The most probable state does not always determine the least harmful action.
By the end: Derive a decision threshold and separate probability estimates, consequences, and the value of further information.
Write the loss table first
Suppose an item is faulty with probability p. Acting on a faulty item prevents all modeled loss at no loss cost; acting on a sound item costs five loss units. Waiting costs 100 if it is faulty and zero otherwise. Expected loss of acting is five times one minus p; expected loss of waiting is 100 times p.
Act when the first expected loss is smaller. The threshold is 5 divided by 105, about 4.76%, not 50%. At p = 10%, acting has expected loss 4.5 and waiting ten. The less probable state matters because the consequences are asymmetric. These losses are stipulated values, not empirical treatment effects.
Derive the threshold from the two state branches
A decision starts with actions that are available before the unknown state becomes known. Imagine choosing whether to isolate a component. The state is faulty or sound, and the action is act or wait. These form four different combinations. A probability belongs to a state given current evidence. A loss belongs to an action in a state. Both action comparisons use the same fault probability given current evidence.
Use a new probability of 6% with the original losses. Among a reference population of 1,000 comparable components, the model expects sixty faulty and 940 sound components. Acting on all of them gives zero loss on the sixty faulty components and five units on each of the 940 sound ones. The expected total is 4,700 units, or 4.7 per component. Waiting gives 100 units on each of sixty faults, totaling 6,000 units, or six per component. These are weighted expectations, not promised batch counts.
The comparison can also be made without a reference population. Acting contributes five units with probability 94%, so its average is 0.94 times five, or 4.7. Waiting contributes 100 units with probability 6%, so its average is six. Both calculations include the zero-loss branch, even though multiplying by zero leaves no visible contribution. Dropping a branch because its contribution is zero is legitimate arithmetic. Dropping a nonzero branch because it is inconvenient would change the decision model.
Weight both state branches before comparing actions
The strip has width proportional to state probability: 36 of 600 drawing units are faulty, and 564 are sound. Equal heights make areas proportional too. Expected counts use 1,000 items; losses are totals for that reference size, not guaranteed outcomes.
To locate the switch, let the fault probability vary while the losses stay fixed. At a tie, five times the sound probability equals 100 times the fault probability. The sound probability is one minus the fault probability. Moving the five fault-probability units to the other side gives five equal to 105 times the fault probability. Dividing by 105 gives about 4.76%. This threshold is a consequence of the loss table. It is neither the detector's sensitivity nor a probability required for scientific proof.
Check the direction using the endpoints. With zero probability of fault, waiting has zero expected loss and acting has five. With certainty of fault, acting has zero and waiting has 100. Increasing fault probability therefore moves the preference toward acting. At the exact threshold the model gives equal averages, not equal consequences in every state. A tie can require another stated criterion, such as avoiding an irreversible action. That criterion adds information about preferences rather than changing the probability calculation.
Now compare classification with action. A rule that labels a component faulty only above 50% may minimize the number of wrong labels when both kinds of mistake count equally. It does not minimize this loss table. At 6%, sound is the more probable state, yet acting has the smaller expected loss. No contradiction appears once the question is explicit. Predicting the most likely label and choosing the least costly action use the same probabilities but attach different consequences to mistakes.
The action boundary is where expected losses meet
Horizontal position is fault probability from 0% to 10%; vertical position is expected loss in units per component. The falling blue path is 5 × the sound probability. The rising gold path is 100 × the fault probability. The dashed path marks 5 / 105, approximately 4.76%.
Separate facts from values
Probability estimates describe uncertainty under evidence and assumptions. Losses encode consequences and their valuation. Changing either can change the preferred action. If a missed fault costs only twenty units, the same false-alarm cost five gives a threshold of 20%.
The simple rule assumes two exhaustive states, two available actions, known conditional losses, and willingness to minimize expected loss. Budget constraints, risk aversion, unequal effects on people, irreversible harm, and uncertain intervention effectiveness may require a richer decision model. A calibrated probability cannot decide those value questions by itself.
