The Gold Standard of Weather Forecasting | Combining Rain-Occurrence Probabilities Across Time Windows

Imagine a fixed observation point and four successive, nonoverlapping one-hour intervals. Define E1 as measurable rainfall meeting a stated hourly criterion during hour one. Define E2, E3 and E4 in exactly the same way for the other hours. We will choose the criterion for the exercise rather than assume that every weather provider uses the same one.

A statement such as P(E1) = 0.25 concerns whether that event occurs. It does not say that rain lasts fifteen minutes, reaches one quarter of the surrounding area, or delivers one quarter of a millimetre. Those are different quantities with different observation rules.

The National Weather Service’s explanation of probability of precipitation attaches its example to a point, an amount threshold and a valid period. That is the institutional definition being illustrated there; it is not a universal interpretation of every app’s icon. For a real product, read its own definition.

Before calculating, ask a learner to finish this sentence: “An outcome counts as yes if…” If the learner cannot finish it without looking at a percentage, the event needs more work.

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For our exercise, let U mean that at least one of E1, E2, E3 or E4 occurs. Mathematicians call this the union of the four events. A day with rain in all four hours counts once towards U. A day with rain in just one hour also counts once.

This counting rule is the first important representation. We are counting qualifying outcomes, not adding the number of wet hours. If one fictional day has the pattern wet, wet, dry, dry, it contributes one to the number of days with any qualifying hour and two to the total number of qualifying hours.

Write a separate column for each quantity. That small separation prevents a later sum from quietly changing its meaning. The total number of wet-hour marks can exceed the number of days, while the number of days satisfying U cannot.

Our question is P(U), conditional on the forecast information available at one specified issue time. Keeping that common information fixed matters just as much as keeping the location fixed. We will return to updated forecasts after the basic counting is visible.

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Suppose an hourly event means at least 1 mm in that hour. Four hours can each receive 0.3 mm. None of the four hourly events occurs, yet the four-hour total is 1.2 mm. Therefore “at least one hour reaches 1 mm” and “the four-hour total reaches 1 mm” are different events.

With nonnegative rainfall and the same numerical threshold, any qualifying hour makes the total qualify. The reverse implication fails. The union of hourly threshold exceedances is contained in the total-threshold event, but it need not equal it.

This distinction becomes easy to miss when a display places hourly percentages beside a daily percentage. A longer product may use an accumulated amount, a different threshold or a different spatial definition. The formula for a union cannot repair a mismatch in event definitions.

For a practical evidence check, write both event descriptions above the calculation. Underline the threshold and circle the interval. If the words are different, explain whether one event contains the other, whether they overlap, or whether the supplied information leaves the relationship unknown. Do that before substituting numbers.

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Let F0 denote everything used for a forecast issued at the start of the exercise. Our four values are P(E1 given F0), through P(E4 given F0). They describe four future events using one common information set.

Now imagine that after hour one, radar observations become available and a new forecast is issued. P(E2 given the new observations) is a different conditional probability. It should not automatically replace the original second-hour value inside an expression intended to represent the original four-hour forecast.

There is no contradiction if a forecast changes when new evidence arrives. The problem is an undocumented mixture of issue times. A table assembled from screenshots taken throughout the afternoon may look like one forecast while actually combining several different states of knowledge.

A reproducible worksheet records issue time separately from valid time. Issue time says when the information was available. Valid time says when the forecast event could occur. Both should include a time zone where ambiguity is possible. The calculation below assumes these fields have already been aligned; it does not infer alignment from a row’s position on a screen.

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Represent each possible four-hour outcome with four digits. A one means the hourly event occurred; a zero means it did not. Thus 1000 means only the first hour qualified, while 1111 means all four qualified.

There are sixteen possible binary patterns. A complete probability model assigns a nonnegative weight to every pattern, with weights summing to one. Some patterns may have zero weight. The hourly probability is obtained by adding the weights of patterns containing a one in that hour’s position.

The window probability is obtained by adding the weights of every pattern except 0000. Equivalently, it is one minus the weight of 0000. This is a direct counting identity and does not require the hours to be independent.

Try a quick check with a single pattern. For 1010, mark which hourly columns receive a contribution and whether the window column receives one. Then repeat with 0000 and 1111. Once this representation is stable, the later formulas become summaries of an account the learner can already explain. Memorising a formula before this step often hides what the formula is allowed to count.

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Imagine one hundred equally weighted fictional days. On twenty-five days, all four hourly events occur. On the other seventy-five days, none occurs. The only patterns with positive weight are 1111 and 0000.

