Expected Value Calculator

Expected value calculator: enter each outcome and its probability, get E[X] with every term shown

Type an outcome value and its probability on each row, add as many rows as the situation has outcomes, and the calculator returns the expected value along with the full working — a table of every x × p term and the running total, so you can see which outcome is actually driving the answer. It recalculates as you type, it checks that your probabilities sum to 1 and tells you when they do not, and it runs entirely in your browser: nothing is uploaded and no numbers leave the page.

The inputs are deliberately plain. Each row is a value and a probability. Values can be negative — that is how you represent a loss — and probabilities must be zero or greater. Two example buttons, Fair die (1–6) and Roulette $1 bet, fill the grid with a working scenario. That is the whole interface. There is no continuous distribution mode, no variance or standard deviation output, and no simulation; this tool does one calculation and shows its work.

The formula, and what the weighting is doing

Expected value is a probability–weighted average:

E[X] = Σ xⁱ · pⁱ

Every outcome contributes its own value, scaled down by how likely it is. An outcome worth 1,000 that happens one time in a thousand contributes exactly the same to the total as an outcome worth 1 that happens every time. The word "expected" is a poor translation of what it means — it is the long–run average across many repetitions, not a prediction of what happens next, and it is frequently a number the process can never actually produce.

That last point is worth sitting with, because it is where most misuse starts. The expected value of one roll of a fair six–sided die is 3.5. A die never shows 3.5. The number is telling you about the average of many rolls, not about the roll in front of you.

Worked example: the fair die, arithmetic in full

Six outcomes, each with probability 1/6:

E[X] = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21 / 6 = 3.5

The built–in Fair die example does something slightly different, and it is instructive. It enters 0.1667 for the first five faces and 0.1665 for the sixth, so the probabilities sum to exactly 1.0000 and the tool shows the green "valid probability distribution" banner. The consequence is a result of 3.4995 rather than 3.5:

Nothing is broken. 1/6 is not representable as a short decimal, so rounding it to four places and then patching the last row to make the column add to 1 moves the answer by 0.0005. If you need the exact figure, type the fractions as more decimal places — 0.166667 across all six rows sums to 1.000002, which still passes the tool's tolerance check, and gives 21 × 0.166667 = 3.500007 — displayed as 3.5, since results are rounded to four decimal places. The lesson generalises: your answer inherits the precision of the probabilities you feed it.

Worked example: a negative–EV bet

A single–number bet on an American roulette wheel. There are 38 pockets. You stake $1; if your number comes up you are paid at 35 to 1, otherwise you lose the stake. Two outcomes:

So the bet loses about 5.26 cents per dollar staked, on average, forever. That figure — 2/38 — is the house edge, and it is the same for almost every bet on the American wheel, which is the real reason no betting system helps: a system is just a rule for choosing how much to stake on bets that each carry the same negative edge, and a weighted sum of negative numbers is negative regardless of the weights.

The tool's Roulette $1 bet preset enters 0.0263 and 0.9737 rather than the exact fractions, and returns 35 × 0.0263 + (−1) × 0.9737 = 0.9205 − 0.9737 = −0.0532. Again the rounding shows: −0.0532 against the exact −0.052632, a difference of about 0.0006. On a $1 bet that is noise. Scaled to a million dollars of handle it is $600, which is why the people who do this professionally carry the fractions rather than four–place decimals.

Worked example: a decision, not a game

Expected value is more useful for choices than for gambling. Suppose a $1,200 repair has a 4% chance of being needed within the warranty term, and an extended warranty covering it costs $80 up front.

• Without the warranty, the cost is 1200 × 0.04 + 0 × 0.96 = 48.
• With the warranty, the cost is $80 with certainty.
• The warranty is therefore worth −$32 in expected–value terms.

Enter it as two rows: value -1200, probability 0.04; value 0, probability 0.96. The tool returns −48, and the term table shows the entire loss coming from a single row. That is the pattern to look for — when one row dominates the sum, your answer is only as good as that row's probability estimate, and everything else is decoration.

