The Gold Standard of Probability | First Success, Retry Limits and Unfinished Runs

Imagine a spinner with four equally likely colours. We call green a success and the other colours failures. Reset the spinner after each spin and assume the results are independent. The success probability is p = 0.25 on every attempt; the failure probability is q = 0.75.

The word success is simply a label for the event we are waiting for. It does not express whether the event is desirable. A quality-control exercise could call the first defect a success for counting purposes. Write the event in ordinary language so that the label does not carry the explanation by itself.

Now compare two instructions. In the first, keep spinning until green appears. In the second, stop when green appears or after four attempts, whichever comes first. These are different experiments. The second instruction allows a run to finish without a green result.

Define T as the number of attempts up to and including the first green. Under the unlimited model, T can be 1, 2, 3 and so on. Define N as the number of attempts actually used under the four-attempt limit. Then N = min(T, 4), and N can only be 1, 2, 3 or 4.

A learner who writes these two definitions has already done an important part of the mathematics. T answers a waiting-time question. N answers a resource-use question. Neither definition alone records whether the capped run succeeded. We will need a completion indicator as well.

Before calculating, ask three questions: what counts as success, does the successful attempt count, and what else can stop the activity? Those questions prevent several apparently difficult formula errors.

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For the first green to occur on attempt three, attempts one and two must fail and attempt three must succeed. The sequence is failure, failure, success. Under our assumptions its probability is 0.75 × 0.75 × 0.25 = 0.140625.

We do not multiply by the number of possible positions for green. The position is already specified: it must be third, with no earlier success. A binomial question about exactly one green anywhere in three attempts would count several possible positions and would answer a different question.

In general, P(T = k) = q^(k−1)p for k = 1, 2, 3, … . The k−1 failures come before the successful attempt. The independent multiplication is justified by the spinner model, rather than by the mere presence of several events.

These probabilities describe separate possible first-success times. Their first four terms do not sum to one because a first green can occur later. Keeping that missing probability visible will become essential when we introduce the limit.

This distribution is called geometric when it counts independent attempts through the first success with a constant success probability. Some references instead count failures before the first success. If F is that failure count, F = T−1. A formula can look off by one because its variable is defined differently. Read the definition before deciding that the arithmetic is wrong.

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“On attempt four” means T = 4. “By attempt four” means T ≤ 4. “After attempt four” means T > 4. Each phrase identifies a different group of runs.

The simplest route to the cumulative question is through its complement. To have no green in the first four attempts, all four must fail. That probability is 0.75^4 = 0.31640625. Therefore P(T ≤ 4) = 1−0.31640625 = 0.68359375, or about 68.36%.

Check the answer another way. Add the four first-success probabilities from the previous table: 0.25 + 0.1875 + 0.140625 + 0.10546875 = 0.68359375. The sequence and complement calculations agree because they describe the same event.

Notice how far this is from a guarantee. A four-attempt budget leaves about 31.64% of modelled runs without success. Nothing has gone wrong with a particular run merely because it belongs to that group.

There is another useful distinction at the boundary. T ≥ 4 means the first three attempts failed, so it has probability 0.75^3 = 0.421875. It includes a success on attempt four and every later first success. T > 4 requires four failures and has the smaller probability 0.31640625.

When an answer seems surprising, translate the inequality back into a sequence. How many failures must have happened? Is success on the boundary included? This verbal reconstruction often repairs the exact mistake more effectively than another page of substitutions.

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For this geometric convention, the expected waiting time is E(T) = 1/p. With p = 0.25, the result is four attempts. It is the model’s long-run mean across repeated runs, rather than the time every run needs.

You can understand the relationship without memorising it first. Let E be the expected number of attempts still needed from the start. We always use one attempt. With probability q it fails, leaving the same expected waiting problem again. Thus E = 1 + qE. Rearranging gives pE = 1 and E = 1/p.

This argument depends on the same conditions applying after a failure. If the spinner changes, the device heats up or the player learns, the remaining problem may no longer be identical to the original one.

The median asks a different question: when have at least half the runs finished? For our spinner, success by attempt two has probability 1−0.75^2 = 0.4375. By attempt three it is 1−0.75^3 = 0.578125. So three attempts is a median waiting time, while the mean is four.

The difference is a consequence of the long right tail. Some runs finish immediately; a smaller number take many attempts and pull the average upward. Reporting both a completion probability and a resource limit is more informative than reporting the average alone.

For a practical explanation, say: “The assumed mean wait is four attempts; a four-attempt limit gives approximately 68.36% completion.” The two figures belong together because they answer different questions a reader might otherwise confuse.

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Under the four-attempt limit, N records how many attempts are actually used. A run uses one, two or three attempts only if it first succeeds there. A run uses four whenever the first three attempts fail, whether or not the fourth succeeds.

