Best case, most likely, worst case — and the two answers they give.
Three points, two answers
The mode sits near the floor. Triangular mean 5.00, PERT mean 4.00 — a full day apart on a three-point estimate.
Triangular(2.00000, 3.00000, 10.00000)
Mean = 5.000000
The PERT convention gives 4.000000 instead — 1.000000 lower, which is 12.500% of the whole range. Both are standard, and they are computing different things.
Triangular mean
5.000000
(a + c + b)/3
PERT mean
4.000000
(a + 4c + b)/6
Median
4.708497
the 50% point
Mode
3.000000
your most likely value
Standard deviation
1.779513
√variance
Skewness
0.532375
right-tailed
P(X ≤ 5.0000)
0.553571
cumulative
Gap between conventions
-1.000000
12.500% of the range
The density is a straight line up to the mode and a straight line down from it — that is the whole distribution. Its peak is always 2/(b − a), whatever the mode, because the area must be 1.
Two standard conventions, 1.00000 apart on the same three numbers. PERT weights the most likely value four times as heavily, which pulls the estimate toward the mode and away from the midpoint of the range. On this estimate that moves the answer by 12.50% of the total spread. Neither is wrong; they are different models of the same three inputs, and a project plan that mixes them is adding numbers that do not belong together.
The mean is never the most likely value unless the estimate is symmetric. With a mode of 3.00000 the mean is 5.00000— the long side of the triangle pulls it. This is why three-point estimates that look optimistic produce schedules that are not: the expected duration is further from the optimistic end than the “most likely” figure suggests.
Use it when you have three numbers and no data. The triangular distribution exists to turn a bounded expert judgement into something you can simulate, and it makes exactly two assumptions: the value cannot fall outside the range, and the density is linear on each side. Both are wrong in detail and neither is unreasonable. Where you have actual observations, fit a distribution to them instead.
What this tool shows
On a 2 / 3 / 10 estimate the triangular mean is 5.00 and the PERT mean 4.00. A full unit apart on an eight-unit range, from the same three numbers. Both conventions are standard and they are computing different things — the tool prints both, with the gap as a percentage of the range.
Mean, median, mode, variance and skewness from three points
The PERT mean beside the triangular one, and the gap between them
The CDF at any value, for turning an estimate into a probability
The density plotted with both mean estimates marked
A warning when the two conventions differ by enough to change a plan
Validation that the mode really lies between the other two points
Three-point PERT gap shown Median exact CDF included
Updated 12 September 2026 · Works in any browser, no installation
A triangular distribution turns three numbers — a minimum, a most-likely value and a maximum — into a distribution you can compute with. Its mean is the plain average of the three, (a + c + b)/3, which is NOT the most likely value unless the estimate is symmetric. The PERT convention weights the mode four times instead, and gives a different answer.
At a glance
Formula shown
The density rises linearly from a to the mode c and falls linearly from c to b, peaking at 2/(b − a) whatever the mode. The mean is (a + c + b)/3 and the variance (a² + c² + b² − ac − ab − cb)/18. The median is a + √((b−a)(c−a)/2) when the mode is above the midpoint and b − √((b−a)(b−c)/2) when it is below. The PERT mean is (a + 4c + b)/6, a beta approximation rather than the triangular mean, and the two coincide only when c is exactly central.
Scenario support
Project and task duration estimates, cost ranges from expert judgement, Monte Carlo inputs where only bounds and a likely value are known, risk registers, and any “optimistic / likely / pessimistic” triple.
Educational estimate
Planning support from the values you enter — not professional advice.
Two standard conventions, one unit apart
Three numbers go in and two different expected values come out, depending on which textbook you are following.
The triangular mean is (a + c + b)/3. It is the actual expected value of the actual distribution the three points describe. On 2 / 3 / 10 that is 5.00.
The PERT mean is (a + 4c + b)/6. It weights the most likely value four times, and it is the mean of a BETA distribution fitted to the same three points rather than of the triangular one. On the same estimate it gives 4.00.
A full unit apart on a range of eight, or 12.5% of the whole spread. Across a hundred tasks that is not a rounding difference; it is a different schedule.
The gap reverses with the skew. With an optimistic estimate PERT comes out lower; with a pessimistic one (2 / 8 / 10) PERT gives 7.33 against the triangular 6.67 and is the higher of the two. The direction depends on which side of centre the mode sits.
They agree exactly when the mode is central, which is why the disagreement goes unnoticed: the symmetric worked example in every tutorial is the one case where it does not arise.
Neither is wrong; mixing them is. Adding a PERT mean for one task to a triangular mean for another produces a total that corresponds to no model at all.
The mean is not the most likely value
This is the point of using a distribution at all, and it is routinely lost when the three numbers are collapsed back into one.
With a mode of 3 and a range of 2 to 10, the mean is 5. The long right side of the triangle pulls it two units above the value the estimator called most likely.
Which is why optimistic-looking estimates produce late projects. People quote the mode, plan against it, and the expected duration is further toward the pessimistic end than the number they wrote down.
The median sits between the two. At 4.708 on that estimate — closer to the mean than the mode, and the right figure if you want a 50/50 date rather than an average.
The three only coincide on a symmetric estimate. The tool prints all three so the asymmetry is visible as a spread of numbers rather than assumed away.
What the distribution assumes, and what that costs
The triangular distribution is a deliberate simplification, and being explicit about the two assumptions is the honest way to use it.
It assumes the value cannot fall outside the range. Hard bounds, no tail. That is usually false — a task can overrun the pessimistic estimate — and it is exactly the failure mode that matters, because the overruns are what hurt.
