An expectation is a probability-weighted average, not necessarily a possible or typical result.
By the end: Compute means, use linearity correctly, and distinguish expected money from decisions about risk.
Weight each outcome by its probability
A payoff of zero with probability 0.9 and 100 with probability 0.1 has expectation ten. No individual play pays ten. Expectation is a property of the model’s full distribution. It can describe a long-run average under suitable repeated-trial assumptions, but it is not a guarantee for a single decision.
An indicator is a variable equal to one when an event occurs and zero otherwise. Its expectation equals that event’s probability. Adding indicators counts events, so expected counts are sums of individual event probabilities. If twenty items each have a 5% fault chance, the expected number faulty is one.
Build a weighted average from possible outcomes
Suppose a delivery takes two hours with probability 0.6 and seven hours with probability 0.4. The expected delivery time is four hours. To find it, first multiply two hours by 0.6, giving 1.2 hours. Then multiply seven hours by 0.4, giving 2.8 hours. Adding the contributions gives four hours. The contributions are weighted amounts, not additional possible delivery times. Under this model, a delivery takes either two hours or seven hours, never four.
Imagine allocating a total probability weight of one across those two outcomes. The shorter delivery receives six tenths of the weight, and the longer delivery receives four tenths. The mean balances the weighted distances on either side. Two hours lies two hours below the mean, contributing 0.6 times two, or 1.2. Seven hours lies three hours above it, contributing 0.4 times three, also 1.2. This balance explains why the mean lies closer to the more probable outcome.
Probability weights and the mean delivery time
The partition width represents probability, not hours. Below it, positions represent hours. The mean at four hours balances weighted distances, 0.6 × 2 = 0.4 × 3, but is not an available delivery time.
For a hypothetical collection of ten deliveries with these same marginal probabilities, the expected counts are six short and four long. Their expected total duration is six times two plus four times seven, or forty delivery-hours. Dividing by ten recovers four hours per delivery. This calculation does not say the collection must contain six short deliveries. Nor does forty delivery-hours mean forty elapsed hours if several deliveries happen simultaneously. The quantity being averaged needs its unit and its definition.
An indicator turns a question about an event into an ordinary number. Define a late-delivery indicator to equal one for a seven-hour delivery and zero for a two-hour delivery. Its expectation is zero times 0.6 plus one times 0.4, which is 0.4. Thus an expected indicator equals the probability of its event. For ten deliveries, adding the ten indicators gives the number late. Adding their expectations gives four, even before specifying how the deliveries depend on one another.
Now distinguish two variables that sound similar. Total delivery time adds the hours attached to every delivery. Number late adds only the zero-or-one indicators. Their expectations are forty hours and four deliveries, respectively. The same probability model supports both answers because the outcomes receive different numerical values. A probability distribution alone does not determine an expectation until we say what numerical quantity each outcome produces. Counting delays and measuring their durations answer different operational questions.
As a paper exploration, hold the two durations fixed and move the long-delivery probability to 0.2. The expected time becomes 0.8 times two plus 0.2 times seven, or three hours. Restore probability 0.4, but lengthen the long delivery to twelve hours. The mean becomes 1.2 plus 4.8, or six hours. The first change alters how often the long case occurs. The second alters its magnitude. Either can change an average, but they describe different mechanisms.
The same delivery outcomes support different averages
Quantity
Short case value
Long case value
Expected total for ten
Duration per delivery
2 hours
7 hours
10 × 4 = 40 delivery-hours
Late indicator
0
1
10 × 0.4 = 4 late deliveries
Long chance changed to 20%
2 hours
7 hours
10 × 3 = 30 delivery-hours
Long duration changed to 12 hours
2 hours
12 hours
10 × 6 = 60 delivery-hours
Baseline probabilities are 60% short and 40% long. Only the named changed assumption differs in each contrast. These are expected totals for ten deliveries, not promised counts. Delivery-hours sum durations, not elapsed time for parallel deliveries.
Linearity does not require independence
The expected sum equals the sum of expectations even for dependent variables, provided the expectations exist. Thus the expected faulty count remains one whether the twenty items fail independently or share a fault that affects the entire batch with probability 5%. Their distributions of total loss are very different.
Products are different: the expectation of a product need not equal the product of expectations. The probability of all items working also depends on dependence. Do not transfer the independence-free rule for sums to every operation involving averages.
Keep the marginal chances separate from the joint outcomes
Consider two jobs, each with a 50% chance of being late. Let the late count add one for each late job. Under independence, the four possible pairs are equally likely. Neither is late in one pair, exactly one is late in two pairs, and both are late in one pair. The expected count is zero times one quarter plus one times one half plus two times one quarter, which equals one. This calculation uses the full distribution of the count.
A different mechanism can give the same marginal probabilities. Suppose the jobs share a vehicle, and a vehicle delay makes both late. Let this happen with probability one half, with neither late otherwise. The expected count is still one because half of two is one. A third mechanism schedules exactly one job late, choosing which job with equal probability. Each individual job is again late half the time, but the total count is now always one. Equal individual chances permit quite different joint outcomes.