Therefore P(N = 4) is 0.75^3 = 0.421875. Assigning it only P(T = 4) = 0.10546875 would omit all the runs that use four attempts and remain unfinished.

The probabilities total one. Adding the final column gives E(N) = 2.734375 attempts. The capped activity uses fewer attempts on average than the unlimited activity, but it also leaves some runs unsuccessful. The smaller number does not establish an improvement in completion.

There is a useful second check. Every run needs attempt one. A fraction q needs attempt two, q^2 needs attempt three and q^3 needs attempt four. The expected total is 1 + q + q^2 + q^3 = 2.734375.

More generally, with a positive integer limit L, E[min(T,L)] = (1−q^L)/p. This follows from the finite geometric sum. It describes expected attempts used, including the work consumed by failures at the cap. Keep the completion probability 1−q^L beside it whenever comparing budgets.

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Suppose a class records only runs in which green appeared before the cap. Its spreadsheet contains first-success times of one through four and no unfinished rows. The resulting average describes successful runs, not every run the class started.

For the model, the conditional mean among successful capped runs is [1(0.25) + 2(0.1875) + 3(0.140625) + 4(0.10546875)] ÷ 0.68359375. That is 1.46875 ÷ 0.68359375, or approximately 2.149 attempts.

Now we have three valid means with different meanings: four attempts for unlimited first-success waiting, 2.734 attempts actually used under the cap, and 2.149 attempts among capped runs that succeeded. A report can become misleading simply by giving the last figure the first figure’s label.

An unfinished capped run is informative. We know its first success was not observed in the first four attempts, so T > 4. We do not know T = 4, and we do not know its eventual first-success time. In waiting-time analysis this is a simple example of a right-censored observation.

Keep two columns: attempts used and success observed. Add the cap and the rule version to the record. A row with four attempts and “no” is different from a row with four attempts and “yes”. Replacing both with a single completion time destroys that distinction.

This habit transfers to many data tasks. An incomplete search, an unresolved request and a timed-out operation should remain visible in a suitable status field. A shorter average created by quietly removing slow or unfinished cases is a change in the population being described.

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Suppose a fictional planning exercise asks for at least 95% probability of observing green within a fixed number of attempts. We need 1−0.75^L ≥ 0.95, or 0.75^L ≤ 0.05.

You can test successive integers without logarithms. Ten attempts give 1−0.75^10 ≈ 0.943686, or 94.37%. Eleven give 1−0.75^11 ≈ 0.957765, or 95.78%. Therefore eleven is the smallest integer cap meeting this modelled target.

With logarithms, L ≥ log(0.05)/log(0.75). Because log(0.75) is negative, dividing by it reverses the inequality. The result is slightly above ten, so round upward to eleven. Rounding to the nearest whole number would violate the stated threshold.

The target is a choice within the exercise. It is not a universal standard for an acceptable service, a recommended classroom duration or an assurance about an actual device. Different consequences and budgets call for different criteria.

Also ask where p came from. Here it is supplied by the spinner model. In an application it may be estimated with uncertainty. A completion calculation based on an optimistic p can give a reassuring answer while the input itself is poorly supported.

A clear planning note preserves the event, the probability assumption, the cap and the desired completion level. After observing real results, compare them with the original note. That lets you investigate a model rather than rewriting the promise around whatever happened.

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After four failures, a student may feel that green must be coming soon. Under the stated independent spinner model, the next attempt still succeeds with probability 0.25. The spinner has no mechanism that compensates for an earlier sequence.

Conditioning on T > 4, the probability of success in the next two attempts is still 1−0.75^2 = 0.4375. The reasoning restarts because the next two results are independent of the first four under this model.

That does not erase the resources already spent. At that point the expected additional unlimited wait is four attempts, and the expected total conditional waiting time is eight. At the original start, E(T) was four. These statements concern different information and starting points.

Now change the experiment. Draw green and non-green cards from a finite bag without replacement. Each failure removes a non-green card, changing the next probability. The constant-p independent model no longer applies. A memoryless conclusion copied into this new situation would be a model-selection error.

Change it again: a student practises a skill between attempts. Success probability may change because the student learns, becomes tired or receives feedback. The spinner model should not be used to turn a child’s learning into a fixed success rate.

The useful question is whether anything relevant changes between attempts. “We did the activity again” does not establish independence or a constant probability. Explain the mechanism that supports those assumptions, or select a model that represents the changes.

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A fictional digital task succeeds with probability 0.25 per attempt in an intentionally simplified test model. Its configuration permits three retries after the first attempt. That gives four attempts in total, not three.

If attempts truly follow our independent constant-p model, the completion probability is the same 68.36%. If someone interprets “three retries” as three total attempts, the calculated probability becomes 1−0.75^3 = 57.8125%. The difference comes from the counting contract before any software detail enters.