Change the consequence model without changing the evidence
Zero loss for a correct action is a simplifying assumption, not a definition of correctness. Suppose isolation always costs two units, whether the component is faulty or sound. Keep an additional five-unit disruption loss when it is sound. Suppose isolation prevents only 80% of the 100-unit fault loss on average, leaving twenty units when a faulty component is isolated. The revised act losses are twenty-two on faulty components and seven on sound ones. Waiting still has losses 100 and zero. These consequences are separate from fault probability.
At 6% fault probability, the revised expected act loss is 0.06 times twenty-two plus 0.94 times seven. The contributions are 1.32 and 6.58, totaling 7.90 units. Waiting still averages six, so the preferred action reverses without any new evidence about fault. To derive the revised threshold, write act loss as seven plus fifteen times fault probability. Equating it with 100 times fault probability gives seven equal to eighty-five times fault probability. The switch is about 8.24%, above the original threshold.
Partial effectiveness must describe the consequence of intervention, not merely an association with receiving it. Components selected for isolation might already be more damaged. Comparing their outcomes with unisolated components would then mix intervention effects with initial differences. Random assignment, where feasible, can make the groups comparable before action. Random sampling instead helps make an evaluated group representative of a target population. Neither operation substitutes for the other. A representative observational comparison can remain confounded, and a randomized experiment can study an unrepresentative population.
Consequences change the threshold
Model
Act if faulty
Act if sound
Threshold
Original
0
5
5 / 105 = 4.76%
Partial prevention plus cost
22
7
7 / 85 = 8.24%
Loss units per component. Waiting loses 100 if faulty and zero if sound in both models. At 6% fault probability, original act loss is 4.70; revised act loss is 7.90. Wait loss is 6 in both.
Expected loss also differs from a guarantee against severe harm. Consider two stipulated policies at the same decision point. One loses six units for certain. The other loses zero with probability 99% and 500 units with probability 1%. The second policy has expected loss five, so minimizing expected loss prefers it. Someone unable to survive a 500-unit loss may reject it. That disagreement concerns how losses represent consequences or whether a hard constraint applies. It does not make the multiplication of probability and loss incorrect.
Constraints can prevent applying the individual rule to everyone. Suppose only ten components can be isolated today. Under the original loss table, isolating one component at fault probability 6% reduces expected loss by six minus 4.7, or 1.3 units. At 12%, the reduction is twelve minus 4.4, or 7.6 units. With equal isolation effort and no effects between components, prioritize larger reductions until capacity is used. Unequal effort, shared shutdown costs, or interacting faults would require comparing whole feasible policies rather than isolated items.
Keep a consequence comparison beside a probability comparison when reviewing recommendations. First hold the probability fixed and vary plausible effectiveness or disruption losses. Then hold consequences fixed and vary the fault estimate. These check different assumptions. A narrow fault interval cannot settle who should bear disruption. Agreement about losses cannot make a poorly measured probability reliable. The revised table and the original table show why an apparently small change in operational assumptions can move the action boundary.
A smaller average can include a larger possible loss
Policy
Loss outcomes
Expected loss
Chance loss exceeds 100
Certain loss
6 with probability 100%
6
0%
Rare severe loss
0 with 99%; 500 with 1%
5
1%
All losses use the same units. These are discrete probability masses. An expected-loss preference for the second policy is not a guarantee of avoiding a severe outcome.
Work through the mechanism
1. Compare expected losses
p is fault probability. Five is the false-alarm loss; 100 is the missed-fault loss.
2. Solve for the action threshold
Equality gives a tie under this exact loss model. The threshold is not a property of the test alone.
Compare actions under the same probability
Lower expected loss: Act. Threshold: 4.76%.
False-alarm loss is five; correct actions have zero modeled loss. A loss preset keeps the current probability; probability controls keep the current loss assumption. All choices update immediately. The selected action minimizes this constructed expected loss, not an unstated real-world utility.
