Why Equally Likely Outcomes Give a Probability Ratio

Why a Count Can Give a Probability

Suppose a fair six-sided die is rolled once. There are six possible face results, and three of them are even. It is tempting to say that the chance of an even result is three out of six. But a count becomes a probability only when the possible outcomes have the right structure: the outcomes must be mutually exclusive, must cover every possibility, and must be equally likely.

The earlier material on sample spaces treats an experiment’s possible outcomes as a set, and an event as a subset of that set. Here, the key question is how the probabilities of the individual outcomes relate to their counts. When all outcomes are equally likely, counting them is enough. When they are not, the count alone can mislead.

Definition: Let a finite sample space \(\Omega\) contain \(N\) elementary outcomes, where \(N\) is a positive integer. An event \(A\) is a subset of \(\Omega\). If every elementary outcome has the same probability, then the outcomes are equally likely. The elementary outcomes in \(A\) are called the favorable outcomes for event \(A\).

“Favorable” does not mean good or desirable; it means that the outcome belongs to the event being considered. For example, if \(A\) is “the die shows an even number,” the favorable outcomes are the even face results. The probability ratio rule compares the number of favorable elementary outcomes with the total number of elementary outcomes.

Theorem: If a finite sample space \(\Omega\) has \(N\) equally likely elementary outcomes and event \(A\) contains \(k\) of them, then \(P(A)=k/N\). Equivalently, \(P(A)=\text{number of outcomes in }A \,/\, \text{total number of outcomes in }\Omega\).

Why the Ratio Works

The rule follows from two basic requirements of a probability model. First, the probabilities of all outcomes in the sample space add to one, because some outcome in that space must occur. Second, probabilities of mutually exclusive outcomes add when we consider the chance that any one of them occurs. Distinct elementary outcomes in a sample space are mutually exclusive: on a single run of the experiment, exactly one of them occurs.

Give each outcome a common probability.
There are \(N\) equally likely outcomes, so write \(q\) for the probability of each one.

Use the total probability.
All \(N\) outcomes together have probability one. Adding their equal probabilities gives \(Nq=1\), so \(q=1/N\).

Add the probabilities in the event.
If \(A\) contains \(k\) outcomes, its probability is the sum of the probabilities of those \(k\) mutually exclusive outcomes, or \(kq=k/N\).

The same argument can be written as a short derivation. It makes clear which assumption does the work: every elementary outcome has the same probability.

\begin{aligned} 1 &= P(\Omega)=\underbrace{q+q+\cdots+q}_{N\ \text{outcomes}}=Nq,\\ q &= \frac{1}{N},\\ P(A) &= \underbrace{q+q+\cdots+q}_{k\ \text{outcomes in }A} = kq=\frac{k}{N}. \end{aligned}

The numerator and denominator must count outcomes from the same sample space. The denominator counts every elementary outcome, not only those that seem relevant, and the numerator counts the outcomes belonging to the event. If \(A\) is empty, \(k=0\) and the rule gives probability zero. If \(A=\Omega\), then \(k=N\) and the rule gives probability one.

Worked Example: An Even Result on a Fair Die

A standard fair six-sided die is rolled once. Find the probability of an even face result.

Check the model. The sample space is \(\Omega=\{1,2,3,4,5,6\}\). We assume the die is fair, so each of its six faces is equally likely to appear. The outcomes are mutually exclusive and exhaustive: exactly one face appears on the roll.

The event is \(A=\{2,4,6\}\). It contains \(k=3\) favorable outcomes, while the sample space contains \(N=6\) outcomes. Therefore,

P(A)=\frac{k}{N}=\frac{3}{6}=0.5.

The probability of an even result is \(0.5\), or 50%. This is a probability for the stated fair-die model; it does not follow just from the fact that three of the six face labels are even.

Worked Example: Winning a Fair Raffle

Five different people each hold one ticket in a raffle. One ticket is selected at random, and the person holding it wins. Find the probability that either Lina or Omar wins.

Check the model. Use the ticket holders as the elementary outcomes: \(\Omega=\{\text{Lina},\text{Omar},\text{Priya},\text{Sam},\text{Tess}\}\). The raffle is assumed to select each of the five tickets with the same chance, and exactly one ticket is selected. Thus, there are five mutually exclusive, exhaustive, equally likely outcomes.