Each hour is wet on twenty-five of the hundred days, so every hourly probability is 25%. The four-hour union also occurs on only twenty-five days. Its probability is 25%, not 100% and not approximately 68%.

The hourly events coincide in this construction. Once you know that the first hour qualifies, you know that the other three qualify as well within this invented model. There is no new set of qualifying days added by the later hourly columns.

This is a mathematical extreme, not a claim that real showers behave exactly like this. Its purpose is to prove that four 25% hourly probabilities can coexist with a 25% window probability. One valid counterexample is enough to disprove a proposed universal conversion. A physical prediction would require a fitted weather model and evidence, which this classroom construction does not supply.

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Keep the same hundred equally weighted fictional days. Divide them into four groups of twenty-five. The first group has pattern 1000, the second 0100, the third 0010 and the fourth 0001.

Every hourly probability is still 25%. Each hourly column has twenty-five qualifying days. But now every day contains one qualifying hour, so the four-hour union probability is 100%.

Compare this record with the previous chapter. The marginal hourly probabilities are identical. The relationship among the hours has changed. In the persistent record, the four columns identify the same twenty-five days. Here, they identify four different groups that together cover all hundred days.

That contrast establishes non-identification: the four hourly percentages alone do not select a unique window probability. It is stronger than saying that one particular formula might be inaccurate. It shows that no formula using only those four marginals can recover the missing relationship in every case.

A helpful learner question is, “Which days did the second hour add?” It directs attention to overlap, the information that the four separate percentages did not preserve.

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Now impose a new assumption: the four events are mutually independent, conditional on F0. Each hour has probability 0.25 of qualifying and 0.75 of not qualifying. Under that assumption, the probability that all four fail is 0.75 multiplied by itself four times.

The all-dry probability is 0.31640625. Therefore P(U) = 1 − 0.31640625 = 0.68359375, or about 68.36%. This is a valid result for the independent model. It is not implied by the hourly probabilities alone.

An equally weighted list of sixteen patterns would not represent this particular model, because patterns with different numbers of wet hours have different probabilities. A pattern with k ones has weight 0.25 raised to k, multiplied by 0.75 raised to 4 − k. Adding all sixteen weights gives one.

When reporting the answer, keep the assumption beside it: “About 68.36% under conditional mutual independence.” Omitting those final words changes a conditional calculation into an unsupported forecast claim. More decimal places cannot replace the dependence information that the input table lacks.

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Let N be the number of qualifying hours. It can be zero, one, two, three or four. Its expectation is the sum of the four hourly probabilities, so E(N) = 1 in all three models above.

This sum does not require independence. Each qualifying hour contributes one to N, and expectations add even when the contributions are dependent. The same average count can be produced by rare four-hour episodes, one-hour episodes on every day, or a mixture of patterns.

The persistent model has N = 4 on one quarter of days and zero otherwise. Its mean is one. The separate-possibilities model has N = 1 on every day, also with mean one. Their probabilities of any qualifying hour are 25% and 100% respectively.

Notice that N counts hourly threshold events. It still does not measure exact wet duration within those hours. A qualifying hour might contain a short intense shower or sustained lighter rain, depending on the threshold rule. Converting E(N) into expected rainfall duration would need a further model of what happens inside each interval. Keep that extra question separate.

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The union contains each individual hourly event. Consequently, its probability cannot be smaller than the largest hourly probability. It also cannot exceed the sum of the hourly probabilities or exceed one.

For any finite set of events, max(p1, …, pn) ≤ P(U) ≤ min(1, p1 + … + pn). These are bounds, not a choice of one preferred estimate. For our four 25% values, the interval is 25% to 100%.

The two explicit records already reach these endpoints. That is why the interval cannot be narrowed using only these marginals in this example. A narrower answer would require additional information or additional assumptions.

The upper bound follows from an overcount: adding all hourly probabilities counts outcomes several times when several hourly events occur together. Removing that duplication cannot increase the sum. The lower bound follows from containment: every outcome in the largest hourly event is already in the union.

This explanation helps a learner remember both directions of the inequality. Before doing arithmetic, ask whether a proposed answer lies below an event that it contains, or above one. Either failure is a useful immediate error signal.

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For two events A and B, the exact identity is P(A or B) = P(A) + P(B) − P(A and B). The intersection removes the overlap counted twice by the first two terms.

Suppose two hourly probabilities are 0.20 and 0.40, and their intersection probability is known to be 0.10 under the same information set. The union is 0.20 + 0.40 − 0.10 = 0.50.

If independence were imposed instead, the intersection would be 0.08 and the union 0.52. That is a different model. The fact that the numerical answers are close does not make the assumptions interchangeable.