Action
Faulty
Sound
Act
0
5
Wait
100
0
Constructed loss units. The idealized action fully prevents the fault loss and has no loss when correctly applied.
Explore the decision display as a controlled comparison
The decision display varies fault probability and the loss of a missed fault. It keeps false-alarm loss at five and correct-action losses at zero. Start with the severe missed-fault preset and choose the 5% probability preset. Before reading the result, compare five units paid on the 95% sound branch with 100 paid on the 5% faulty branch. These averages are 4.75 and five. The small difference of 0.25 favors acting, even though most components are expected to be sound.
Switch only to the smaller missed-loss preset. The probability remains 5%, so act loss remains 4.75. Wait loss becomes twenty times 0.05, or one. The action changes because waiting is now less harmful in the faulty state. The probability has not fallen, and the evidence has not weakened. This controlled comparison isolates a consequence assumption. At the 20% probability preset, the smaller missed-loss model gives four units for either action. The displayed tie follows from equality, not from uncertainty about which number is larger.
With the severe loss restored, use the slider to compare 4% and 5%. At 4%, acting averages 4.8 and waiting four. At 5%, acting averages 4.75 and waiting five. The threshold lies between them. The slider uses whole percentages, so it does not select the exact 4.7619% switch. The threshold falls between selectable settings. This is a display resolution limit, not a gap in the probability scale or a region where neither action applies.
Hold probability fixed while changing missed loss
Fault probability
Missed loss
Act average
Wait average
Choice
5%
100
4.75
5
Act
5%
20
4.75
1
Wait
20%
20
4
4
Tie
Loss units per item under the display’s original consequence structure. Changing missed loss does not change the fault probability or the false-alarm loss of five.
Suppose external evidence supports fault probabilities somewhere between 3% and 7%, without providing probability weights within that range. Under severe missed loss, waiting is preferred at 3% and acting at 7%. The evidence range therefore crosses the decision boundary. It does not tell you to average its endpoints or treat every probability in the range as equally likely. A confidence interval describes a procedure's coverage under repeated sampling. It does not automatically supply posterior weights for possible fault probabilities in this particular decision.
If a Bayesian analysis did supply a posterior distribution over the fault probability, the expected-loss calculation would average over it. For the original table, both action losses are linear in that probability. A posterior mean of 6% therefore gives act loss 4.7 and wait loss six, whatever the posterior's shape. This statement uses the posterior mean as an assumed result; it does not derive it from an interval. If the probability model is continuous, posterior probability belongs to areas under its density, not to density heights at single values.
Linearity does not mean parameter uncertainty never matters. It can matter for the value of additional information, hard safety limits, or uncertain losses that vary with fault probability. A posterior mean alone also hides whether a recommendation would reverse under plausible alternative models. Use the display to compare the endpoints of a justified range, then describe whether the choice changes. That is a sensitivity analysis using existing controls. The display does not fit a posterior, estimate intervention effectiveness, or represent the revised consequence table from the preceding section.
An input range can cross the action boundary
Position uses a common 0–10% fault-probability scale. The blue segment marks only an assumed range of 3–7%, not a uniform distribution or posterior density. Its left endpoint favors waiting and its right endpoint favors acting under the original loss table.
Further information has a price
Perfect information can let you choose the better action in each state. Its expected value is the current minimum expected loss minus the expected loss with perfect information. In this zero-loss-correct-action table, perfect information removes all loss, so at p = 10% its value is at most 4.5 units.
An imperfect test cannot be valued just by its accuracy. Work out how each possible result would change the action, calculate losses over all branches, and include the test’s cost and risks. If no possible result would change the action under the model, that test has no decision value here, though it might serve a different purpose.
