Why Counting Errors Change Probabilities
A counting-based probability can look like a simple fraction: count the outcomes of interest, then divide by the number of possible outcomes. The fraction is reliable only if the outcomes being counted match the random process. In particular, the sample space must be complete, the outcomes used in the calculation must be equally likely, and the numerator and denominator must use the same counting convention.
Many mistakes arise because a description groups several elementary outcomes into one category. Other mistakes arise because a calculation counts arrangements in one place and selections in another. Even when the outcomes are listed correctly, adding event counts without checking for overlap can count some outcomes twice.
The phrase equally likely applies to the elementary outcomes used in the calculation, not automatically to every category someone might name. For example, the possible numbers of heads in two coin tosses are 0, 1, and 2, but those three categories do not have equal probabilities. Each category combines a different number of elementary results.
Worked Example: Categories Need Not Be Equally Likely
Worked Example: Categories Need Not Be Equally Likely
Suppose a fair coin is tossed twice. Assume the tosses are independent and that each toss has probability \(1/2\) of heads and \(1/2\) of tails. What is the probability of getting exactly one head?
Represent the outcomes in order, first toss followed by second toss. The four elementary outcomes are \(HH, HT, TH,\) and \(TT\). Under the stated model, each has probability \(1/4\). Exactly one head occurs for \(HT\) or \(TH\), so
A tempting but incorrect approach is to list the possible numbers of heads as 0, 1, and 2, then assign probability \(1/3\) to exactly one head. That treats the three categories as equally likely. They are not: 0 heads comes from \(TT\), 1 head comes from \(HT\) and \(TH\), and 2 heads comes from \(HH\). The probabilities of the categories are therefore \(1/4, 1/2,\) and \(1/4\), respectively.
The distinction is between counting the equally likely elementary outcomes and counting broader categories. A category containing more elementary outcomes has a greater probability in this model.
Ordered and Unordered Outcomes Must Match
An ordered outcome records the positions or sequence of selected objects. For instance, \(AB\) and \(BA\) are different ordered outcomes. An unordered outcome records only which objects were selected, so the selections \(\{A,B\}\) and \(\{B,A\}\) are the same outcome.
Either convention can work for a probability calculation, provided it matches the random process and is used consistently in both the total count and the favorable count. If every unordered selection is equally likely, converting every selection into all of its possible orders multiplies both counts by the same factor. Counting unordered selections in the numerator but ordered outcomes in the denominator does not.
Worked Example: Counting a Two-Person Committee
A committee of two is selected uniformly at random from five people: Ana, Bao, Cam, Dev, and Eli. Find the probability that the committee includes Ana or Bao, or both. Assume each of the \(\binom{5}{2}\) possible committees is equally likely.
Using unordered committees, the total number is
There are three committees with neither Ana nor Bao, since both members must then be chosen from Cam, Dev, and Eli. Thus the favorable count is \(10-3=7\), and
The same probability can be found with ordered pairs. There are \(5\cdot4=20\) ordered pairs of distinct people. Each of the seven favorable committees has two orders, so there are \(7\cdot2=14\) favorable ordered pairs. The ordered calculation gives \(14/20=0.700\), as it should.
A common error is to use the seven favorable committees as the numerator but the 20 ordered pairs as the denominator, producing \(7/20\). That fraction mixes two kinds of outcomes: each committee represents two ordered pairs in the denominator but only one count in the numerator. Use either \(7/10\) for unordered committees or \(14/20\) for ordered pairs.
Overlapping Events and Double-Counting
Two events overlap if at least one outcome belongs to both. When counting outcomes in “\(A\) or \(B\),” an outcome in the overlap should appear only once in the union. Simply adding the counts for \(A\) and \(B\) counts each shared outcome twice.
The correction follows directly from counting. The outcomes in \(A\) but not \(B\) are counted once in \(|A|\); those in \(B\) but not \(A\) are counted once in \(|B|\). Outcomes in both sets are counted once in each of those two counts, so they have been counted twice. Subtracting \(|A\cap B|\) leaves every outcome in the union counted exactly once:
Worked Example: Correcting for an Overlap on a Die
A fair six-sided die is rolled once. Let \(A\) be the event that the result is even, and let \(B\) be the event that the result is greater than 3. Find the probability that the result is even or greater than 3.
