can only happen as a 6 and a 6. A 7 can happen as 1 and 6, as 2 and 5, or as 3 and 4. That is three ways against one, so a 7 is three times as likely as a 12.

The second person says: the two dice are different objects. "A 1 and a 6" is really two things that can happen, because either die can be the one that shows the 1. Counted that way a 7 can happen in six ways and a 12 still in only one, so a 7 is six times as likely.

Both people counted correctly. They disagree about *what to count*, and no amount of arithmetic will decide between them. Deciding takes an idea about what a probability is, and by the end of this lesson we will have that idea in the exact form mathematicians use.

## The same dispute with two coins

Dice have six faces, which makes for a lot of cases. The same disagreement already shows up with the simplest random objects there are, so we settle it there first.

Flip two coins. The first person lists three things that can happen: two heads, two tails, or one of each. Three possibilities, so each should happen a third of the time. The second person says "one of each" is two possibilities, since either coin can be the head, so there are four possibilities and "one of each" should happen half the time.

**Pause and try.** Before looking at the figure below, decide which you believe: does "one of each" happen about a third of the time or about half the time? If you have two coins in reach, flip the pair twenty times and tally the three results.

**Answer.** About half. In twenty real flips expect "one of each" around 10 times, with two heads and two tails around 5 each, give or take a few.

We can ask the coins. The figure flips a simulated pair 400 times and keeps a tally.

[Figure 2, animation: A simulation of 400 flips of a pair of fair coins. The number in each bar counts how often that result has occurred. If the three results were equally likely, all three bars would settle on the dashed line at 1/3. They do not: "one of each" settles near one half.]

In this run of 400 pairs, "one of each" came up 201 times, two tails 103 times and two heads 96 times. The three-way split predicted about 133 for each. The four-way count predicted 200, 100 and 100. The coins side with the second person.

To see why, make the two coins impossible to confuse. Say one is a penny and the other is a dime. Then there is no doubt that "penny heads, dime tails" and "penny tails, dime heads" are different things that can happen: if you photographed the table, the two photographs would differ. Listing what each coin does separately gives a table with four cells.

[Figure 3, image: Every way a penny and a dime can land. Each coin has two faces, so the table has four cells, and "one of each" covers two of them.]

Now ask which list consists of things that are equally likely. For the four cells there is a reason to believe it. The penny is fair, so its two columns each get half of all flips. The dime does not know what the penny did, so within each column its two rows split the flips evenly again. Half of a half is a quarter, for every cell. For the list of three there is no such argument, and the table shows why it fails: "one of each" is two cells, and the other two results are one cell apiece.

So the first person's list was a perfectly good list of what can happen. The error was the unspoken step from "there are three possibilities" to "each has chance one third". Counting possibilities gives probabilities only when the possibilities are equally likely, and whether they are is a fact about the coins, not about the list.

## Thirty-six outcomes for two dice

The dice dispute is the same dispute. Take one blue die and one orange die, so that "blue 1, orange 6" and "blue 6, orange 1" are visibly different. Each die has six faces. By the argument we used for the coins, every pairing of a blue face with an orange face is as likely as every other: the blue die is fair, so each column of the table below gets a sixth of the rolls, and the orange die splits each column into six equal parts. That gives $6 \times 6 = 36$ equally likely pairs.

[Figure 4, steps: The 36 outcomes for a blue die and an orange die. Each cell is one pair of faces and shows the total of that pair. Cells with the same total lie along a diagonal.]

  - step: Blue 5 with orange 6 and blue 6 with orange 5 are different cells. Both total 11.
  - step: A total of 12 needs a 6 on both dice. That is one cell.
  - step: A total of 7 is a full diagonal: six cells.
  - step: Every total is a diagonal. The number of cells climbs from 1 to 6 and falls back to 1.
**Pause and try.** Use the grid to count the cells with a total of 8. Then answer the opening question: how many times as likely is a 7 as a 12?

**Answer.** A total of 8 is the diagonal $(2,6), (3,5), (4,4), (5,3), (6,2)$: five cells, so its probability is $5/36$. A 7 has six cells and a 12 has one, so a 7 is six times as likely as a 12. Note that $(4,4)$ is a single cell. A double is the one kind of pair that has no mirror image, and this is exactly why 12 loses to 7 by a factor of six and not three.

