An AI probability audit separates an expected success count from the chance of at least one success. Use a fictional three-trial model to save eight path weights, a total-mass control and two routes to 57.8125%, with the assumptions kept visible.
What assumptions define this probability exercise?
An AI probability audit must distinguish a supplied assumption from an observed success rate. This fictional exercise gives three trials, each with success probability one quarter. Success is an abstract outcome marked 1; a non-success is marked 0. It does not mean a correct model answer, a conversion or a reliable business operation.
A trial with these two outcomes is called a Bernoulli trial. The three trials are stipulated to be mutually independent and identically distributed: after every possible prior history, the next success probability remains one quarter. This is a strong source condition, not something established by giving trials different names or placing them in separate rows.
Copy the boundary: “Practice V1; three binary trials; p=1/4 after every prior history; no physical or model experiment performed.” The source contains neither an observed sample nor a reason to believe a real process satisfies those assumptions. If independence is unknown in practice, leave it unresolved rather than importing this calculation as a forecast.
The artificial intelligence guide covers the broader tools. Here the work product is a text file containing eight weighted terminal patterns, a total-probability control and two routes to the chance of at least one success. A fluent explanation does not validate the assumptions on which its arithmetic depends.
Why do eight terminal patterns need different weights?
Write the outcomes in their order. Pattern 001 means two initial non-successes followed by success; 100 means success followed by two non-successes. Both have one success, but they are different ordered histories. Retain each pattern separately so another reader can reconstruct the product attached to it.
The non-success probability is three quarters. Under the stipulated history-independent contract, multiply the three branch probabilities along a terminal path. For 001 the product is 3/4×3/4×1/4=9/64. For 111 it is 1/4×1/4×1/4=1/64. The existence of eight patterns does not make each one equally likely.
| Pattern | Success count | Path probability | At least one? |
|---|---|---|---|
| 000 | 0 | 27/64 | No |
| 001 | 1 | 9/64 | Yes |
| 010 | 1 | 9/64 | Yes |
| 011 | 2 | 3/64 | Yes |
| 100 | 1 | 9/64 | Yes |
| 101 | 2 | 3/64 | Yes |
| 110 | 2 | 3/64 | Yes |
| 111 | 3 | 1/64 | Yes |
These are possible histories, not eight observed records. Two different terminal histories cannot both be the complete result of one three-trial run, so the terminal patterns are mutually exclusive. Together they exhaust the possible binary histories. That is why their probabilities can be added when selecting a group of terminal outcomes.
Success on trial one and success on trial two are different events: they can occur together, as pattern 110 shows. Do not confuse mutually independent trial outcomes with mutually exclusive terminal histories. The AND OR grouping audit evaluates fixed known flags; it attaches no probabilities to those inputs. This exercise instead supplies a stochastic contract and weights every possible history.
How do total mass and the complement cross-check agree?
First add the eight path numerators: 27+9+9+3+9+3+3+1=64. Their common denominator is 64, so the total is 64/64=1. Also verify that every three-digit binary pattern appears once. A matching total is useful but does not by itself rule out a missing pattern accidentally offset by a duplicate.
At least one success includes histories with one, two or three successes. Only 000 fails that condition. Add the other seven masses: 9+9+3+9+3+3+1=37, giving 37/64. Converting to a percentage gives 57.8125%. Retain the fraction as the exact calculation evidence rather than replacing the source with a rounded display.
Now check the same event through its complement. No success means the single history 000. Its mass is 3/4×3/4×3/4=27/64. Since all possibilities have total probability one, at least one success has probability 1−27/64=37/64. The path-sum and complement routes must agree under the same source assumptions.
| Control | Calculation | Result |
|---|---|---|
| All terminal paths | 64/64 | 1 |
| Seven qualifying path masses | 37/64 | 57.8125% |
| No success | 27/64 | 42.1875% |
| Complement cross-check | 1−27/64 | 37/64 |
The two-list distinct count audit counts supplied identifiers after tracing shared membership. This file is not a registry union or a count of real people. Seven qualifying patterns carry unequal probability masses, so their number alone cannot answer the question. Keep that distinction explicit in the handoff label.
Which three reporting claims need correction or restraint?
Claim one says: “Three trials at 25% mean a 75% chance of at least one success.” Correct that event probability to 37/64, or 57.8125%, under the written assumptions. The sum 0.25+0.25+0.25=0.75 is not universally meaningless: it is the expected number of successes, a different quantity from the probability of at least one.
