Equally Likely Outcomes: A Modeling Pitfall

Why Counting Can Give the Wrong Probability

Counting is a useful probability method, but it depends on a modeling condition: the outcomes being counted must be equally likely. A common error is to list several possible results, assume each result has the same chance, and divide the number of favorable results by the total. That calculation may look reasonable while ignoring that some listed results represent many more ways for the experiment to occur than others.

The key distinction is between elementary outcomes, which describe the experiment at a sufficiently detailed level, and broader categories formed by combining those outcomes. Categories can have different probabilities because they may contain different numbers of elementary outcomes—or because the elementary outcomes themselves have different probabilities.

Definition: A finite sample space is a set $\Omega=\{\omega_1,\ldots,\omega_N\}$ of possible elementary outcomes. The outcomes are equally likely when each has probability $1/N$. An event $A$ is a subset of $\Omega$. If the elementary outcomes are equally likely, then $P(A)=|A|/N$. If they are not equally likely, calculate $P(A)$ by adding the probabilities assigned to the outcomes in $A$.

The counting formula is justified for every event when the elementary outcomes in the sample space are equally likely. In a nonuniform model, a count-based answer can still match the true probability for a particular event by coincidence. That coincidence does not make counting divided by the total a generally valid method for the model.

Categories Can Have Unequal Multiplicities

Suppose a fair experiment has $N$ equally likely elementary outcomes, and a reported result groups those outcomes into categories. A category containing $m$ elementary outcomes has probability $m/N$. Thus, the reported categories have equal probabilities only when they contain equal numbers of elementary outcomes.

This gives a reliable repair when the experiment is uniform at the elementary level: expand each category into the elementary outcomes it represents, count those outcomes, and divide by the total. If the elementary outcomes are not equally likely, multiplicities alone are not enough; use the actual probability assigned to each outcome.

Worked Example: A Sum of Two Fair Dice

Two fair six-sided dice are rolled, and the reported result is their sum. Find the probability that the sum is at least 10. Explain why treating the possible sums as equally likely is not valid.

Method. Use the 36 ordered pairs of die faces as the elementary sample space. Each ordered pair has probability $1/36$, because the dice are fair and the face on one die can pair with any face on the other die.

The possible reported sums are 2 through 12, or 11 categories. The incorrect calculation treats those 11 sums as equally likely:

\frac{\text{number of sums at least }10}{\text{number of possible sums}} =\frac{3}{11}.

This is wrong because, for example, the sum 7 can be made in six ordered ways, while the sum 12 can be made in only one. Count the favorable elementary outcomes instead. A sum of 10 occurs as $(4,6),(5,5),(6,4)$, giving 3 outcomes. A sum of 11 occurs as $(5,6),(6,5)$, giving 2 outcomes. A sum of 12 occurs as $(6,6)$, giving 1 outcome. Therefore the number of favorable ordered pairs is $3+2+1=6$, and

P(\text{sum}\geq 10)=\frac{6}{36}=\frac{1}{6}.

Check. The six listed pairs are all distinct, and $6/36=1/6$. The method avoids the common mistake of giving each possible sum the same probability merely because there are 11 different sums.

Repairing a Coarse Sample Space

A sample space may be too coarse for a counting argument. It is not necessarily wrong to describe an experiment using categories such as colors, sums, or types of outcomes. The problem occurs when a calculation assumes those categories are equally likely without checking. Refining the space means listing the more detailed elementary outcomes that generate the categories.

For an experiment that selects one object uniformly from a collection, the individual objects are often the appropriate elementary outcomes. A category such as “red” is then an event containing all the red objects. This approach makes multiplicities visible and helps avoid counting categories as if each were a single equally likely object.

Worked Example: Colored Tokens With a Number Property

A container holds six distinct tokens: three red tokens numbered 1, 2, and 3; two blue tokens numbered 2 and 4; and one gold token numbered 1. One token is selected uniformly at random. Find the probability that it is red or has an even number. A report lists only the color-and-parity categories that occur; explain why counting those categories would be misleading.

Method. Each of the six individual tokens has probability $1/6$. The observed color-and-parity categories are red odd, red even, blue even, and gold odd. Their multiplicities are respectively 2, 1, 2, and 1, for a total of 6 tokens.

The event “red or even” includes the two red-odd tokens, the one red-even token, and the two blue-even tokens. The gold-odd token is not included. Thus the favorable token count is $2+1+2=5$, so

P(\text{red or even})=\frac{5}{6}.

If the four observed color-and-parity categories were mistakenly treated as equally likely, the first three categories would be counted as favorable, giving $3/4$. But the categories have different multiplicities: red odd and blue even each represent two tokens, whereas red even and gold odd each represent one. They are not equally likely. Weighting each included category by its multiplicity gives $(2+1+2)/6=5/6$, agreeing with the direct token count.

Check. There are exactly five favorable tokens—red 1, red 2, red 3, blue 2, and blue 4—out of six. The method avoids treating reported categories as equally likely when they represent different numbers of objects.

When the Elementary Outcomes Are Not Equally Likely

Refining a sample space does not automatically make outcomes equally likely. If the experiment assigns different probabilities to its elementary outcomes, the correct event probability is the sum of those assigned probabilities. A count may happen to produce the same value for one event, but probabilities—not outcome counts—are the general rule.