Linearity works because every joint outcome contributes its probability weight to each quantity being added. Group those contributions by the first job and you obtain its expectation. Group the remaining contributions by the second job and you obtain the second expectation. Nothing in that regrouping requires equal weights or independence. However, the chance that at least one job is late does require the joint model. It is 75% under independence, 50% with the shared delay, and 100% with exactly one late.
Three joint models with the same individual late chances
Columns give job A and rows give job B, with 0 meaning on time and 1 meaning late. Cell text gives probability mass; equal cell areas are outcome positions, not equal chances. Each job is late with probability one half in every panel.
Products expose the difference. Multiply the two late indicators instead of adding them. The product is one only when both jobs are late, and zero otherwise. Its expectation therefore equals the probability both are late. That probability is one quarter under independence, one half with the shared delay, and zero with exactly one late. Multiplying the two individual expectations gives one quarter in every case. It matches the expected product only in the independent case among these three examples.
The collection process supplies reasons for choosing between these models. Separate jobs do not automatically have independent outcomes if weather or dispatch software affects both. Conversely, a scheduling rule can create dependence that prevents simultaneous delay. Two recorded dates merely establish that observations occurred twice. They do not establish independence, either unconditionally or after fixing a relevant condition. Independence given weather would need to hold within each weather state, not merely across the collection as a whole.
Use the three distributions below to make a capacity decision without changing the arithmetic. A team can handle one late job but needs extra staff for two. The expected late count of one does not reveal the chance that extra staff are needed. Read the probability attached to count two instead. Then consider a team that pays one fixed call-out charge whenever any job is late. Its expected charge depends on the chance of at least one late job, another quantity that linearity alone does not supply.
Adding indicators and multiplying them answer different questions
Joint model
Expected A + B
Expected A × B
Expected A × expected B
Independent
1
1/4
1/4
Shared delay
1
1/2
1/4
Exactly one late
1
0
1/4
A and B here are numerical late indicators. Their sum counts late jobs; their product indicates that both jobs are late. Only the independent model factors the expected product into the product of expectations in this comparison.
Work through the mechanism
Weight each payoff by its probability
The two probabilities sum to one. Ten is an average, not an available payoff.
Payoff
Probability
Contribution to mean
0
0.9
0
100
0.1
10
Total
1
10
The mean is ten even though no single outcome is ten.
Separate an expected total from concentration around it
Return to a rare payout, but use a different scale. A ticket pays forty units with probability one quarter and zero otherwise. Its expected payout is ten units. Two independent tickets have an expected combined payout of twenty. Their actual combined payout is zero, forty, or eighty. The probabilities are nine sixteenths, six sixteenths, and one sixteenth. The mean payout per ticket can therefore be zero, twenty, or forty, although its expectation remains ten. Even averaging two plays does not produce the expected value reliably.
These probabilities come from the branches. Two losses have probability three quarters times three quarters, or nine sixteenths. A win followed by a loss and a loss followed by a win each have probability three sixteenths. Their sum is six sixteenths. Two wins have probability one sixteenth. Weighting the total payouts gives forty times six sixteenths plus eighty times one sixteenth, or twenty. Dividing each total by two changes the average's expectation to ten, without changing which branches can occur.
For many independent tickets with the same payout distribution, the average tends to concentrate around ten. Each new ticket adds variation, but the average divides the total by the growing ticket count. Independence prevents every deviation from moving together. Concentration is a statement about the distribution of possible averages, not a rule that forces later wins to compensate for earlier losses. After ten losses, the next ticket still wins with probability one quarter under the stipulated independent model.
Now imagine that every ticket in a bundle uses one shared draw. Each ticket still has probability one quarter of paying forty. A hundred tickets still have expected total payout one thousand. Yet their average is zero with probability three quarters and forty with probability one quarter, exactly like one ticket. Buying more copies does not narrow this average's distribution. This contrast separates linearity, which gives the expected total in both cases, from assumptions that make an average concentrate near its expectation.
There is also a difference between the average of a transformed quantity and a transformation of the average. Suppose a service pays a penalty of one hundred units whenever at least one of two tickets loses. Under independent tickets, both win with probability one sixteenth, so the penalty applies with probability fifteen sixteenths. Its expectation is 93.75 units. The average payout alone cannot determine that penalty. The penalty rule needs the joint outcomes and a separate weighted calculation.
Before using an expected total as a planning number, specify what event creates an unacceptable shortfall. For the two independent tickets, a total below twenty occurs precisely when both lose, with probability nine sixteenths. A total at least twenty occurs whenever there is a win, with probability seven sixteenths. The expected total of twenty is neither a typical guaranteed receipt nor the boundary dividing equal probability halves. The diagram keeps possible totals visible so that the average cannot replace the underlying distribution.
The possible totals do not include their expected total
Two independent tickets each pay forty units with probability one quarter. Horizontal position is total payout; point height is probability mass. The dashed reference at twenty is the expected total, not another outcome. Average payouts per ticket are half the displayed totals.
The best decision needs consequences
Expected net money is one criterion, not a universal measure of welfare. A rare loss that causes bankruptcy may matter more than its contribution to average money. Utility represents how outcomes matter to a decision maker. Expected utility uses the same weighted-average arithmetic on utilities rather than on money.