The intersection itself has constraints. It cannot exceed the smaller marginal probability, and it cannot be less than max(0, P(A) + P(B) − 1). A proposed overlap of 0.30 would be impossible when P(A) is only 0.20.

Draw two overlapping sets or make a four-row outcome table: neither, A only, B only and both. All four probabilities must be nonnegative and sum to one. This checks the supplied relationship instead of trusting a plausible-looking percentage.

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Consider three hourly events. Assign equal probability to the four patterns 000, 011, 101 and 110. Each marginal probability is one half. Every pair occurs together with probability one quarter, so each pair is independent.

Yet the probability of at least one event is three quarters, because 000 has probability one quarter. Mutual independence would give 1 − 0.5 cubed = seven eighths. Pairwise independence did not determine the three-way relationship.

Now assign equal probability to 111, 100, 010 and 001. The marginals and pairwise intersections remain the same, but the union probability is one. Again, all three events together behave differently.

These are exact discrete constructions. They show why checking all pairs does not automatically validate a product across three or more intervals. Higher-order overlap can matter even when every pair passes an independence check.

A useful transfer question is to ask what observation distinguishes the two records. The frequency of 000 or 111 would do so, while another table of the same marginals would not. Choose evidence that identifies the missing structure, rather than collecting more copies of a summary that already agrees.

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There is another valid product formula, but its inputs are different. Let q1 be the chance of E1. Let q2 be the chance of E2 conditional on no earlier event, and q3 the chance of E3 conditional on neither earlier event occurring. Continue in this way.

The probability of remaining event-free is the product of 1 − qj. This follows from the chain rule for conditional probability; it does not require the marginal hourly events to be independent.

For three supplied conditional hazards 0.10, 0.20 and 0.30, the no-event probability is 0.90 × 0.80 × 0.70 = 0.504. The union probability is 0.496. Here the later percentages describe progressively restricted groups of possible outcomes.

They are not interchangeable with ordinary hourly marginal probabilities. A displayed 20% second-hour chance generally cannot be relabelled as “20% if no rain occurred earlier” without supporting documentation or a model.

Write the conditioning words beside every q. If an earlier no-event condition has zero probability, the corresponding conditional probability is not defined by the usual ratio. At that point the union is already certain and later hazards are unnecessary for this account.

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Suppose a model provides four complete three-hour scenarios with weights: 000 has 0.40, 100 has 0.20, 011 has 0.10 and 111 has 0.30. The weights sum to one and no scenario has negative weight.

The hourly marginals are 0.50, 0.40 and 0.40. The union probability is 0.60 because every scenario except 000 qualifies. Multiplying the three marginal dry probabilities would instead give 0.50 × 0.60 × 0.60 = 0.18, implying a union of 0.82. That result contradicts the supplied joint scenario model.

When complete scenarios exist, calculate the requested event within each scenario, then combine the event indicators using the scenario weights. This preserves temporal relationships. Reconstructing a sequence by independently drawing each hourly column discards those relationships.

For an ensemble, member counts can illustrate this operation, but the fraction of members is not automatically a calibrated real-world probability. Member dependence, unequal weighting, model limitations and calibration remain separate questions. First preserve the model’s joint information; then assess whether that model represents observed weather well enough for its intended interpretation.

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If a one-hour event is contained in the four-hour union, the four-hour probability must be at least as large. Similarly, extending a union from four hours to five cannot reduce its probability when the event definition, place and information set remain fixed.

Suppose the same model assigns 70% to a qualifying event in hour two and 50% to any qualifying hour among hours one through four. Those statements cannot both describe the defined nested events. Every outcome counted by hour two must also be counted by the larger union.

Do not immediately declare a real forecast provider inconsistent. First check whether thresholds, spatial locations, valid periods or issue times differ. The longer product might concern a different accumulated amount. Rounding can also create small apparent discrepancies near a boundary.

A coherence check identifies a question for investigation. It does not tell you which number to change. Repair requires the original event definitions and forecast construction. Averaging the two percentages would hide the logical problem while producing a third unsupported number. Preserve the original records so another reader can inspect the mismatch.

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Imagine A covers hours one and two, while B covers hours two and three. Their union covers hours one through three, provided both events mean any qualifying hour within their respective windows.

P(A) + P(B) double counts outcomes that satisfy both window events. Such outcomes need not consist solely of rain in hour two. Rain in hour one and hour three also makes both A and B occur. Therefore the intersection of the window events is not simply the shared clock interval’s event.

This is a subtle distinction between overlap in time and intersection in outcome space. A diagram of two time bars shows which intervals share hours. It does not, by itself, describe all rainfall patterns satisfying both events.

