Define the experiment before assigning probabilities
An outcome, an event, and a probability model answer different questions.
By the end: Specify a sample space, distinguish uncertainty from observed frequency, and check whether outcomes really are equally likely.
This course assumes arithmetic and percentages, not prior statistics or calculus. It builds models before interpreting evidence or choosing actions. Numerical examples are invented unless identified otherwise. Probabilities depend on assumptions; no model here certifies a real detector, medical decision, or research finding.
What can happen?
A probability model starts with an experiment and its possible outcomes. The sample space is the set of those outcomes. An event is a subset: all outcomes that satisfy a condition. Rolling two dice produces ordered pairs such as first die 2, second die 5. The event “sum seven” contains several pairs.
Probabilities are nonnegative weights that sum to one over mutually exclusive, exhaustive outcomes. An impossible event has probability zero; the full sample space has probability one. The complement of A means all outcomes outside A, so its probability is one minus the probability of A. These rules constrain a model; they do not tell you whether it describes the world.
Specify what one observation records
Suppose a delivery service wants a probability for a late parcel. Before assigning a number, define the experiment. Select one parcel from a stated week's deliveries, record its promised arrival date, and compare its actual arrival with that date. An outcome might record the parcel's identity, route, and delay. The event 'late' contains every outcome with a positive delay. Changing the week, selecting only express deliveries, or defining late as more than one day overdue changes the question.
Outcomes must be mutually exclusive at the level used for calculation. One selected parcel cannot simultaneously have a delay of zero days and two days. They must also be exhaustive, so every permitted result belongs somewhere. Categories 'early' and 'late' leave out arrivals exactly on time. Categories 'on time or earlier' and 'late' avoid that omission. Lost parcels need an explicit rule too. Excluding them silently would change a reliability calculation into a calculation only about parcels that arrived.
For a constructed population of 100 parcels, suppose 15 arrive early, 65 exactly on time, and 20 late, with no losses. Select each parcel with equal probability. The early event has weight 15/100, the exactly-on-time event 65/100, and the late event 20/100. Their sum is 100/100. Combining early and exactly on time gives 80/100 because those categories do not overlap. That combined event is the complement of late within this specified experiment.
A complete partition of 100 parcels
Constructed fixed population, with no losses. Width represents parcel count at 6.2 SVG units per parcel. Under uniform parcel selection, the same widths represent probabilities.
Now suppose two of the late parcels belong to the same recipient. Choosing a recipient uniformly and then choosing one of their parcels is a different procedure from choosing a parcel uniformly. A recipient with one parcel gives that parcel all of the recipient's selection weight. A recipient with many parcels divides their weight across several parcels. The list of possible parcels can stay unchanged while its probability weights change. A sample space alone therefore does not fully specify a probability model.
An event can be defined before or after a record is collected, but its meaning must stay fixed when comparing probabilities. 'Late relative to the original promise' differs from 'late relative to the latest revised promise.' Updating promises after delays can make the second event much rarer without improving delivery. In a probability calculation, such a change is a change in what counts as success. It is not evidence that the same event has become less likely.
To audit a reported 80% delivery success rate, reconstruct the denominator and the success rule. Does the denominator contain dispatched parcels, completed deliveries, or customers? Do cancellations disappear? For the constructed parcel model, the denominator is all 100 parcels and success means arrival no later than originally promised. This identifies the population probability exactly for uniform selection from that list. Applying 80% to a future parcel adds an assumption that future conditions resemble those represented in the list.
A different selection unit changes the weight
Selection procedure
Weight of A's parcel
Weight of each B parcel
Late probability
Uniform among four parcels
1/4
1/4
1/4 = 25%
Uniform recipient, then parcel
1/2
1/6
1/2 = 50%
A separate four-parcel example. Recipient A has one late parcel. Recipient B has three on-time parcels. Choosing a recipient first and then one of their parcels does not choose parcels uniformly.
