From an Experiment Description to a Probability Law
A probability calculation begins before any arithmetic: we must decide what the possible outcomes are and what assumptions describe how the experiment produces them. Saying that a die is “fair,” that a ball is “randomly selected,” or that a device chooses equally among its settings is not just background language. Each phrase supplies a modeling assumption that can determine the point probabilities.
The important discipline is to make the assumptions explicit. If a set of outcomes is symmetric and the model treats the outcomes symmetrically, equal point probabilities follow. If objects are selected uniformly, categories receive probabilities according to how many objects belong to each category. If a mechanism first chooses among options and then produces a reported result, the reported probabilities depend on how the mechanism maps its options to results.
The space \(\Omega\) should describe outcomes at a level that supports the assumptions. For example, when a box contains individually distinguishable balls, the physical balls may be the elementary outcomes of a selection. “Red” and “blue” are then categories grouping those elementary outcomes. Treating the categories as the elementary outcomes instead can hide the effect of their different numbers of objects.
Symmetry as a Justification for Equal Probabilities
A statement such as “all outcomes are equally likely” needs justification. One precise route is a symmetry of the experiment: a transformation rearranges the outcomes without changing the probability law. If the symmetry can carry any outcome to any other, the point probabilities must all be equal.
Proof. Fix any two outcomes \(\omega,\eta\). By transitivity, there is a probability-preserving transformation carrying \(\omega\) to \(\eta\), so \(P(\{\omega\})=P(\{\eta\})\). Thus all \(m=|\Omega|\) point probabilities have a common value \(c\). Since they sum to one, \(mc=1\), hence \(c=1/m\). \(\square\)
The theorem makes the assumption doing the work visible: the probability law must be unchanged by the relevant transformations. A collection of outcomes that looks symmetric in its labels is not enough if the physical mechanism favors some outcomes. Conversely, a symmetry may apply only within groups of outcomes, in which case it justifies equality within each group, not across the whole space.
Worked Example: A Die Under a Fairness Assumption
Model one roll of a six-sided die with \(\Omega=\{1,2,3,4,5,6\}\). Assume the rolling procedure and die are symmetric under relabeling the faces, so that the probability law is unchanged when any face is exchanged with another. These transformations act transitively on the six outcomes.
By the symmetry result, the point probabilities are equal. Since their sum is one, \(P(\{i\})=1/6\) for each \(i\in\{1,\ldots,6\}\).
For the event \(A=\{2,4,6\}\), the event probability is the sum of its three point probabilities: \(P(A)=1/6+1/6+1/6=3/6=1/2\).
The value \(1/2\) depends on the assumed fair, symmetric model. The event has three labels, but its probability is not \(3/6\) merely because it has three labels; that calculation is justified here because the six point probabilities are equal.
Uniform Selection Applies to Objects, Not Automatically to Categories
“Select one at random” is incomplete unless we know what is being selected uniformly. Uniform selection from a finite collection means that each object in that collection has the same probability. It does not generally mean that each category represented in the collection has the same probability.
Proof. Each object has probability \(1/N\), by uniformity and the fact that the \(N\) point probabilities sum to one. The event \(C\) consists of \(k\) distinct objects, so additivity gives \(P(C)=k(1/N)=k/N\). The same argument applies to any union of categories by counting all objects in that union. \(\square\)
Worked Example: Selecting a Ball Uniformly
A container holds four individually distinguishable balls: two red, \(R_1,R_2\), and two blue, \(B_1,B_2\). One ball is selected uniformly. Take the four physical balls as the outcomes, each with probability \(1/4\).
The event “red” contains \(R_1\) and \(R_2\), so \(P(\text{red})=P(\{R_1\})+P(\{R_2\})=1/4+1/4=1/2\). Likewise, \(P(\text{blue})=1/2\). The equal category probabilities follow here because the two categories contain the same number of balls.
Now suppose the container instead holds three red balls and one blue ball, still with uniform selection of a physical ball. There are four equally likely ball outcomes, but \(P(\text{red})=3/4\) and \(P(\text{blue})=1/4\). Uniformity of selection has not changed; the number of objects in each category has.
If we had named only “red” and “blue” as outcomes and assigned each probability \(1/2\), we would have imposed a different assumption: uniformity over colors. That is not the same as uniformity over the four physical balls.
Building Probabilities From a Mechanism
Sometimes an experiment has a simple underlying randomization and a separate rule for reporting its result. To find the point probabilities of the reported outcomes, first assign probabilities to the underlying possibilities, then combine the probabilities that lead to the same report. This is especially useful when different underlying outcomes are grouped together.
Proof. The report is \(t\) exactly when the underlying outcome belongs to the set \(\{s\in S:r(s)=t\}\). The sets of underlying outcomes for distinct reports are disjoint. By additivity, the probability of this event is the sum of the point probabilities of all underlying outcomes in that set, as claimed. \(\square\)
Worked Example: A Mechanism That Produces Unequal Report Probabilities
A device selects one of four tickets, \(HH, HT, TH, TT\), with each ticket having probability \(1/4\). It reports “success” if the selected ticket contains at least one \(H\), and “failure” otherwise.