The last remark in that answer explains what the first person got wrong in a way that goes beyond "you forgot the order". Carried through for every total, the first person's method lists unordered pairs of faces, and there are 21 of them: six doubles and fifteen pairs of two different numbers. The six doubles are each one cell of the grid, while each of the fifteen mixed pairs is two cells ($6 + 2 \times 15 = 36$). So the 21 items on that list are not equally likely: a mixed pair such as "a 3 and a 4" has twice the chance of a double such as "a 6 and a 6".

## Asking the dice

An argument that the 36 cells are equally likely is still only an argument, so we check it against rolls as we did for the coins. The figure rolls a simulated pair of dice 3600 times. For each total it marks two predictions: the green mark is the number of cells divided by 36, and the red mark is the first person's count of unordered pairs divided by 21.

[Figure 5, animation: A simulation of 3600 rolls of two dice. Each gray bar is the fraction of rolls so far with that total. Green marks are the predictions from 36 equally likely ordered pairs, red marks the predictions from treating the 21 unordered pairs as equally likely. As the rolls accumulate the bars settle on the green marks.]

Early on, the bars jump around and prove nothing. By a few thousand rolls they have settled, and they have settled on the green marks. In this run a total of 7 came up 596 times in 3600 and a total of 12 came up 101 times. The ordered count predicts 600 and 100. The unordered count predicts about 514 and 171, which is far outside anything the dice produced.

## Sample spaces, events and probability

We have now done the same three things twice, and each has a name.

First we wrote down every way the experiment could come out, at a level of detail fine enough that the items were equally likely: four penny and dime results, 36 blue and orange pairs. Each item is an **outcome**, and the set of all of them is the **sample space**.

Second, we took the thing we cared about ("the total is 7") and identified it with the set of outcomes in which it happens (a diagonal of six cells). Such a set is an **event**.

Third, we divided: six cells out of 36.

**Definition.** Let an experiment have finitely many possible outcomes, exactly one of which occurs each time the experiment is run. The set of all outcomes is the **sample space**, written $\Omega$. An **event** is a subset $A \subseteq \Omega$; we say $A$ *occurs* when the outcome of the experiment is an element of $A$. If the outcomes are **equally likely**, the **probability** of the event $A$ is

$$
P(A) = \frac{|A|}{|\Omega|},
$$

where $|A|$ denotes the number of elements of $A$.

Each clause is something we met. "Exactly one of which occurs" is what makes the list a legitimate sample space: the outcomes must not overlap and must not leave anything out. The three-item coin list passes that test, and so does the four-item list. "Equally likely" is the hypothesis that licenses the formula, and it is the one that separated the two lists. The definition does not tell you whether it holds. That has to come from a physical argument (a symmetric die, coins that do not influence each other) or from data.

For two dice, $\Omega = \{(a, b) : a, b \in \{1, 2, \dots, 6\}\}$, with $|\Omega| = 36$. The event "the total is 7" is $A = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$, and $P(A) = 6/36 = 1/6$.

Because an event is a set, a question about probability becomes a question about a region of the grid. Here is one where the picture prevents a mistake.

**Pause and try.** Two dice are rolled. What is the probability that at least one of them shows a 6? Many people reason: "the chance of a 6 on the blue die is $1/6$, the chance on the orange die is $1/6$, so the answer is $1/6 + 1/6 = 1/3$." Count cells in the grid and decide whether that is right.

**Answer.** The event is the blue-6 column together with the orange-6 row. Each has 6 cells, but they share the corner cell $(6, 6)$, so the region has $6 + 6 - 1 = 11$ cells and the probability is $11/36$, a little less than $1/3 = 12/36$. Adding the two sixths counts the double six twice.

[Figure 6, image: The event "at least one die shows a 6" is the last column together with the last row: 11 cells of the 36. The remaining cells form a 5 by 5 block of 25 outcomes with no 6.]

The picture also shows a quicker route to 11. The cells with *no* 6 form a block with five columns and five rows, 25 cells, and every cell is in the block or in the event but not both. So the event has $36 - 25 = 11$ cells.

## Two rules that follow from counting

Both observations about that figure are general, and they are the two rules used most in practice.