Equal labels do not imply equal chances
Two independent fair six-sided dice give 36 equally likely ordered pairs. Their sums run from 2 through 12, but these eleven labels are not equally likely. Only one pair gives 2; six give 7. Compressing outcomes into labels must preserve their total weights.
Fairness and independence are assumptions about the dice and rolling procedure, not consequences of calling an object a die. A biased die can use the same six labels with different weights. A model of tomorrow’s rain needs evidence about weather and how the event is defined; counting “rain” and “no rain” does not establish a half chance.
Preserve weights when combining outcomes
Equally likely elementary outcomes make counting convenient, but equal likelihood is an extra property. Consider a spinner with three labeled sectors occupying one half, one third, and one sixth of its circle. Assume the pointer's final direction is uniform around the circle and the boundary rule does not affect the calculation. The labels have probabilities 1/2, 1/3, and 1/6. There are three labels, but assigning each one third would discard the stated geometry and the uniform-direction assumption.
The spinner can instead be described using six equal angular sectors. Label three sectors A, two B, and one C. Each small sector has weight 1/6, so A has 3/6, B has 2/6, and C has 1/6. This finer description gives the same label probabilities as the unequal-sector description. Refining or compressing a sample space is legitimate when every event keeps its total weight. The number of labels need not remain the same between descriptions.
Six equal angles give three unequal label weights
Uniform final direction is assumed. Each small sector spans 60 degrees and has probability 1/6. A combines three sectors, B two, and C one. Circle area and angle both preserve these proportions; text positions do not encode probabilities.
A useful contrast uses two tosses of a biased coin. Assume each toss has head probability 3/4 and that the tosses are independent. Head then head has probability 3/4 times 3/4, which is 9/16. Head then tail and tail then head each have probability 3/16. Tail then tail has probability 1/16. The four sequence labels are exhaustive, but they are not equally likely. Counting two mixed sequences out of four would incorrectly assign mixed results probability one half.
Adding the two mixed weights gives 6/16, or 37.5%. Combining the matching sequences gives 9/16 plus 1/16, or 62.5%. These two events are complements, and their weights add to one. The multiplication used to obtain the sequence weights depended on independence. The later addition did not. Once disjoint outcome weights are available, addition combines them regardless of whether an independent mechanism produced those weights. This separates an accounting rule from a claim about how observations arise.
Imagine instead a device that tosses once and displays that same result twice. Keep the single-toss head probability 3/4. Only head-head and tail-tail can occur, with weights 3/4 and 1/4. Mixed displays now have probability zero. Each display separately still shows heads 75% of the time under the model, but the joint outcomes differ. Collecting or showing a value twice does not justify multiplying its probability by itself. The procedure determines whether the second observation supplies fresh variation.
Before using favorable outcomes divided by possible outcomes, ask whether every counted outcome has the same weight. If that claim is doubtful, write the separate weights and add the ones inside the event. For the spinner, angular size supplies the weights under a uniform-direction model. For the biased coin, stipulated head probability and independence supply them. Neither model follows merely from naming the outcomes. Evidence about a real spinner or coin would be needed to judge whether these mathematical assumptions fit it.
Independent tosses and duplicated displays differ
Pair
Two independent tosses
One toss displayed twice
Head, head
9/16
3/4 = 12/16
Head, tail
3/16
0
Tail, head
3/16
0
Tail, tail
1/16
1/4 = 4/16
Total
16/16
16/16
Head probability is 3/4 for each individual observation in both procedures. Independence permits multiplying the two toss weights. A guaranteed copy instead repeats the first observation, so mixed pairs cannot occur.
Work through the mechanism
Find the equally likely pairs
Selected: 6/36 ≈ 16.67%.
Every cell is one ordered pair with probability 1/36 under independent fair dice. The row is the first die, the column the second. Changing the event replaces the selection immediately; it does not reroll the dice.