To calculate an exact combined probability, enumerate complete patterns or obtain the appropriate intersection probability. For example, pattern 101 satisfies both two-hour unions even though the shared middle hour is dry. That one pattern is a useful test of an incorrect shortcut.

The same lesson applies to overlapping school sessions, service windows or observation periods: define the events before treating a time diagram as a probability model.

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Suppose probabilities are available for hours one, two and four, but hour three is absent. The union of the three observed hourly events is a different event from the union of all four hours.

If the known values are 0.10, 0.20 and 0.30, the known-hours union lies between 0.30 and 0.60. Adding an entirely unconstrained missing hour leaves a full-window lower bound of 0.30 and an upper bound of one. A missing value is not a zero.

If independent evidence bounds the missing-hour probability by 0.15, the full-window upper bound becomes min(1, 0.10 + 0.20 + 0.30 + 0.15) = 0.75. Dependence is still unknown, so the interval remains a bound rather than an estimate.

Record why a value is missing. It might be unavailable, outside a product’s coverage, suppressed or not a separate hourly quantity at all. Filling the cell by interpolation changes the evidence. If interpolation is used as a modelling choice, label it and examine how the resulting answer depends on that choice.

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A displayed 30% may represent a rounded value. If values are rounded to the nearest ten percentage points under a documented rule, an interior display of 30% could correspond to a range around 25% to 35%, with endpoint conventions determined by that rule.

You should not assume that rule simply because the display uses multiples of ten. The provider may use categories, truncation or another reporting convention. First establish what the displayed number represents.

When valid probability intervals are known, propagate them through containment and union bounds. For two marginals bounded by [0.25, 0.35] and [0.45, 0.55], a conservative union range without dependence information is 0.45 to 0.90. That range includes uncertainty in both marginals and overlap.

A more precise central value can be reported only with additional assumptions. Keeping six decimal places from an independence calculation does not make rounded inputs exact. A useful written answer distinguishes an exact result inside a stipulated classroom model from a real-product interpretation limited by reporting resolution and unknown dependence.

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If the question is how often rain occurs at least once across a window, historical complete-window records are more informative than separate hourly counts alone. Retain which hours belong to the same date, location and forecast issue.

For verification, also preserve the issued probabilities before the outcome is known. An analysis assembled only from later updated forecasts can accidentally use information unavailable at the original decision time.

Missing observations require an explicit rule. A window with an unobserved hour may still be known to satisfy the union if another observed hour qualifies. A window with no observed qualifying hour and one missing hour cannot automatically be classified as dry. This asymmetry follows directly from the union definition.

Once the event is correctly observed, forecast reliability can be evaluated over suitable repeated cases. That is a separate task from the algebra of combining events. The existing guide to reliability, sharpness and decision thresholds develops that evaluation rather than treating one outcome as proof that a probability was right or wrong.

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Set the earlier numbers aside. Two nonoverlapping intervals have qualifying-event probabilities 0.35 and 0.55, based on the same issue-time information. You are not told their dependence.

First write the broadest valid bounds for the probability of at least one event. Then suppose a complete record establishes an intersection probability of 0.20. Calculate the exact union. Finally calculate what an independence assumption would have implied and identify which input changed.

For an extension, replace the intersection information with the conditional statement P(second event given no first event) = 0.40. Recover the union by following the no-event branch. Explain why this 0.40 is not the original marginal value of 0.55.

Before looking at the next chapter, make a four-row probability table for the intersection version. Label neither, first only, second only and both. Check that all entries are nonnegative and add to one. This representation tests whether the supplied figures can coexist, and gives a second route to the answer without relying on recall of a formula.

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The marginal-only bounds are 0.55 to 0.90. With intersection 0.20, the union is 0.35 + 0.55 − 0.20 = 0.70. The four outcome weights are 0.30 for neither, 0.15 for first only, 0.35 for second only and 0.20 for both.

Under independence, the intersection would be 0.35 × 0.55 = 0.1925, producing a union of 0.7075. The exact intersection record and the independent model are close numerically here, but they remain different accounts.

For the conditional extension, the no-first probability is 0.65. The probability of neither is 0.65 × 0.60 = 0.39, so the union is 0.61. If the first marginal stays 0.35, that conditional statement cannot simultaneously be combined with an arbitrary second marginal without checking compatibility. For instance, retaining second marginal 0.55 would require an intersection of 0.29, which is feasible.

The extension therefore supplies a new relationship, not permission to multiply the original marginal complements. The conditioning words determine which branch of the outcome table receives the number.