An expected count is a mathematical average count across the stipulated distribution of repeated experiments. One actual three-trial history contains zero, one, two or three successes, not three quarters of a success. Preserve the label “expected count” beside 0.75. Converting that number into a 75% at-least-one probability changes the question without justification.
Claim two says: “Seven of eight patterns qualify, so the probability is 87.5%.” Reject it. Pattern 000 has mass 27/64 while 111 has mass 1/64. Dividing seven by eight would assign equal weights that contradict the supplied one-quarter success probability. Use the mass ledger as evidence rather than a verbal warning alone.
Claim three says: “This proves that retrying a real AI prompt three times achieves the same improvement.” Mark it unsupported. No real model success rate, history dependence, identical distribution or retry behavior has been measured. The classification ledger also avoids turning fictional labels into a product-quality claim. Here no model is run at all, and no retry strategy is recommended.
What do the official sources establish, and what stays unknown?
OpenStax distinguishes independence from mutual exclusivity and explains multiplying joint probabilities when independence holds. Its tree-diagram chapter illustrates products along branches and sums of distinct terminal outcomes. These two pages belong to the same textbook authority, not two independent institutions.
The textbook does not establish one quarter as the success rate of a real tool. That value, the number of trials and the full history-independence contract come from this exercise’s source card. We are applying documented probability principles to declared teaching assumptions, not estimating those assumptions from data.
Keep an unresolved question in the file: “What evidence would establish each trial’s probability and its dependence on earlier outcomes in the real process?” The card does not answer it. Starting a fresh chat, changing an identifier or repeating an action later does not itself prove statistical independence. Do not silently replace that missing evidence with a confident calculation.
The segment-rate audit works with observed-style supplied group counts and changes in mixture. This exercise compares neither groups nor treatments and estimates no causal improvement. It is not a financial, medical, safety or reliability recommendation. The accepted conclusion remains conditional: given this specific probability model, the weighted path sum and complement both yield the same event probability.
What must the saved probability note contain?
On Windows, open Start, type Notepad and launch the text editor. Head the document “At least one probability audit V1.” Add p=1/4, three trials and the explicit condition that the success probability remains fixed after every prior history. State that these are synthetic assumptions rather than measured rates.
Copy all eight terminal patterns. Beside each, retain its three branch factors, exact product and whether it includes at least one success. Add total mass one, qualifying mass 37/64 and the complement calculation 1−27/64. Include the three claim decisions and their evidence before saving, together with the unresolved real-independence question.
Press Ctrl+S, choose Documents and save as probability-audit-V1.txt. Close the editor and reopen the saved file from that folder. Check that all eight unique patterns remain, 000 and 111 keep their different weights, and 57.8125% remains attached to at least one success. Confirm that 0.75 is labelled expected count, not an event probability.
For an optional second opinion, supply only this fictional card to an assistant you already use. Ask: “Use three mutually independent Bernoulli trials with p=1/4 after every prior history. List all eight ordered patterns and their branch products. Check total mass one and calculate at least one success by path sum and complement. Distinguish 3p expected count from event probability. Make no real model retry claim.” If no comparison was performed, record that explicitly.
A suggested fifteen-minute plan gives four minutes to assumptions, four to path weights, three to claims and four to saving and reopening. It is a cognitive planning allowance, not a measured participant result or speed guarantee. In earlier public writing about perfection and effectiveness, I described delaying projects while pursuing every detail. Here that principle means completing one reproducible calculation note, not claiming a personal probability experiment or a customer outcome.
Frequently asked questions
Why is three times 25% not the event probability?
The sum 0.75 is an expected number of successes. Under the stipulated model, the probability of at least one is 1−(3/4)³=37/64, or 57.8125%.
Why is seven qualifying patterns out of eight insufficient?
The terminal patterns have unequal weights. For example 000 has probability 27/64 while 111 has probability 1/64, so equal counting is invalid here.
Was independence demonstrated by this exercise?
No. Fixed conditional success probability after every prior history is explicitly stipulated. A real process needs separate evidence for that assumption.
Does this recommend retrying a real AI task?
No. No model was executed and no real success probabilities were estimated. The calculation applies only to the supplied fictional probability model.