Value a test through the actions its results permit
Consider a separate information purchase at a prior fault probability of 6%. Under the original losses, acting without a test costs 4.7 expected units and waiting costs six, so the best current action is act. Suppose a proposed test has sensitivity 80% and false-positive rate 10%. Treat those rates as known for this calculation. Its positive-result likelihood is 0.80 under fault and 0.10 under sound. Those likelihoods describe results given states, not fault probability given a positive.
Use expected frequencies among 1,000 comparable items. Sixty faulty items produce forty-eight positives and twelve negatives. The 940 sound items produce ninety-four positives and 846 negatives. A positive therefore has posterior fault probability forty-eight divided by 142, about 33.80%. A negative has probability twelve divided by 858, about 1.40%. The first exceeds the original action threshold and the second falls below it. Thus the test permits a result-dependent policy that acts on positives and waits on negatives, rather than acting on everyone.
Evaluate that policy before paying for information. The ninety-four false positives each lose five units, totaling 470. The twelve false negatives each lose 100, totaling 1,200. Correct decisions contribute zero. The expected classification loss is 1,670 per 1,000 items, or 1.67 per item. Relative to acting immediately at 4.7, the expected reduction is 3.03 per item before test cost. A test priced at 0.50 per item leaves expected total loss 2.17, still below 4.7. No realized batch must reproduce these branch counts.
This calculation weights possible results before the result is known. The probability of a positive is 142 divided by 1,000, and of a negative 858 divided by 1,000. Within positives, acting has expected loss five times ninety-four divided by 142. Within negatives, waiting has expected loss 100 times twelve divided by 858. Multiplying each conditional loss by its result probability recovers 0.47 plus 1.20. Conditioning rescales branch weights; averaging over results restores them.
Information value depends on the full result policy
A schematic branch tree for the separate 6%-prior example; lengths are not probabilities. Leaf counts are expected frequencies per 1,000 items. Acting on positives and waiting on negatives gives 470 + 1,200 = 1,670 loss units before test price. This is 1.67 per item, versus 4.70 for acting without the test.
A test can provide information yet have no value for this particular action. Suppose another test's likelihood ratios are 1.2 for a positive and 0.8 for a negative. Starting odds are six to ninety-four. The resulting fault probabilities are about 7.11% and 4.86%. Both remain above 4.76%, so the best action stays act for either result. Before test cost, the policy is unchanged and its expected loss remains 4.7. These ratios follow from positive probabilities 0.6 under fault and 0.5 under sound.
A follow-up test requires its own conditional model. Two results need not have independent errors. If the second measurement shares contamination or repeats a stored reading, its likelihood after the first result can differ from its marginal likelihood. To value a two-test policy, branch on both results using their joint probabilities under each state. Only conditional independence under both fault and sound justifies multiplying the separate likelihoods. Better evidence about dependence can change information value even when each test's advertised sensitivity and false-positive rate stay the same.
Exercises
A rare but costly miss
Use the baseline loss table at p = 0.02 and then at p = 0.10. Which action minimizes expected loss in each case?
Hint
Compare both weighted losses, not just p with one half.
Worked answer
Fault probability
Act loss
Wait loss
Choice
2%
4.9
2
Wait
10%
4.5
10
Act
The choices are conditional on the stipulated loss table.
At 2%, acting costs 4.9 expected units and waiting two, so wait. At 10%, acting costs 4.5 and waiting ten, so act. The threshold lies between these probabilities at about 4.76%. Waiting does not mean a fault is impossible; it is the lower expected-loss choice under these values.
Price perfect information
At p = 0.10 under the baseline table, would perfect information be worth paying six units per item?
Hint
Compare its cost with the entire loss remaining under the best current action.
Worked answer
Strategy
Expected total loss
Act now
4.5
Perfect information, free
0
Perfect information, cost 6
6
Information quality does not imply that every price is worthwhile.
No. The best current expected loss is 4.5. Perfect information reduces modeled decision loss to zero, but paying six leaves total loss six, worse than 4.5. Thus 4.5 is an upper bound on the value of any information for this decision, before accounting for other benefits or harms outside the model.
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.