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Consider this invented sentence: “There is a 20% chance each hour for five hours, so there is a 100% chance it will rain, for exactly one hour.” It makes two unsupported conversions.

The sum of the hourly probabilities gives an expected count of one qualifying hour. It does not generally give a 100% union probability. Nor does an expected count establish that every outcome has exactly one qualifying hour, or specify rainfall duration inside an hour.

A repaired explanation might say: “The five hourly event probabilities sum to one expected qualifying hour. Without their temporal dependence, the probability of at least one qualifying hour is bounded between 20% and 100%. Exact duration and rainfall amount require additional information.”

That wording is longer because it preserves three distinct questions. It can be shortened for a particular audience once the distinctions are secure. A short answer should not become simpler by merging different quantities.

For practice, ask the learner to invent two complete records satisfying the original five marginals. One should use an all-wet episode on one fifth of outcomes; another should allocate one distinct qualifying hour to each outcome. Both expose why the original claim was not identified.

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A useful note begins with place, event criterion, interval boundaries and issue-time information. It then lists the supplied marginal probabilities and any joint or conditional information. Keep model assumptions in a separate sentence so they are not mistaken for observations.

State the requested event in words. If it is a union, say “at least one of these specified hourly events.” If it is accumulated rainfall above a threshold, say so and use an amount model rather than silently substituting a union.

Show one checkable calculation and one limitation. For example: “Under stipulated mutual independence, the all-dry probability is the product of the hourly dry probabilities. The supplied marginals alone do not establish independence.” Include a bound when a single estimate is not identified.

For an observed record, retain complete patterns and their weights or counts. Identify missing intervals, unequal weights and exclusions. Another reader should be able to recover the hourly marginals and the window result from the same account. If the two cannot be reconstructed consistently, return to the event labels before adjusting the arithmetic.

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For an early mathematics learner, start with twenty cards. Mark five cards in each hourly colour, first using the same five cards for every colour and then using four separate groups. Count cards with any mark. The physical record makes overlap visible before probability notation appears.

For a secondary learner, translate the cards into binary patterns and derive the two-event intersection identity. Change the group sizes and ask for bounds before an exact answer. Include one deliberately impossible overlap so the learner practises checking compatibility.

For an advanced learner, enumerate weighted three- or four-event patterns in a spreadsheet or short program. Reconstruct marginals, unions and conditional hazards. Test pairwise independence separately from mutual independence. Keep the code’s assumptions in the written explanation.

Repair means returning to a mislabelled event or a double count. Stabilisation means solving several different complete records correctly. Extension means identifying what new joint evidence would narrow an unresolved interval. A learner has transferred the idea when they can explain an unfamiliar window problem without relying on the rain example’s particular percentages.

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Can I add hourly chances of rain?

Their sum gives the expected number of qualifying hourly events when definitions are aligned. It gives a union probability only when the events are mutually exclusive, or when an appropriate overlap correction has been made. A sum above one is allowed for an expected count but not for a probability.

Can I multiply the dry probabilities?

You can do that for mutually independent hourly events under a common information set. You can also multiply a sequence of correctly defined conditional no-event probabilities. Those are different inputs; ordinary marginal hourly chances are not automatically conditional hazards.

Does a longer window always have a higher probability?

For a union of the same events, extending the set cannot lower the probability. Different thresholds, products, locations or issue times may describe different events, so check those definitions before applying the containment rule.

Does one rainy outcome prove that a 25% forecast failed?

No. A probability of 25% allows both occurrence and non-occurrence. Reliability requires suitable repeated forecast-outcome records. The event-combination calculations here do not replace that evaluation.

What if I only have a daily percentage?

It does not generally identify a shorter event-time percentage. You need information about the timing relationship or a product defined for the required interval. Dividing by the number of hours is not a general probability rule.

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Return to the Weather Forecasting subject guide for the broader sequence from observations to models and forecast evaluation. Read displaced rainfall patterns and neighbourhood fractions when the missing relationship concerns spatial displacement rather than temporal overlap.

For interpretation and evidence, pair forecast reliability and decision thresholds with the practical case about confusing rain chance with a share of the day. The case establishes the probability-versus-duration distinction; this guide develops the joint-event calculation beyond that distinction.

The authoritative context is the National Weather Service explanation of probability of precipitation. Its point, threshold and period belong to that described product. The complete patterns, counterexamples, bounds and transfer exercises in this guide are original teaching constructions, not extracted weather observations. Source checked 23 October 2026.

Connect the learning to How X Works and The Gold Standard of Education: learning transfers into life. The transferable skill is to preserve the relationship among events when moving from separate percentages to a combined question.

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