By the end: Distinguish ordered paths, unordered selections, replacement, and overlapping events.
Multiply the available choices
If a choice has three options and each can be followed by four options, there are twelve ordered paths. To convert path counts into probabilities, the paths must be equally likely. Without replacement, the choices change: drawing two distinct cards from five gives five times four ordered paths.
If order does not matter, each unordered pair appears twice among those twenty paths. There are ten pairs. For k items chosen from n, the combination count divides the number of ordered choices by the number of orders of the chosen items. The symbol C(n,k), also written as n above k in parentheses, means this unordered count.
Count the choices that remain at each step
Suppose a lunch order chooses one of two mains and one of three side dishes. An order records both choices. Each main permits all three sides, giving three orders under the first main and three under the second, for six total. Multiplication abbreviates this repeated addition as two times three. It does not establish a probability. If customers favor one main, the six orders can have unequal probabilities even though the count of possible orders remains six.
Now impose a restriction. Main A permits three sides, but main B permits only two because one side is unavailable with it. There are five orders, not six. Drawing branches exposes the difference: count three complete paths under A and two under B, then add. Multiplying the two initial main choices by three would include an impossible path. A product of option counts requires the same number of allowed continuations at that step along every path being counted.
Even equal random choices at each branch need not make complete paths equally likely. Choose the main with probability one half, then choose uniformly among its allowed sides. Each A order receives probability one half times one third, or one sixth. Each B order receives one half times one half, or one quarter. The three A weights sum to one half and the two B weights sum to one half. Dividing one favorable order by five possible orders would give the wrong probability.
Contrast that procedure with putting all five allowed orders on identical tickets and drawing a ticket uniformly. Each order then has weight one fifth. Main A occurs with probability three fifths and main B with probability two fifths because A has more tickets. Both procedures can be described informally as choosing a random lunch, but they define different distributions. For a calculation, specify whether the random choice applies to complete orders or to each successive decision separately.
Branch-uniform and order-uniform choices disagree
Five allowed lunch orders. Main A permits sides 1, 2, and 3; main B permits sides 1 and 2. Branch lengths are schematic. Endpoint weights, not line lengths, encode probabilities. A uniform ticket over complete orders gives each endpoint 1/5 instead.
Order also changes what counts as a distinct path. If two numbered lockers receive different keys selected from four labeled keys, there are four choices for locker one and three remaining choices for locker two. This gives twelve assignments. If instead two keys are selected for an unassigned spare-key pouch, swapping their selection order changes nothing in the final pouch. Each pair appears twice in the twelve ordered paths, so there are six distinct pairs of keys.
Division works there because every final pair has exactly two ordered representations. For the pair A and B these are A then B and B then A, and the same duplication holds for every other pair. If the ordered paths are equally likely, each pair inherits two equal path weights and all pairs are equally likely too. If keys have unequal selection chances, counting six pairs remains correct, but converting a pair count to probability requires its actual combined path weights.
Each unordered key pair has two ordered versions
Pouch pair
Locker 1, locker 2
Reversed assignment
A and B
A, B
B, A
A and C
A, C
C, A
A and D
A, D
D, A
B and C
B, C
C, B
B and D
B, D
D, B
C and D
C, D
D, C
Four distinct keys, labeled A through D, chosen without replacement. Pouch contents disregard order; numbered locker assignments distinguish it. Under uniform ordered selection, each pouch pair receives weight 2/12 = 1/6.
Subtract overlaps once
For two events, adding their probabilities counts their overlap twice. The probability of A or B is the probability of A plus the probability of B minus the probability of both. “Or” includes both unless explicitly stated otherwise.
For two fair dice, first die six has six pairs and second die six has six pairs. Pair 6,6 appears in both lists. At least one six therefore has eleven pairs, not twelve. Alternatively, neither is six has 25 pairs, so the complement has 36 minus 25, also eleven.
Identify the overlap before adding event weights
The addition rule concerns membership in events, not whether the process happened in stages. Consider a uniform choice among twelve numbered tickets. Let A mean the number is even and B mean it is divisible by three. A contains six tickets and B contains four. Tickets six and twelve belong to both. Adding six and four counts each of those two tickets twice. Their union contains eight distinct tickets, so the probability of A or B is 8/12, or two thirds.
Partition the twelve tickets into four disjoint groups to see why subtraction works. Four tickets are even but not divisible by three. Two are divisible by three but not even. Two satisfy both conditions. Four satisfy neither. The union contains the first three groups, with total eight. Adding the original event counts gives four plus two, plus the overlap twice. Subtracting the two overlap tickets once restores exactly one contribution for every ticket belonging to the union.
The word 'or' usually includes outcomes satisfying both conditions. An exclusive-or question instead asks for exactly one condition. Here exactly one means the four even-only tickets plus the two divisible-by-three-only tickets, totaling six. Starting with the sum of the original counts, subtract the overlap twice to obtain six plus four minus four. One subtraction corrects double-counting for the inclusive union. A second subtraction removes the shared tickets entirely when the question excludes them.