The underlying outcomes are the four tickets. The success report comes from \(HH,HT,TH\), so \(P(\text{success})=1/4+1/4+1/4=3/4\). The failure report comes only from \(TT\), so \(P(\text{failure})=1/4\).
The mechanism chooses uniformly among the four tickets, but it does not choose uniformly among the two reports. The success report corresponds to three tickets and the failure report to one. The report probabilities depend both on the uniform-ticket assumption and on the stated reporting rule.
A multi-stage mechanism can be handled in the same careful way. For example, if an experiment first selects one of several options and then produces an outcome conditional on that selection, the probability of a complete path is the probability of its first step multiplied by the conditional probability of its next step. Probabilities for a reported result are then added across all paths that produce it. The assumptions must specify both the selection probabilities and the conditional behavior at each stage.
Worked Example: Two Draws Without Replacement
A container has \(R_1,R_2,B_1,B_2\). Two balls are drawn in order, uniformly at each draw from the balls still present. For any particular ordered pair of distinct balls, the probability is \((1/4)(1/3)=1/12\). There are \(4\cdot3=12\) such ordered pairs, whose probabilities sum to one.
If the reported outcome is the unordered pair of balls, each possible pair can occur in two orders. For example, \(\{R_1,B_1\}\) occurs as \(R_1,B_1\) or \(B_1,R_1\), giving probability \(1/12+1/12=1/6\). The same reasoning applies to each of the six unordered pairs, so each has probability \(1/6\).
If the report records only the colors, the outcome “one red and one blue” corresponds to four unordered pairs: \(\{R_1,B_1\},\{R_1,B_2\},\{R_2,B_1\},\{R_2,B_2\}\). Its probability is \(4(1/6)=2/3\). “Two red” corresponds only to \(\{R_1,R_2\}\), giving probability \(1/6\), and “two blue” also has probability \(1/6\). Thus the three color reports are not equally likely, even though the six unordered pairs are.
The calculation uses the assumption that each remaining physical ball has equal chance at each draw. If the drawing device favors particular balls, neither the ordered-pair probabilities nor these color probabilities are guaranteed.
Which Conclusions Depend on Which Assumptions?
A useful probability model separates the experiment description into three parts: the elementary outcomes, the assumptions assigning probabilities to those outcomes, and the rule grouping them into events or reported results. This separation makes it possible to say precisely what would change if an assumption were altered.
| Assumption | Conclusion it supports | What it does not imply by itself |
|---|---|---|
| A law-preserving symmetry can carry any elementary outcome to any other | All elementary outcomes have equal point probability | That the physical experiment actually has this symmetry |
| Each physical object is selected with equal probability | A category’s probability is its share of the objects | That categories have equal probability |
| Underlying options have specified probabilities and a reporting rule is fixed | Reported probabilities are sums over the options leading to each report | That distinct reports are equally likely |
| A staged mechanism specifies each selection and conditional step | Probabilities of complete paths and their grouped reports can be calculated | That a different mechanism has the same path probabilities |
These are modeling statements, not automatic properties of the words “random,” “fair,” or “symmetric.” For a physical experiment, a model idealizes the mechanism. A fair-die calculation, for example, is conditional on treating the die and rolling procedure as sufficiently symmetric for the intended analysis. The mathematical conclusion follows from the stated assumption; whether the assumption is a good description of a particular die is a separate question.
Common Mistake: Equal Labels Do Not Mean Equal Chances
A frequent error is to list the possible categories and divide equally among them. For instance, with three red balls and one blue ball, one might reason: “There are two possible colors, so each has probability \(1/2\).” This silently changes the experiment from choosing one of four balls uniformly to choosing one of two colors uniformly.
The correction is to identify the elementary outcomes and the selection mechanism first. If each of the four physical balls is equally likely, red has three elementary outcomes and probability \(3/4\), while blue has one and probability \(1/4\). Equal probability for categories would require an additional mechanism that selects the colors equally; it does not follow from the number of category names.
Check Your Understanding
For each question, state the assumption that supports your answer.
- A symmetric model has eight outcomes, and its probability law is unchanged by transformations that can carry any outcome to any other. What is each point probability?
- A box contains five green objects and three yellow objects. One object is selected uniformly. Find the probabilities of green and yellow.
- A uniformly selected ticket is one of \(A,B,C,D,E\). The report is “group 1” for \(A,B,C\), and “group 2” for \(D,E\). Find each report probability.
- Why does selecting uniformly from physical objects not generally make the represented categories equally likely?
- In the two-draw example, identify the assumption used to assign probability \(1/12\) to a particular ordered pair.