For an event $A$, its **complement** $A^c$ is the set of outcomes not in $A$: the event that $A$ does not occur. The **union** $A \cup B$ is the set of outcomes in $A$ or in $B$ or in both, the event that at least one of the two occurs. The **intersection** $A \cap B$ is the set of outcomes in both, the event that both occur. Two events are **disjoint** when $A \cap B$ is empty, meaning they cannot occur together.

In the equally likely setting both rules are statements about counting. Every outcome lies in exactly one of $A$ and $A^c$, so $|A| + |A^c| = |\Omega|$, and dividing by $|\Omega|$ gives the **complement rule**

$$
P(A^c) = 1 - P(A).
$$

For a union, add $|A|$ and $|B|$. An outcome in exactly one of the two sets has been counted once, and an outcome in both has been counted twice, so subtracting $|A \cap B|$ leaves each outcome of $A \cup B$ counted exactly once. Dividing by $|\Omega|$ gives the **addition rule**

$$
P(A \cup B) = P(A) + P(B) - P(A \cap B),
$$

which for disjoint events reduces to $P(A \cup B) = P(A) + P(B)$. The "at least one 6" computation used both: it is $\frac{6}{36} + \frac{6}{36} - \frac{1}{36}$ by the addition rule and $1 - \frac{25}{36}$ by the complement rule. Here is a second instance, with a larger overlap.

[Figure 7, image: The addition rule on the grid. "Doubles" has 6 cells and "total at least 10" has 6 cells, but the two events share the cells (5,5) and (6,6), shown in red, so their union has 6 + 6 - 2 = 10 cells.]

## When outcomes are not equally likely

The definition so far covers fair coins, fair dice and shuffled cards, and it has an obvious limit: it says nothing when the outcomes are not equally likely. We do not have to look far for an example. Suppose all you record is the total of the two dice. Then the possible results are the eleven numbers $2, 3, \dots, 12$, exactly one of which occurs, so $\{2, 3, \dots, 12\}$ is a legitimate sample space. Its outcomes are far from equally likely: 7 has probability $6/36$ and 12 has $1/36$. For such a space, "count and divide" is the wrong recipe.

The repair is to give each outcome its own weight. For a finite sample space $\Omega$, assign to each outcome $\omega$ a number $p(\omega) \ge 0$, with the weights adding up to 1, and define the probability of an event as the total weight of its outcomes:

$$
P(A) = \sum_{\omega \in A} p(\omega).
$$

Equally likely outcomes are the special case in which every weight is $1/|\Omega|$, and then the sum is $|A|/|\Omega|$ as before. For the totals of two dice the weights are $\frac{1}{36}, \frac{2}{36}, \dots, \frac{6}{36}, \dots, \frac{1}{36}$, which are the bar heights from the grid figure.

In 1933 Andrey Kolmogorov proposed taking three properties that all such assignments share as the definition of probability, and that definition is the one in use today.

**Definition.** Let $\Omega$ be a finite sample space. A **probability measure** on $\Omega$ is a function $P$ that assigns a real number $P(A)$ to every event $A \subseteq \Omega$ and satisfies:

- (Nonnegativity) $P(A) \ge 0$ for every event $A$.
- (Normalization) $P(\Omega) = 1$.
- (Additivity) If $A$ and $B$ are disjoint events, then $P(A \cup B) = P(A) + P(B)$.

The pair $(\Omega, P)$ is called a **probability space**.

The weighted sum above satisfies all three: a sum of nonnegative weights is nonnegative, the weights total 1, and the outcomes of two disjoint events are summed separately with none shared. Conversely, on a finite space any $P$ satisfying the axioms comes from weights, namely $p(\omega) = P(\{\omega\})$, because additivity lets you build any event up one outcome at a time. (For infinite sample spaces axiom 3 is strengthened to cover infinite sequences of disjoint events. We will not need that here.)

The axioms say nothing about where the numbers come from: symmetry, data and judgment are all acceptable sources. They guarantee that, whatever the source, the rules we found by counting still hold.