Classify every ticket before adding
Twelve uniformly selected tickets. A means even; B means divisible by three. Each circle is one ticket, weight 1/12. Horizontal positions separate membership categories, not numerical intervals. The union has eight tickets; exactly one condition has six.
These are probability identities even when tickets have unequal weights. Suppose ticket twelve has weight 1/2 and every other ticket has weight 1/22. The six even tickets then have total weight 16/22. The four multiples of three have weight 14/22. Their overlap, six and twelve, has weight 12/22. The union has weight 18/22 after subtracting that overlap once. Counting eight favorable labels out of twelve would fail because the labels no longer carry equal weights.
A complement can make a union easier to calculate. Neither even nor divisible by three means the four tickets one, five, seven, and eleven. Under uniform selection their total probability is 4/12, so the union is one minus 4/12. Under the weighted model their total is 4/22, so the union is 18/22. The complement rule is unchanged; only the weights change. It applies because the event and its complement divide the same complete experiment without gaps or overlap.
For an unfamiliar problem, write what the overlap actually contains before using the formula. If two machine alarms can fire together, adding their separate rates may overcount joint alarms. If the protocol forbids joint activation, the overlap is empty and ordinary addition is enough. Neither case establishes independence. The union identity permits any overlap consistent with the model. Independence would be an additional statement that the overlap probability equals the product of the separate probabilities, and requires its own justification.
The union rule preserves weights, not label fractions
Event
Uniform tickets
Heavy ticket 12
A: even
6/12
16/22
B: multiple of three
4/12
14/22
Both
2/12
12/22
A or B
8/12
18/22
Neither
4/12
4/22
Total: union plus neither
1
1
Same twelve tickets, different probability models. In the weighted model ticket 12 has weight 11/22 and every other ticket 1/22. The identity P(A or B) = P(A) + P(B) − P(both) applies in either column.
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.
Method
At least one six
Union with overlap
6 + 6 − 1 = 11 pairs
Complement
36 − 25 = 11 pairs
Probability
11/36 ≈ 30.56%
Use the “At least one six” preset to locate the shared pair.
Compare two ways to count the same dice event
Use the dice grid's 'At least one six' preset to inspect a union. The row with first die six contains six outcomes. The column with second die six also contains six. Their intersection is the single cell for six-six. Counting the row and column separately therefore produces twelve entries but only eleven distinct cells. The selected event has probability 11/36. The phrase 'at least one' includes both dice showing six, so the shared cell must remain in the event once.
The unselected region is a five-by-five square where neither die shows six. It contains 25 cells, so the complement calculation gives 36 minus 25, again eleven favorable cells. Under independence, the probability of neither six is 5/6 times 5/6, which is 25/36. Subtracting this from one gives 11/36. The grid explains both the counted region and the product's assumptions. The complement subtraction is always valid; expressing the neither-six probability as a product uses independence.
A nearby question asks for exactly one six. Remove the six-six cell from the selected event. Five outcomes have first die six and second die below six. Another five have second die six and first die below six. These groups are disjoint, so the probability is 10/36, or 5/18. The difference between 'at least one' and 'exactly one' is only one cell here, but the distinction changes whenever both conditions can occur together in the same experiment.
The corner distinguishes at least one from exactly one
Independent fair dice, one ordered pair per cell. The selected row and column share the six-six corner. Exactly one excludes the cell labeled 2; at least one includes it once. The zero-labeled square is the complement of at least one.
For a contrast, imagine a different device that rolls one fair die and displays its value twice. Each display separately shows six with probability 1/6, just as each die did in the original grid. But both displays are six with probability 1/6 rather than 1/36. At least one six also has probability 1/6, and exactly one six is impossible. The individual six rates cannot distinguish these devices. Their joint behavior, including the overlap, determines the union probability.
Return to 'Matching dice' and compare its six cells with a paper event, 'first die is one.' The latter is an entire row of six cells. Their intersection contains one-one, exactly one cell. Each event has probability 1/6 and their joint probability is 1/36, equal to 1/6 times 1/6. These events are independent in this model even though their definitions both involve the first die. Independence is a property of probabilities, not a ban on sharing words or variables.
To explain that last result without a formula, restrict attention to the first row. Exactly one of its six equally likely cells is a match. The fraction matching remains one sixth, the same as in the whole grid. This comparison prepares the denominator change used in conditional probability. The current widget does not perform that restriction automatically; inspect the row on paper or in the visible grid. Preset changes only choose among the three supplied highlighted events and never alter the underlying dice model.
Equal individual rates do not determine the overlap
Quantity
Independent dice
One die copied
First display six
1/6
1/6
Second display six
1/6
1/6
Both six
1/36
1/6
At least one six
1/6 + 1/6 − 1/36 = 11/36
1/6 + 1/6 − 1/6 = 1/6
Exactly one six
10/36
0
Compare two independent fair dice with one fair die copied to a second display. No inference of independence follows from having two displays or from matching individual probabilities. The inclusive union subtracts the actual overlap in each model.