Note what the −$32 does not settle. If a surprise $1,200 bill would genuinely hurt, paying $80 to remove the possibility can be entirely rational. Expected value prices the average; it does not price your tolerance for the tail. Insurance is negative–EV by construction — the insurer needs a margin — and people still buy it for good reasons.

Positive expected value is not the same as a good bet

This is the failure mode worth more than everything else on the page. Expected value is linear and additive; your bankroll is multiplicative. Those are not the same arithmetic, and a strategy can have a positive expected value and still bankrupt you with near–certainty.

Take a wager that returns +50% of whatever you stake with probability 0.5, and −40% with probability 0.5. Per round, the expected return is:

0.5 × (+0.50) + 0.5 × (−0.40) = 0.25 − 0.20 = +0.05

A clean +5% per round in expectation. Now stake your entire bankroll each time. Over two rounds, one win and one loss — the typical case — multiplies your money by 1.5 × 0.6 = 0.9. You are down 10%. The per–round growth rate you actually experience is the geometric mean, √0.9 = 0.9487, which is a loss of about 5.13% per round. Repeat it long enough and you converge on zero, while the expected value stays resolutely positive — because it is being held up by an ever–shrinking probability of an ever–larger fortune.

Two practical consequences. First, when the answer here is positive, ask how much of your resources any single trial risks. Second, expected value says nothing about spread: a certain +1 and a one–in–a–million shot at +1,000,000 both have E[X] = 1, and they are not remotely the same proposition. This calculator will not tell you which you are holding — for that you want a dispersion measure, and the standard-deviation-calculator on this site is the natural next step.

The probability–sum check, and what it will and will not catch

Above the result the tool shows the sum of your probability column. Within 0.01 of 1 it turns green and calls the distribution valid; outside that it turns amber and tells you to add the missing outcomes or recheck your numbers.

Read the amber banner as a warning, not as a block. The calculator does not normalise your probabilities — it computes Σ x · p from exactly what you typed. Two things follow:

• If your probabilities sum to less than 1, you have forgotten an outcome, and its value is being treated as if it were zero. In the warranty example above, dropping the "no repair needed" row entirely gives a sum of 0.04 and an answer of −48 — the same number, but only because that outcome really was worth 0. Where the missing outcome has a non–zero value, the answer is simply wrong.
• If you type frequencies instead of probabilities, the answer scales. Entering the six die faces with a "probability" of 1 each gives a sum of 6 and a result of 21, not 3.5. Divide your counts by the total before entering them.

A few other behaviours worth knowing:

Getting the model right before you get the arithmetic right

The arithmetic here is trivial and the calculator will never get it wrong. Almost every mistake is upstream, in how the situation was turned into rows.

• Make the outcomes mutually exclusive and exhaustive. Every possible result must appear on exactly one row. Overlapping rows double–count; missing rows silently disappear.
• Decide whether values are net or gross and stay consistent. In the roulette example the win is entered as +35, not +36, because the $1 stake is already on the table — you are paid 35 to 1 on top of it. Enter +36 and you have quietly modelled a different game with a positive edge. If you prefer gross figures, use +36 and −0, but then the stake has to be subtracted separately.
• Use the same units on every row. Mixing dollars with percentages, or per–month with per–year, produces a number that means nothing.
• Check that the dominant term is the one you understand best. Sort the x · p column mentally. If the largest contribution comes from the outcome whose probability you were least sure about, the precision of the final figure is illusory.
• Do not chain expected values through a non–linear step. The expected value of a function is generally not the function of the expected value. If your real question involves a threshold, a cap, a ratio or compounding, computing E[X] first and then applying the transformation gives the wrong answer.

Used within those limits, this is a useful and honest instrument: it turns a scattered set of "this might happen, that might happen" statements into one number you can compare against another one number, and it shows you exactly which assumption that number rests on.