**Theorem.** Let $P$ be a probability measure on a finite sample space $\Omega$, and let $A$, $B$ be events. Then:

- $P(A^c) = 1 - P(A)$, and in particular $P(\emptyset) = 0$.
- If $A \subseteq B$ then $P(A) \le P(B)$. Consequently $P(A) \le 1$ for every event.
- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

**Proof.** We may use only the three axioms.

- $A$ and $A^c$ are disjoint and their union is $\Omega$. By additivity and normalization, $P(A) + P(A^c) = P(\Omega) = 1$. Taking $A = \Omega$ gives $P(\emptyset) = 1 - 1 = 0$.
- Let $B \setminus A$ denote the outcomes in $B$ but not in $A$. If $A \subseteq B$, then $B$ is the union of the disjoint events $A$ and $B \setminus A$, so $P(B) = P(A) + P(B \setminus A) \ge P(A)$, using nonnegativity for the last step. Since every event is a subset of $\Omega$, $P(A) \le P(\Omega) = 1$.
- The event $A \cup B$ is the union of the disjoint events $A$ and $B \setminus A$, so $P(A \cup B) = P(A) + P(B \setminus A)$. The event $B$ is the union of the disjoint events $A \cap B$ and $B \setminus A$, so $P(B) = P(A \cap B) + P(B \setminus A)$. Solve the second equation for $P(B \setminus A)$ and substitute into the first.

Each step is one of the counting steps from the last section with "number of outcomes" replaced by "probability". That is the sense in which the axioms capture what counting was doing.

## What the number means

One question has been waiting since the coin simulation. We say a total of 7 has probability $1/6$. The next roll will either be a 7 or it will not. What does the number $1/6$ say about the world?

The answer most people use, and the one our simulations relied on, is about repetition: if the experiment is run many times under the same conditions, the fraction of runs in which the event occurs will be close to its probability, and closer the more runs there are.

[Figure 8, animation: The same simulated 3600 rolls, in order. The purple curve is the fraction of rolls so far that totaled 7 and the gold curve the fraction that totaled 12. Both wander at first and then stay close to 1/6 and 1/36.]

That the fractions behave this way is a theorem, the **law of large numbers**, first proved by Jacob Bernoulli, in a book published in 1713, for exactly this kind of repeated experiment. We are citing it here and not proving it; [the lesson on the law of large numbers](@probability/law-of-large-numbers) does. Two cautions belong with it even now. It is a statement about fractions over a long run, and it promises nothing about any single roll or any short streak. And the dice have no memory: the fraction settles because a few thousand new rolls outweigh whatever happened early, with no tendency for later rolls to make up for earlier ones.

**Key idea.** To find a probability by counting, list outcomes that are equally likely, mark the ones where the event occurs, and divide. The list is the sample space, the marked set is the event, and "equally likely" is a claim about the world that has to be justified.

## Worked examples

**Worked example.** **Three coins: the probability of exactly two heads.**

We need outcomes that are equally likely, so we keep track of each coin separately. Writing the three coins in a fixed order, the sample space is

$$
\Omega = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\},
$$

with $2 \times 2 \times 2 = 8$ outcomes, each of probability $1/8$ by the same halving argument as before. The event "exactly two heads" is $\{HHT, HTH, THH\}$: the single tail can be in any of the three positions. So the probability is $3/8$.

*Check.* By the same count, "exactly one head" is also $3/8$, and "three heads" and "no heads" are $1/8$ each. These four events are disjoint and cover $\Omega$, and $\frac{3}{8} + \frac{3}{8} + \frac{1}{8} + \frac{1}{8} = 1$.

**Worked example.** **Two dice: the probability that the larger of the two numbers is 4.**

(If both dice show 4, the larger number is 4.) Listing these outcomes directly is error-prone, so we look for a region whose size is easy. The event "both dice show at most 4" is a $4 \times 4$ block of 16 cells. The event "both dice show at most 3" is a $3 \times 3$ block of 9 cells inside it. The larger number is exactly 4 when the outcome is in the first block and not in the second, which leaves $16 - 9 = 7$ cells, an L-shaped strip. The probability is $7/36$.

*Check.* The same reasoning gives $m^2 - (m-1)^2 = 2m - 1$ cells for "the larger number is $m$": that is 1, 3, 5, 7, 9, 11 cells for $m = 1, \dots, 6$, and these add up to 36, as they must.

## Common mistakes

**Common mistake.** **Counting possibilities that are not equally likely.** "There are 11 possible totals, so each has probability $1/11$." "Either it rains tomorrow or it does not, so the chance is 50%." Both arguments list the possibilities correctly and then divide as if the list were symmetric. The formula $|A|/|\Omega|$ has a hypothesis. Before using it, ask what reason there is to think each item on the list is as likely as every other. If there is none, refine the list until there is (totals into ordered pairs), or accept that the weights are unequal and must be found another way.