Make the sampling procedure explicit
A bag contains two red and three blue tokens. Drawing two with replacement gives a red-red probability of 2/5 times 2/5. Without replacement it is 2/5 times 1/4, since the first red has left the bag. “Choose two tokens” is incomplete until replacement and randomization are specified.
A useful audit is to count a small instance in two ways. Ordered paths and unordered groups must give the same probability when their weights are handled correctly. Real sampling may not give equal weights: a convenience survey is not a uniform draw from everyone who could answer.
Track labeled objects before grouping by color
Consider a bag containing three green tokens and one yellow token. Label the greens G1, G2, and G3, even though the question concerns color. Assume every remaining physical token is equally likely to be drawn. Labels let us count complete paths without confusing several green objects with one green outcome. On the first draw, green has probability 3/4 and yellow 1/4. Those two color labels are unequal because they combine different numbers of equally likely physical selections.
Draw twice without replacement. There are four choices first and three second, giving twelve equally likely ordered token paths. Green-green has three choices of first green and two choices of second green, giving six paths and probability 6/12, or one half. Green-yellow has three paths and yellow-green has three paths. Yellow-yellow has none because there is only one yellow token and it cannot be drawn again. These color-event weights sum to twelve twelfths, as a complete enumeration should.
The same green-green probability follows by updating the available choices. The first green chance is 3/4. After a green leaves, two of the three remaining tokens are green, so the second green chance on that branch is 2/3. Multiplying gives 6/12. After a yellow first draw, all three remaining tokens are green, so that branch's second green chance is one. A second-draw probability can depend on the first outcome even when each physical draw is uniformly randomized.
With replacement, return the first token and mix before drawing again. Under an independent uniform-draw model there are sixteen ordered token paths, including repeated identities. Green-green now contains nine paths and has probability 9/16. Green-yellow and yellow-green each have three paths; yellow-yellow has one. The first-draw green probability is unchanged, but the second green probability after green remains 3/4 instead of falling to 2/3. Replacing the object changes the experiment, not merely the counting notation used to describe it.
Replacement adds the identity-repeat paths
Three labeled green tokens and one yellow. Each draw is uniform over the available physical tokens. With replacement, draws are also assumed independent. Cell area counts paths, not probabilities across panels: allowed cells have weight 1/12 on the left and 1/16 on the right.
For the without-replacement experiment, unordered counting gives six equally likely pairs of distinct physical tokens. Three contain two greens, namely G1 with G2, G1 with G3, and G2 with G3. The other three contain yellow with one green. Thus the green-green probability is three out of six, again one half. Each unordered pair combines exactly two equally likely ordered paths. Grouping by color afterward is safe because the physical-token weights have been preserved through both descriptions.
With replacement, unordered identity pairs do not have equal probabilities. G1 with G2 combines two orders, whereas G1 with itself has only one order. There are ten unordered identity pairs, but treating them as ten equally likely outcomes would lose that difference. Six unordered pairs contain only greens, yet their probability is not six tenths. Three repeated-green pairs contribute one path each and three distinct-green pairs contribute two each, totaling nine of sixteen paths. Count representations and their weights before dividing by a total.
Unordered pairs have unequal multiplicities with replacement
Unordered pair type
Number of pairs
Orders per pair
Ordered paths
Probability
Same green twice
3
1
3
3/16
Two different greens
3
2
6
6/16
One green and yellow
3
2
6
6/16
Yellow twice
1
1
1
1/16
Total
10
Varies
16
1
Uniform independent draws with replacement from G1, G2, G3, Y. There are ten unordered identity pairs but sixteen equally likely ordered paths. Green-green combines the first two rows, with nine paths and probability 9/16.
Exercises
Choose a committee
Three people are chosen uniformly from five, without roles. How many committees are possible? Why is 5 × 4 × 3 too large?
Hint
Each committee appears in every ordering of its three people.
Worked answer
Object counted
Count
Ordered role assignments
60
Orders of each committee
6
Unordered committees
10
Decide whether exchanging positions changes the outcome.
There are 60 ordered selections. Each group of three appears in 3 × 2 × 1 = 6 orders. There are 60/6 = 10 committees. If chair, secretary, and treasurer were distinct roles, those 60 assignments would instead be distinct outcomes.
Two red tokens
Find the chance of two reds from two red and three blue tokens, without replacement. Check by unordered counting.
Hint
There are ten pairs of distinct tokens.
Worked answer
Procedure
Red-red probability
Without replacement, sequential
2/5 × 1/4 = 1/10
Without replacement, unordered
1/10
With replacement
4/25
Replacement changes the second draw’s available tokens.
The sequential probability is 2/5 × 1/4 = 1/10. There is one pair containing both red tokens out of C(5,2) = 10 equally likely pairs. With replacement, 4/25 would be a different experiment.
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.