State the time horizon, costs, and who bears the loss. Buying insurance can reduce expected money while protecting against an unacceptable downside. That is not a contradiction. The later decision lesson uses explicit losses so that probability estimates and value judgments remain separate.
Apply the weights to the consequences that matter
Suppose a stall costs thirty units to open for a day. Revenue is twenty with probability 0.5, sixty with probability 0.3, and one hundred with probability 0.2. Expected revenue is ten plus eighteen plus twenty, or forty-eight units. Subtracting the fixed cost gives expected net money of eighteen units. Equivalently, the possible net amounts are negative ten, thirty, and seventy. Weighting those amounts gives negative five plus nine plus fourteen, again eighteen. Subtracting a fixed cost commutes with taking an expectation.
A positive expected net amount does not mean a loss is unlikely. Here the stall loses money half the time because the lowest revenue is below its opening cost. Changing the low revenue from twenty to zero would make that loss larger while leaving its probability at one half. Changing its probability would be a different alteration. A business choosing whether to open needs both the net distribution and its available cash, not merely a comparison between expected revenue and cost.
Subtract the opening cost in every possible state
Probability
Revenue
Net money
Weighted net
0.5
20
−10
−5
0.3
60
30
9
0.2
100
70
14
Total
Expected: 48
Expected: 18
−5 + 9 + 14 = 18
The fixed opening cost is thirty money units. Net loss occurs half the time even though expected net money is positive. Weighted contributions sum to an expectation; they are not extra possible outcomes.
Insurance makes the distinction between money and consequences concrete. Imagine initial wealth of one thousand units and a loss of eight hundred with probability 0.1. Without insurance, final wealth is one thousand with probability 0.9 and two hundred with probability 0.1. Expected wealth is nine hundred twenty. Full insurance costs one hundred units and leaves final wealth of nine hundred in either state. The premium exceeds the expected uninsured loss of eighty, so insurance reduces expected wealth by twenty units.
A person can still prefer insurance if low wealth has especially harmful consequences. To make that preference explicit, assign utility one hundred to wealth one thousand, utility ninety-five to wealth nine hundred, and utility zero to wealth two hundred. Utility is a numerical representation of how the person ranks uncertain consequences, not a probability or another currency. Expected uninsured utility is ninety, while insured utility is ninety-five. The preference follows from those stated utility values; it does not follow from probability theory alone.
Do not infer the effect of buying a policy merely because insured people have smaller observed losses. Insured and uninsured people may face different hazards or choose different protections. The calculation above stipulates a contract that fully reimburses the specified loss. In a real application, exclusions, delayed payments, and insurer default would alter its branches. Evidence about intervention effects belongs in the consequence model. Random sampling can help estimate population risks, while random assignment addresses a separate causal comparison between actions.
For a guided contrast, keep the loss probability and utility values fixed but raise the premium to two hundred. Insurance then leaves wealth eight hundred, whose utility has not been specified. Expected money is easy to calculate, but expected utility cannot yet rank the choices. Ask for the missing utility value rather than inventing it. If that value exceeds ninety, insurance wins by expected utility; if it is below ninety, remaining uninsured wins. Probabilities supply the weights, while consequences and preferences supply what receives those weights.
Money and utility can rank the same choices differently
Choice
No loss, 90%
Loss, 10%
Expected value
Uninsured wealth
1,000
200
920 money units
Insured wealth
900
900
900 money units
Uninsured utility
100
0
90 utility units
Insured utility
95
95
95 utility units
Insurance with premium 200
800 wealth
800 wealth
Utility at 800 is needed
Initial wealth is one thousand, loss is eight hundred with probability 0.1, and full insurance costs one hundred. Utility values are stipulated preferences. The insurance consequence is a contract assumption, not a causal effect inferred from observed policyholders.
Exercises
Correlated defects
Twenty parts each have a 5% chance of failure. Compare the expected faulty count if failures are independent with a batch defect that makes all twenty fail together 5% of the time.
Hint
Expectation is linear even when the outcomes share a cause.
Worked answer
Model
Expected count
Chance all twenty fail
Independent
1
0.05²⁰
Shared batch fault
1
0.05
The extreme-tail probabilities differ despite identical marginal rates and means.
Both expected counts are 20 × 0.05 = 1. In the shared-batch model, count is zero with probability 0.95 and twenty with probability 0.05. Independence gives a spread of possible counts instead. Equal expected count does not imply equal catastrophe risk.
Include the entry price
A game pays 30 with probability 0.2 and zero otherwise. Entry costs seven. What is expected net money, and does a negative value make winning impossible?
Hint
Subtract the cost from either the mean payout or every outcome.
Worked answer
Outcome
Net
Weighted net
Win
23
4.6
Lose
−7
−5.6
Expected net
−1
This calculation evaluates money, not an unstated utility function.
Expected payout is 0.2 × 30 = 6, so expected net is −1. A win still gives net 23 with probability 0.2. A loss gives −7 with probability 0.8. The expectation describes their weighted average, not each play.
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.