Probability rule for a finite model: If the elementary outcomes are $\omega_1,\ldots,\omega_N$, with assigned probabilities $p_1,\ldots,p_N$, then $p_i\geq 0$, $\sum_{i=1}^N p_i=1$, and for any event $A$, $$ P(A)=\sum_{\{i:\,\omega_i\in A\}}p_i. $$ When every $p_i=1/N$, this sum reduces to $|A|/N$.

Worked Example: A Biased Four-Outcome Device

A hypothetical device produces one of four results, $A,B,C,D$, with probabilities $0.10, 0.20, 0.30, 0.40$, respectively. Find the probabilities of the events $\{A,D\}$ and $\{D\}$. Compare each with the count divided by four, and state what the comparison teaches.

Method. First verify the model: $0.10+0.20+0.30+0.40=1.00$, and all four probabilities are nonnegative. Since the elementary outcomes have unequal probabilities, add the assigned probabilities for each event.

For $\{A,D\}$, the assigned probabilities give

P(\{A,D\})=0.10+0.40=0.50.

There are two outcomes in this event out of four, so the count ratio is also $2/4=0.50$. This equality is a coincidence for this event; it does not show that the four outcomes are equally likely.

For $\{D\}$, the assigned probability is $P(\{D\})=0.40$, while the count ratio is $1/4=0.25$. The count ratio is wrong for this event. As a further check, the complementary outcomes $A,B,C$ have total probability $0.10+0.20+0.30=0.60$, and $0.60+0.40=1.00$.

Check. The first event demonstrates that a count-based answer can match by coincidence in a nonuniform model; the second shows why that match cannot justify using the method generally. The common mistake is to treat one successful numerical coincidence as evidence that the outcomes are equally likely.

Why the Multiplicity Formula Holds

The multiplicity calculation follows directly from finite additivity. Suppose the underlying sample space has $N$ equally likely elementary outcomes, and a reported category $C$ contains $m$ of them. The category is the union of those $m$ distinct elementary outcomes. Each has probability $1/N$; because the outcomes are disjoint, their probabilities add:

P(C)=\underbrace{\frac{1}{N}+\cdots+\frac{1}{N}}_{m\text{ terms}} =\frac{m}{N}.

This justification uses both ingredients: the elementary outcomes are equally likely, and the category consists of $m$ such outcomes. If either ingredient is missing, the multiplicity formula need not apply. In a nonuniform model, the corresponding calculation is the sum of the probabilities of the elementary outcomes in the category.

Describe the actual experiment.
Identify what is selected, rolled, generated, or observed before deciding which outcomes to count.

Choose the elementary outcomes.
Use a level of detail at which the possible results of the experiment are represented without merging outcomes that may behave differently.

Check likelihoods.
Determine whether the elementary outcomes are equally likely. Do not infer equal likelihood merely because the outcomes are listed in a symmetric-looking set.

Calculate the event probability.
For equally likely elementary outcomes, use favorable count divided by total count. Otherwise, add the assigned probabilities of the favorable outcomes.

Common Exam Traps

Counting reported values instead of elementary outcomes. The possible sums on two dice are not equally likely simply because each sum is one listed value. Expand each sum into the ordered die-face pairs that produce it.
Assuming a uniform model from a list. Writing down four possible outcomes does not establish that each has probability $1/4$. The experiment or its specified probability model must justify equal likelihood.
Ignoring multiplicity. If several objects share a category, that category may represent more probability than a category containing only one object. Count the underlying equally likely objects, not just the labels.
Overstating the counting rule. In a nonuniform model, count divided by total is not generally justified. It may nevertheless equal the correct probability for a particular event by coincidence, as in the event $\{A,D\}$ above. Verify the model and use the assigned probabilities.
Assuming finer detail fixes unequal likelihood. A detailed list helps expose the structure, but if the elementary outcomes still have different probabilities, add their probabilities rather than their counts.

Key takeaway: Counting favorable outcomes divided by the total is generally justified for every event when the elementary outcomes are equally likely. Otherwise, use the probabilities assigned by the model. When categories conceal different numbers of equally likely outcomes, refine the sample space and account for their multiplicities.

Check Your Understanding

For each problem, identify the relevant elementary outcomes and avoid assuming that reported categories are equally likely.

A fair coin is tossed three times. A report records only the number of heads, from 0 through 3. Find the probability of exactly two heads, and compare it with the result from treating the four reported counts as equally likely. (Answer: $3/8$; treating the four counts as equally likely would give $1/4$.)
A bag contains eight distinct, equally likely tokens: five green and three orange. One is selected, and a separate report records only its color. Find the probability of green and the error in assigning equal probability to the two colors. (Answer: $5/8$, not $1/2$.)
A six-outcome device has probabilities $0.05,0.10,0.15,0.20,0.20,0.30$ for outcomes 1 through 6. Find the probability of an outcome at most 3 and compare it with the count ratio. (Answer: $0.30$, whereas the count ratio is $3/6=0.50$.)
An integer is selected uniformly from 1 through 10. A summary reports three categories: low (1–3), middle (4–8), and high (9–10). Find the probability of low or high, and compare it with treating the three categories as equally likely. (Answer: $5/10=1/2$, not $2/3$.)
A device has four elementary outcomes $W,X,Y,Z$ with probabilities $0.15,0.25,0.25,0.35$. Find the probability of $\{W,Z\}$ and of $\{X,Y\}$; explain why both can equal their count ratios without validating equal likelihood for all events. (Answer: $0.50$ for each event; each count ratio is $2/4=0.50$, but the outcomes have unequal probabilities.)

What You'll Learn