Event
Ordered pairs
Probability
Sum 2
1
1/36
Sum 7
6
6/36 = 1/6
Any sum
36
1
Two independent fair dice. Sums are groups of outcomes, not equally weighted outcomes.
Read the dice grid as a complete distribution
In the dice grid, the row specifies the first die and the column specifies the second. An individual cell is an ordered pair, even though its printed number is the sum. Row two, column five and row five, column two therefore represent different outcomes. Under independent fair dice each cell has probability 1/36. The grid is a full list of alternatives for one experiment, not a record of 36 rolls or a promise that each cell appears once in 36 rolls.
First select 'Sum seven' and locate its diagonal. Moving one row downward increases the first die by one, so staying on sum seven requires moving one column left. There are six such pairs, giving 6/36. Compare sum six on paper. Its pairs run from first die one with second die five through first die five with second die one. There are five pairs, giving 5/36. The extra pair for seven explains the probability difference without treating sums as equal labels.
Grouping every cell by its sum produces counts 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums two through twelve. The counts add to 36. This increasing-then-decreasing pattern comes from the limits on each die. Near the middle, several ways to increase one die and decrease the other remain available. Near an endpoint, almost all such alternatives would require a face below one or above six, so those pairs are unavailable.
Sum probabilities count the pairs behind each label
Independent fair dice. Each dot is the probability mass at one integer sum; vertical position is the number of pairs divided by 36. Stems are guides, not density areas. Counts above the dots total 36.
Now select 'Matching dice.' The six selected cells sit on a different diagonal, where row and column agree. Matching and sum seven have the same probability, 6/36, but contain entirely different pairs. They cannot happen together because doubling an integer gives an even sum, never seven. Equal event probabilities do not mean identical events. They also do not imply independence. Here learning that the dice match rules out sum seven, although sum seven had positive probability before that information.
For a further paper comparison, consider the event 'both dice at most three.' It occupies the upper-left three-by-three square, so it has nine cells and probability 9/36, or 25%. Its complement contains 27 cells and means at least one die exceeds three. That complement does not mean both dice exceed three. The latter event is only the lower-right nine-cell square. The remaining eighteen cells have one die in each range, and belong to the complement too.
Use the unchanged grid to check the interpretation of each calculation. Selecting a preset highlights a set without changing any cell's weight. There is no need to generate rolls to learn the exact probability under this model, because every pair is already listed. A record of actual rolls would answer a different question about observed frequencies and possible model fit. If the dice were biased or coupled, the cell locations could remain, but counting highlighted cells alone would no longer recover event probabilities.
A complement includes every outcome outside the event
Independent fair dice, one cell per ordered pair and weight 1/36 per cell. The complement of both dice at most three includes both mixed-range rectangles as well as the lower-right square.
A model is not a promise about a short run
In repeated independent trials with stable probability, observed proportions tend to concentrate near that probability as sample size grows. They need not equal it in a short run. Five heads do not force the next fair toss to be tails. Past imbalance is not a debt that future tosses must repay.
Probability can describe repeatable variation or uncertainty about an unknown state. A statement such as “there is a 20% chance this component is faulty” needs a reference population or an explicit evidential model. Once a component has a fixed state, uncertainty about that state can still be meaningful. Always ask what information the probability conditions on.
Distinguish a probability from a realized fraction
A stable probability describes a model before the outcome is known. An observed fraction describes a particular completed set of trials. Suppose six independent tosses each have head probability one half. Each toss doubles the number of sequences, giving 64 equally likely sequences. Exactly three heads occur in twenty sequences, so an exactly half-heads result has probability 20/64, or 31.25%. Even with a fair coin and a valid independent model, a six-toss record is more likely to be unbalanced than exactly balanced.
Across those sequences, head counts zero through six have multiplicities 1, 6, 15, 20, 15, 6, and 1. Each count's probability is its multiplicity divided by 64. To find the average head count, multiply each count by its multiplicity. The resulting contributions are 0, 6, 30, 60, 60, 30, and 6. Their sum is 192, which divided by 64 is three. This expected count summarizes the distribution; it does not select the outcome of a particular six-toss experiment.