The event \(A\) consists of Lina winning or Omar winning, so it contains two outcomes. The total number of outcomes is five. Substituting these counts gives

P(A)=\frac{\text{number of outcomes in }A} {\text{total number of outcomes}} =\frac{2}{5}=0.4.

The probability that Lina or Omar wins is \(0.4\), or 40%, under the stated fair-selection model. The result is not based on there being two names in the event alone: it also depends on there being five equally likely ticket outcomes in the sample space.

Why Equal Likelihood Is Essential

If outcomes have different probabilities, knowing how many belong to an event is not enough to determine the event’s probability. Two events can each contain one outcome but have different probabilities, or one event can contain more outcomes and still have a smaller probability. In such cases, the probability of an event is found by adding the probabilities of its constituent outcomes; the ratio of counts does not generally apply.

An important subtlety is that the way outcomes are defined can affect whether they are equally likely. Outcomes should be detailed enough to distinguish the elementary results that have equal probabilities. Grouping several elementary outcomes under one label can create categories with different probabilities.

Worked Example: Why the Sum of Two Dice Is Not an Equally Likely Outcome

Assume two fair dice are rolled, with each ordered pair of face results equally likely. What is the probability that their sum is 7? Why would it be wrong to count the possible sums as if they were equally likely?

Choose elementary outcomes. An elementary outcome records the result on the first die and the result on the second die, in that order. For example, \((2,5)\) means that the first die shows 2 and the second shows 5. There are \(6\times6=36\) ordered pairs. Under the stated model, these 36 pairs are mutually exclusive, exhaustive, and equally likely.

The event that the sum is 7 contains the six ordered pairs \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), and \((6,1)\). Thus,

P(\text{sum is }7)=\frac{6}{36}=\frac{1}{6}\approx 0.1667.

Now consider using only the sum as the outcome. The possible sums are 2 through 12, giving 11 sum labels. These labels are not equally likely. For instance, the sum 2 comes only from \((1,1)\), whereas the sum 7 comes from six different ordered pairs. Counting 7 as one of the 11 possible sums would give \(1/11\), which is not the probability:

\frac{1}{11}\approx 0.09091 \qquad\text{but}\qquad P(\text{sum is }7)=\frac{1}{6}\approx 0.1667.

The count ratio works with the ordered pairs because they are equally likely; it does not work with the sum labels because those labels have unequal probabilities. The lesson is not simply to count possible descriptions, but to count equally likely elementary outcomes.

Using the Rule Carefully

Before applying a favorable-to-total ratio, identify the sample space and check the model. In particular, ask whether its outcomes are elementary, mutually exclusive, exhaustive, and equally likely. A finite list of possible labels does not by itself establish equal likelihood. A fair die or a fair raffle supplies a reason to use the equal-likelihood assumption; a list of outcomes alone does not.

A probability can also be confused with a proportion in an observed data set. If 12 of 20 recorded rolls were even, then 12/20 is the observed proportion of even results in those 20 rolls. It is not, just from that calculation, proof that the model probability is 12/20. The ratio rule in this tutorial is a statement about a finite probability model whose elementary outcomes are equally likely.

Key takeaway: Use \(P(A)=k/N\) only when the sample space has \(N\) equally likely elementary outcomes and \(A\) contains \(k\) of them. Equal likelihood makes each elementary outcome worth \(1/N\); without it, counts alone do not determine probabilities.

Common Mistake

Wrong reasoning: “There are 11 possible sums when two dice are rolled, and only one of them is 7, so the probability of a sum of 7 is \(1/11\).”

Correction: The 11 sums are not equally likely. The elementary outcomes should be the 36 ordered pairs of die results, which are equally likely under the stated model. Six of those pairs sum to 7, so the probability is \(6/36=1/6\). A count ratio is valid only after equal likelihood has been established for the outcomes being counted.

Check Your Understanding

For each question, identify the relevant outcomes and explain whether a count ratio is justified.

A fair spinner has eight equal sectors numbered 1 through 8. What is the probability of landing on a number greater than 5?
A sample space has 12 equally likely outcomes, and an event contains 3 of them. Find the event’s probability and state which counts are the numerator and denominator.
Why must the favorable outcomes and total outcomes be counted from the same sample space?
In the two-dice example, why are the 11 possible sums not an equally likely sample space?
A box contains four red balls and one blue ball, and one ball is selected at random. Is it valid to count the two colors as equally likely outcomes? Explain what elementary outcomes would make the count ratio work.