It assumes the density is linear on each side. There is no reason for that beyond simplicity. A beta or a lognormal is more flexible and needs more information than three numbers provide.
Its peak is always 2/(b − a). Wherever the mode sits, the height is fixed by the range alone, because the area has to be 1. That is worth knowing when reading the plot: a sharper-looking peak means a narrower range, not a stronger belief.
And its variance is small relative to what people intend. Bounded distributions understate spread, so a Monte Carlo built from triangular inputs will produce a tighter total than reality delivers.
Use it where you have three numbers and nothing else. Where you have observations, fit a distribution to them; the triangular exists precisely for the case where you cannot.
In a Monte Carlo, the errors add up differently
The usual reason to build one of these is to sample from it, and a few properties matter more in aggregate than individually.
Sampling is one line, because the CDF inverts in closed form. No rejection sampling and no numerical inversion, which is why it is the default in spreadsheet risk tools.
Summing many triangular tasks gives something close to normal. The central limit theorem applies, so the total is far more symmetric than any individual estimate — and much tighter than intuition suggests, because the bounded inputs cannot produce a long tail.
Which understates the risk that the plan is for. Real project totals have a long right tail, produced by correlated overruns and by tasks exceeding their stated maximum. Neither is representable with independent triangular inputs.
Correlation matters more than the choice of distribution. If tasks overrun together — the same cause, the same team, the same dependency — the total spread is far wider than independent sampling gives, and no amount of care over the individual shapes fixes that.
Getting the three numbers right
The distribution is only as good as the triple, and the triple is usually elicited badly.
“Best case” and “worst case” are not percentiles unless you say so. Asked casually, people give something near their 10th and 90th percentile rather than true bounds — which makes the fitted range too narrow for a distribution that treats them as absolute limits.
Ask for the extremes first, then the likely value. Starting with the most likely figure anchors the other two toward it, and the resulting range is narrower still.
Ask what would have to go wrong. A pessimistic estimate with no attached scenario is usually the likely value plus a margin, not a genuine bound.
And record who gave it. Three-point estimates vary enormously between estimators, and the spread between people is often larger than the spread within any one person’s triple.
Reporting a three-point estimate
Four things, and the first is the one this page exists for.
Say which mean you used. “Expected duration 5 days” does not say whether that is the triangular mean or the PERT one, and on this example they differ by a day.
Report all three inputs, not just the answer. The triple is the estimate; the mean is a summary of it, and one that discards the asymmetry that mattered.
Quote a percentile when you need a commitment date. A mean is a 50/50 proposition at best and often worse on a skewed estimate. The CDF above turns any date into a probability directly.
And say that the bounds are treated as hard. Anyone reading a triangular-based total should know that the model assigns exactly zero probability to overrunning the pessimistic estimate, which is the assumption most likely to be wrong.
Sources and methodology
References for the triangular distribution and three-point estimation.
Method. The mean, variance, median and skewness are the exact closed forms for the triangular distribution, with the median taking the correct branch depending on which side of the midpoint the mode falls. The PERT mean is computed alongside rather than substituted for it, and the suite pins the gap numerically on the shipped example — 5.00 against 4.00 — and verifies that the two coincide exactly when the mode is central. A mode outside the range returns no result rather than a number, the density is confirmed to peak at 2/(b − a) regardless of the mode, and the CDF is checked to reach exactly 1 at the upper bound. That engine is verified on every change against 75 assertions. The count and the per-case breakdown are published on the formula verification page.
Related calculators
Where this goes next:
Uniform DistributionHandles the continuous and discrete uniform as the separate distributions they are — P(X = x) is zero in one and 1/n in the other, and the variances differ for identical endpoints.
Beta DistributionTakes raw successes and failures and runs the conjugate update, so 5 out of 5 returns 6/7 rather than the 100% a plain proportion claims — with a credible interval that stays honest at zero.
Normal DistributionProbabilities under a normal curve in all four directions with the region shaded — and the empirical rule given exactly, because two standard deviations is 95.45% and the 95% everyone quotes sits at 1.96σ.
Expected ValueReports the spread, the chance of any gain and the single most likely outcome beside the average — because on a roulette number bet the expected value is −2.70% and the modal result is losing, 36 times in 37.
Central Limit TheoremComputes exact sampling distributions instead of simulating them, so you can see what a nominal 95% interval really covers — and that the two-sided figure hides a tail 36% too large.
Weighted AverageEach value carries the weight you give it, with every item's share of the total shown as a percentage so you can see what is actually driving the answer.
An educational tool. A triangular distribution treats the optimistic and pessimistic estimates as hard limits and assumes a linear density on each side — both are simplifications, and a total built from independent triangular inputs will be tighter than a real one, because it cannot represent correlated overruns or the tail beyond the stated maximum.
Published the three-point estimate with both standard conventions computed side by side. On a 2/3/10 estimate the triangular mean is 5.00 and the PERT mean 4.00 - a full unit apart on an eight-unit range, from the same three numbers.
The gap reverses with the skew: on a pessimistic 2/8/10 estimate PERT gives 7.33 against the triangular 6.67 and is the higher of the two. They agree exactly only when the mode is central, which is why the disagreement goes unnoticed in every symmetric tutorial example.
The median takes the correct branch depending on which side of the midpoint the mode falls, and is printed alongside - it is the right figure for a 50/50 commitment date where the mean is not.
The density peaks at 2/(b-a) whatever the mode, verified, because the area must be 1 - so a sharper-looking peak means a narrower range rather than a stronger belief.
A mode outside the range returns no result rather than a number, and the page states plainly that the model assigns zero probability to overrunning the pessimistic estimate, which is its weakest assumption.
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