**Common mistake.** **Adding probabilities of events that overlap.** The chance of at least one 6 in two rolls is not $\frac{1}{6} + \frac{1}{6}$. Carry the same reasoning to seven rolls and it gives $\frac{7}{6}$, a "probability" greater than 1, which is impossible. Probabilities add only for disjoint events. When events overlap, subtract the overlap, or count the complement instead.

## Stepping back

The calculations here were easy. Both people in the opening counted correctly, and what separated them was a hypothesis that neither had said aloud. Making that hypothesis explicit is what the sample space is for: you are forced to say what the outcomes are, and then to defend the claim that they are equally likely, or else to supply weights.

Once that is done, probability questions about dice, coins and cards become questions about counting regions, and the two rules for regions (complement and addition) carry over unchanged to any probability measure. [The next lesson](@probability/conditional-probability-bayes) asks what happens to these numbers when you learn something partway through, for example that the blue die has already come up 6.

**Check yourself.** Two fair dice are rolled. What is the probability that the total is 4?
Answer: $1/12$ The cells are $(1,3)$, $(2,2)$ and $(3,1)$: three of 36 equally likely outcomes, and $3/36 = 1/12$.

**Recap.** - A sample space $\Omega$ is a list of outcomes, exactly one of which occurs. An event is a subset of $\Omega$.
- If the outcomes are equally likely, $P(A) = |A| / |\Omega|$. Whether they are equally likely must be argued from symmetry or checked against data.
- Two dice have 36 equally likely ordered outcomes. A total of 7 has probability $6/36$ and a total of 12 has $1/36$.
- A probability measure is any assignment satisfying $P(A) \ge 0$, $P(\Omega) = 1$, and $P(A \cup B) = P(A) + P(B)$ for disjoint events.
- From those axioms: $P(A^c) = 1 - P(A)$, and $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
- A probability is a prediction about long-run frequency, and it can be tested by repeating the experiment.

## Problems

Draw the grid for the dice problems. The first two repeat the lesson's moves, the next three combine them, and the last two go a step further.

**Problem 1.** Two fair dice are rolled. Find the probability that the total is 5, and the probability that the total is at least 11.

**Solution.** Use the 36 equally likely ordered pairs. A total of 5 occurs in $(1,4), (2,3), (3,2), (4,1)$, four cells, so the probability is $4/36 = 1/9$. A total of at least 11 means 11 or 12: the cells $(5,6), (6,5), (6,6)$, three in all, so the probability is $3/36 = 1/12$.

**Problem 2.** Three fair coins are flipped. Find the probability of at least one head. Do it twice: by listing, and with the complement rule.

**Solution.** The sample space has 8 equally likely outcomes: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$. *Listing:* every outcome except $TTT$ contains a head, so 7 outcomes, probability $7/8$. *Complement:* "at least one head" is the complement of "no heads", which is the single outcome $TTT$ with probability $1/8$. So the answer is $1 - 1/8 = 7/8$.

**Problem 3.** Two fair dice are rolled. Find the probability that the total is even or greater than 8.

**Solution.** Let $A$ be "total is even" and $B$ be "total is greater than 8". The even totals $2, 4, 6, 8, 10, 12$ have $1 + 3 + 5 + 5 + 3 + 1 = 18$ cells. The totals $9, 10, 11, 12$ have $4 + 3 + 2 + 1 = 10$ cells. The overlap is the totals 10 and 12, with $3 + 1 = 4$ cells. By the addition rule, $|A \cup B| = 18 + 10 - 4 = 24$, so the probability is $24/36 = 2/3$. *Check:* the complement is "odd and at most 8", the totals 3, 5, 7, with $2 + 4 + 6 = 12$ cells, and $1 - 12/36 = 2/3$.

**Problem 4.** A fair die is rolled four times. What is the probability of at least one 6? Describe the sample space you are using.