Suppose each attempt occupies two seconds and the test model ignores pauses and overhead. Expected active attempt time is 2 × 2.734375 = 5.46875 seconds. The maximum active attempt time is eight seconds. Neither figure includes a waiting interval between retries, network delay outside the defined attempt, or cleanup after a timeout.

A real service outage can make retries strongly dependent. A request that failed because a shared service is unavailable may fail again for the same reason. Repeating an arithmetic formula does not repair the underlying outage. Operational design also needs to consider whether an earlier request might already have taken effect before a response was lost.

The transfer is therefore precise and limited: count the initial attempt, name the success event, preserve failed runs and define the time boundary. The probability result transfers only when the new process supports the original model assumptions.

For a different application of geometric waiting, the existing guide to skip lists and random heights shows how repeated random decisions can control a data structure. Its height convention and stopping event should be read separately from this retry example.

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Use a paper exercise or a teacher-approved random simulation. Write the success rule and cap at the top of the sheet. For every started run, record its identifier, attempt sequence, attempts used and whether success was observed. Preserve the unfinished runs.

Begin by asking for the fraction completed within four attempts. The denominator is all started runs. Next ask for average attempts used across those runs. Finally ask for mean first-success time among completed runs. These are three separate summaries of the same records.

If the observed fractions differ from the model, start by checking the definitions and records. Were all started runs included? Was the spinner reset? Did anyone continue after success? Were unfinished rows marked as missing and accidentally excluded by a spreadsheet formula?

After those checks, consider random variation and whether the model fits. A small class exercise need not reproduce theoretical percentages exactly. One short sequence neither validates the model nor proves the spinner is biased.

Change the cap from four to six in a supplied full sequence. Recompute the completion indicator and attempts used. Ask which summaries can be recalculated from the saved record. If only successful times were retained, some of the new question may be unanswerable.

That is the evidence lesson behind the calculation. A good record keeps enough meaning to support the next reasonable question. It does not compress success, duration, failure and uncertainty into a single convenient number.

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Exactly third or within three? For the same spinner, first success on attempt three has probability 0.75^2 × 0.25 = 0.140625. Success within three has probability 1−0.75^3 = 0.578125. The first excludes earlier successes; the second includes them.

Two attempts, with every run counted With cap two, P(N=1)=0.25 and P(N=2)=0.75. Thus E(N)=1(0.25)+2(0.75)=1.75 attempts. Completion probability is 0.4375. Assigning P(N=2)=0.1875 would leave out two-failure runs.

Change the success probability Let p = 0.5 and cap L = 3. Completion probability is 1−0.5^3 = 0.875. Expected attempts used are 1+0.5+0.25 = 1.75. The unlimited mean is two attempts. Name each quantity when reporting it.

Change the mechanism A bag contains one green and three non-green cards, drawn without replacement. After three non-green draws, the next draw must be green. This differs from the independent reset spinner. The original geometric formula is unsuitable because the probability changes after each draw.

Find the reporting mistake A summary says “average completion time: four attempts” after coding all unfinished four-attempt runs as completed at four. The repair is to separate attempts used from completion status. The censored runs establish T > 4; they do not establish a first success at four.

Explain each answer in words before checking the number. That reveals whether an error came from the event, the stopping rule, an assumption or the arithmetic.

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Does an average of four attempts mean four is a sensible cap? It gives one piece of information. Under this model a cap of four completes about 68.36% of runs. Choose a cap using the intended completion level and resource constraints, with the model’s limits visible.

Is an unfinished run a failure forever? No. It means success was not observed before the stopping point. The eventual waiting time is unknown unless the run continues and is recorded appropriately.

Can I estimate p from a list of successful waiting times alone? That list is selected by completion. Ordinary summaries of it do not represent all started runs. Retain the cap, censoring information and all runs before selecting an appropriate statistical method.

Why not keep retrying until the task works? An unlimited mathematical model is useful for learning. Real activities have budgets, changing conditions and consequences. A stopping rule should reflect the real task rather than treating eventual success in an ideal model as an operating plan.

Sources and scope The formal first-success distribution and the attempts-including-success convention can be checked in OpenStax: Geometric Distribution. The capped tables, conditional averages, planning calculations and fictional retry comparison above are original worked examples derived from the stated assumptions. No measured service reliability or learning rate is claimed.

Probability foundations Return to The Gold Standard of Probability for sample spaces, conditional information, independence and distributions. Continue into How Conditional Probability Works for the meaning of “among”, or How Expected Value Works for weighted averages.

Applications and wider learning Read skip lists and random heights for a different first-stopping application. The Gold Standard of Education: Learning That Transfers into Life explains why changing the context is a useful understanding check. The How X Works hub connects the wider subject library.

Choose one waiting-time claim and write what ends the run. Then check whether unsuccessful runs are still present in its denominator. Those two moves often reveal what the headline average has left unexplained.

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