The sample space is \(\{1,2,3,4,5,6\}\), with all six outcomes equally likely. The events are \(A=\{2,4,6\}\) and \(B=\{4,5,6\}\). Their overlap is \(A\cap B=\{4,6\}\). Applying the union formula,
The favorable outcomes are indeed \(\{2,4,5,6\}\), four of the six die results. Adding \(3/6+3/6\) without subtracting the overlap would give \(6/6\), incorrectly counting results 4 and 6 twice. The word “or” does not imply that the events cannot happen together.
Make the Sample Space Complete
A sample space must include every possible elementary outcome allowed by the rules of the random process. Leaving out an outcome changes the total count and may also change the favorable count. One useful check is to describe the process precisely before listing outcomes: Are repeats allowed? Does order matter? Are there restrictions on which results can occur?
A list can be internally consistent and still be wrong if it leaves out a valid result. For small problems, listing the possibilities is often the clearest check. For larger problems, counting methods can verify that a list or table has the expected number of outcomes.
Worked Example: Repeated Codes Must Be Included
A two-digit code is formed by independently choosing each digit from 1, 2, and 3. Repeats are allowed, and each of the nine codes is equally likely. Find the probability that the code contains at least one 3.
Because order matters and repeats are allowed, the complete sample space is
The codes containing at least one 3 are \(13,23,31,32,\) and \(33\). Therefore,
If the code \(33\) is accidentally omitted, a calculation based on the remaining list gives \(4/8=1/2\). That is a probability for a different set of possibilities, not for the stated code-generating process. The rule allowing repeated digits specifically makes \(33\) possible, so it must be included.
A Practical Error-Checking Routine
Before accepting a counting-based probability, check each part of the setup. These checks are especially useful when a problem involves multiple stages, selections, or events described with “or.”
Record the rules that determine what can happen, including whether order matters and whether an object can be selected more than once.
Choose outcomes fine enough to represent the process correctly. Do not assume that grouped categories are equally likely just because they are easy to name.
Verify that no allowed outcome is missing and that the outcomes counted in a simple favorable-over-total fraction are equally likely.
Count both total and favorable outcomes as ordered outcomes or both as unordered selections.
When combining events, identify outcomes that belong to more than one event so they are not counted more than once.
Common Mistakes and Why They Fail
- Treating named categories as equally likely. A category may combine several equally likely elementary outcomes. Count the elementary outcomes or add their probabilities instead of assigning equal probability to each category.
- Mixing ordered and unordered counts. A numerator of selections cannot be paired with a denominator of arrangements unless the difference is accounted for. Use the same unit of counting on both sides of the fraction.
- Adding overlapping event counts without a correction. The outcomes in the intersection then appear twice. Subtract the overlap once when calculating a union.
- Forgetting outcomes permitted by the rules. If repeats, particular labels, or certain sequences are allowed, include them. A shortened list describes a different sample space.
- Assuming a favorable-over-total ratio always works. That ratio requires equally likely elementary outcomes. If outcomes have different probabilities, a raw count does not represent their probability contributions equally.
Check Your Understanding
For each question, identify the counting issue before deciding what a correct calculation would require.
- Two fair coins are tossed. Why is it incorrect to treat 0, 1, and 2 heads as three equally likely outcomes?
- A problem counts 12 favorable unordered pairs out of 60 ordered pairs. What should be checked before using \(12/60\) as the probability?
- On a fair die, let \(A=\{1,2,3\}\) and \(B=\{3,4\}\). Which outcome is in both events, and how does it affect the count for \(A\) or \(B\)?
- A code uses two digits from 0, 1, and 2, with repetition allowed. Explain why a sample space that omits 00 is incomplete.
- When is favorable outcomes divided by total outcomes a valid probability calculation?