Six fair tosses need not give three heads
Head count
Number of sequences
Probability
Count × multiplicity
0
1
1/64
0
1
6
6/64
6
2
15
15/64
30
3
20
20/64
60
4
15
15/64
60
5
6
6/64
30
6
1
1/64
6
Total
64
1
192
All 64 independent fair-toss sequences have weight 1/64. Multiplicity groups the sequences by head count. The expected count is 192/64 = 3, while the probability of realizing exactly three heads is only 20/64.
If the first five tosses are heads, independence still gives head probability one half on the sixth. Among the two possible continuations of that particular prefix, one is heads and one tails, with equal weights. Neither continuation restores a half-heads fraction for the six-toss record. Longer runs can approach one half without undoing earlier results because later observations enlarge the denominator. A shrinking proportional imbalance does not require the absolute difference between head and tail counts to return to zero.
For example, a record with 55 heads in 100 tosses has five more heads than its expected count of fifty. Appending 100 tosses containing fifty heads produces 105 heads in 200, a fraction of 52.5%. The excess over the new expected count is still five. The proportion moved closer to one half because the total grew. This is one possible continuation, not a prediction that the next hundred tosses must divide evenly or reduce the discrepancy.
Uncertainty can also concern a fixed but hidden state rather than a future randomizing device. Suppose a sealed drawer contains one document, and exactly three of ten equally plausible candidate documents are signed. A 30% probability of a signed document describes uncertainty relative to that candidate model. The document is not partly signed, nor does it alternate between states. Opening the drawer resolves this uncertainty. Repeatedly reading the same signature would not create independent draws from the ten candidates.
Evidence can change a probability about that fixed state, but the probability rules alone do not say how persuasive the evidence is. A handwriting report requires a model of what reports occur for signed and unsigned documents, plus the starting candidate weights. Later lessons develop that update. Even a well-supported probability does not by itself choose an action. Paying to inspect the document depends on consequences and costs as well as uncertainty. Keep the event definition, evidence model, and decision consequences as separate inputs.
An unchanged excess can become a smaller fraction
Record
Heads
Expected heads under fairness
Excess heads
Observed fraction
First 100 tosses
55
50
5
55%
Next 100 tosses alone
50
50
0
50%
Combined 200 tosses
105
100
5
52.5%
One possible constructed continuation, not a requirement for future tosses. The second block contains fifty heads in one hundred tosses. The accumulated excess remains five as the denominator doubles.
Exercises
Count a complement
A forecast assigns a 0.3 probability to rain tomorrow. What is the probability of no rain, and does this imply exactly three rainy days in the next ten?
Hint
A complement exhausts the same experiment. Repetition needs additional assumptions.
Worked answer
Same day
Weight
Rain
0.3
No rain
0.7
Total
1
The complement concerns this day and this definition of rain.
No rain has probability 0.7. The forecast alone says nothing about a ten-day joint distribution. Even ten independent days with probability 0.3 would have expected count three, not guaranteed count three.
Reject an equal-label argument
A machine either fails or works. Someone assigns each a probability of one half because there are two outcomes. What is missing?
Hint
The number of labels is not a measurement of their weights.
Worked answer
Possible model
Failure
Works
Rare failure
0.01
0.99
Frequent failure
0.50
0.50
Both rows are valid distributions; evidence must distinguish their applicability.
We need a failure model or relevant evidence, including duration, operating conditions, and what counts as failure. Two exhaustive labels only imply that their probabilities sum to one. Both 0.01/0.99 and 0.5/0.5 obey that rule.
Sources and further reading
The explanations and exercises are complete without these optional links. The references distinguish probability identities from inference procedures and reporting guidance. The examples are constructed teaching models, not empirical estimates or professional decision advice.