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three-point estimate

Also called: three-point estimation

An estimate built from an optimistic, a most likely and a pessimistic value. It is useful only if you keep the spread: reporting just the weighted average hides how late the work could be.

A three-point estimate asks for three numbers per task instead of one: optimistic (O, if everything goes well), most likely (M) and pessimistic (P, if the usual things go wrong). The three numbers carry what a single one hides: how much is still unknown and on which side the risk sits. Ask the owner for the cause of a long pessimistic tail; that cause is often a risk worth managing on its own.

A common way to summarize the three numbers: expected value E = (O + 4M + P) ÷ 6, which weights the most likely case four times, and standard deviation σ = (P − O) ÷ 6, a rough measure of the spread. When the bad side is longer than the good side, as it usually is, E lands above M. These formulas are a convention, not a law of nature: they give a reasonable centre and spread for a skewed range, and nothing more.

For tasks that run one after another, add the expected values and add the variances (σ²), then take the square root of the sum. Adding the standard deviations instead pretends that every task hits a bad day at the same time and overstates the spread. This holds only for independent tasks: if one cause (one late supplier, one overloaded expert) can delay several tasks, treat it as one risk rather than as noise that cancels out.

The point of the method is the spread. Reporting only the expected value throws away the information you collected. Say the expected value, a range at a stated confidence (about two standard deviations either side covers roughly 19 cases in 20 if the total is close to bell-shaped, which is an assumption) and what drives the spread.

Example

Klasna's diagnostic test for physics: three tasks in a row, estimates in working days.

Task
O
M
P
E = (O + 4M + P) ÷ 6
σ = (P − O) ÷ 6
σ²
Write 40 questions (Vira)
3
5
13
(3 + 20 + 13) ÷ 6 = 6
10 ÷ 6 = 1.667
2.778
Scoring and result page (Danylo)
4
6
8
(4 + 24 + 8) ÷ 6 = 6
4 ÷ 6 = 0.667
0.444
Trial with 20 learners and fixes (Danylo)
2
3
10
(2 + 12 + 10) ÷ 6 = 4
8 ÷ 6 = 1.333
1.778
Total
14
16
5.0

The test is expected in 16 working days, with σ = √5.0 = 2.236 days. Two standard deviations either side: 16 ± 4.47, about 11.5 to 20.5 days. The wrong shortcut, 1.667 + 0.667 + 1.333 = 3.667, would almost double the spread. And the "14 days" first given is just the sum of the most likely values: with long bad tails on two tasks, it is missed more often than met.

The message to the sponsor: "Test expected in 16 working days; 21 we are fairly sure of; the questions are the uncertain part, because the diagrams depend on the illustrator."

Common mistakes

  • Reporting only the expected value. The spread is the reason to ask for three numbers; drop it and you are back to a single guess.
  • Adding standard deviations across tasks. Add variances and take the square root; adding deviations overstates the spread.
  • Treating linked tasks as independent. One shared cause, such as a late illustrator, can push several tasks at once; handle it as one risk.
  • A "pessimistic" value that is only slightly worse. Ask what the usual bad case looks like; a too-narrow tail hides the real risk.
  • Promising the sum of the most likely values. With skewed tasks it is usually below the expected total.

Learn it in the course

  • Estimating in ranges · Three-point estimates that keep the spread: the expected value and standard deviation of each task, how to add several tasks, and how to say a date with its confidence.