**Solution.** Take the sample space of all sequences of four faces. There are $6^4 = 1296$ of them, equally likely by the same argument as for two dice: each new roll splits every existing case into six equal parts. Counting the sequences with at least one 6 directly is awkward, because a sequence may contain several. The complement is simple: sequences with no 6 use only five faces in each position, so there are $5^4 = 625$ of them. Hence

$$
P(\text{at least one 6}) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518.
$$

A bet on this event at even odds is slightly favorable, and the gambler the Chevalier de Méré is said to have profited from it in the 1650s.

**Problem 5.** A die is loaded so that the probability of each face is proportional to the number on it. Find the probability of each face, check that these form a probability measure, and find the probability of rolling an even number.

**Solution.** Let $p(k) = ck$ for $k = 1, \dots, 6$. The weights must total 1: $c(1 + 2 + 3 + 4 + 5 + 6) = 21c = 1$, so $c = 1/21$ and $p(k) = k/21$. These are nonnegative and sum to 1, so $P(A) = \sum_{k \in A} p(k)$ is a probability measure, as shown in the lesson. Then

$$
P(\text{even}) = p(2) + p(4) + p(6) = \frac{2 + 4 + 6}{21} = \frac{12}{21} = \frac{4}{7}.
$$

Counting faces would have given $3/6 = 1/2$, which is wrong here because the six faces are not equally likely.

**Problem 6.** Let $P$ be a probability measure on a finite sample space and let $A$, $B$ be events. Using only the axioms and the theorem proved from them, show that $P(A \cup B) \le P(A) + P(B)$, and that $P(A \cap B) \ge P(A) + P(B) - 1$. Then: if $P(A) = 0.7$ and $P(B) = 0.6$, what are the smallest and largest possible values of $P(A \cap B)$?

**Solution.** By the theorem, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, and $P(A \cap B) \ge 0$ by nonnegativity, so $P(A \cup B) \le P(A) + P(B)$. Rearranging the same identity, $P(A \cap B) = P(A) + P(B) - P(A \cup B)$, and $P(A \cup B) \le 1$, so $P(A \cap B) \ge P(A) + P(B) - 1$.

With $P(A) = 0.7$ and $P(B) = 0.6$, the bound gives $P(A \cap B) \ge 0.3$. In the other direction $A \cap B \subseteq B$, so $P(A \cap B) \le P(B) = 0.6$. Both extremes can occur. Take $\Omega = \{1, \dots, 10\}$ with equally likely outcomes and $A = \{1, \dots, 7\}$. With $B = \{5, \dots, 10\}$ the intersection is $\{5, 6, 7\}$, probability $0.3$. With $B = \{1, \dots, 6\}$ the intersection is $B$, probability $0.6$. So the answer is: between $0.3$ and $0.6$ inclusive.

**Problem 7.** Gamblers in Galileo's time noticed that with three dice a total of 10 comes up slightly more often than a total of 9, and were puzzled, because each can be written as a sum of three faces in exactly six ways: $9 = 1{+}2{+}6 = 1{+}3{+}5 = 1{+}4{+}4 = 2{+}2{+}5 = 2{+}3{+}4 = 3{+}3{+}3$ and $10 = 1{+}3{+}6 = 1{+}4{+}5 = 2{+}2{+}6 = 2{+}3{+}5 = 2{+}4{+}4 = 3{+}3{+}4$. Resolve the puzzle and find both probabilities.

**Solution.** The six "ways" are unordered triples, and unordered triples are not equally likely, for the reason unordered pairs were not. The equally likely outcomes are the $6^3 = 216$ ordered triples (one face for each of three distinguishable dice). An unordered triple with three different numbers, such as $1{+}2{+}6$, corresponds to $3 \times 2 \times 1 = 6$ ordered triples. One with exactly two equal numbers, such as $1{+}4{+}4$, corresponds to 3 (choose which die shows the odd one out). One with all three equal corresponds to 1.

For 9: the triples $126, 135, 234$ have distinct numbers (6 each), $144, 225$ have a pair (3 each), and $333$ is a triple (1). Total $6 + 6 + 6 + 3 + 3 + 1 = 25$.

For 10: the triples $136, 145, 235$ have distinct numbers (6 each), and $226, 244, 334$ have a pair (3 each). Total $6 + 6 + 6 + 3 + 3 + 3 = 27$.

So $P(9) = 25/216 \approx 0.116$ and $P(10) = 27/216 = 1/8 = 0.125$. The difference comes entirely from $3{+}3{+}3$, which is a single outcome where the other triples are three or six.