Map outcomes to a probability distribution

Try

Toss two independent fair coins and record their order. The four elementary outcomes are TT, TH, HT and HH, each with probability one quarter. Now define X to be the number of heads. The experiment still has four outcomes, but X has only three possible values. Make a card for each outcome, write its value of X on the card, and group cards with the same value.

A random variable is a rule that assigns a numerical value to each outcome. It is not another random mechanism added afterward. Counting heads, counting tails and awarding a prize for matching coins are different numerical rules applied to the same coin experiment. The question is how the probabilities of elementary outcomes combine when several receive the same value.

Predict

Predict whether X = 0, X = 1 and X = 2 are equally likely. Then predict the average value over many independent repetitions. Could that average lie between possible outcomes? A weighted average can do so; a single repetition must still produce one of the allowed values. Commit to a three-bar graph before reading the table below. Use horizontal positions 0, 1 and 2 and a vertical probability scale from 0 to 1.

Elementary outcome	Value X	Outcome probability
TT	0	1/4
TH	1	1/4
HT	1	1/4
HH	2	1/4

Experiment

The live playground uses two independent uniform numbered spinners instead of coins. Select two sides and predict the distribution of their sum. Reveal the complete outcome-to-value mapping, then change the rule to the larger value or the indicator of a match. An indicator is 1 when its event occurs and 0 otherwise. Keep the spinners unchanged so the experiment isolates the effect of changing the numerical rule.

Map elementary outcomes to values

Two independent uniform spinners each show an integer from 1 to the selected number of sides. Predict which output value is most likely. Changing the value rule changes the random variable, not the elementary experiment.

Faces on each spinner23456Value assigned to each pairSum of the two valuesLarger value1 if equal, otherwise 0

Reveal calculationsReset explorer

Commit to a prediction, then reveal the table and fixed-scale plot. Change one control and explain the difference.

All settings and traces are temporary. This explorer does not save an answer or award mastery; keep your assumptions and reasoning in the written practice.

For one setting, copy the ordered-pair list and circle all pairs that produce a chosen output. Compare their combined count with the probability bar. Next increase the number of sides while keeping the match indicator rule. Explain why a larger elementary sample space can make a particular event less likely. The fixed zero-to-one vertical scale makes such probability changes comparable without magnifying small bars automatically.

Understand

For the worked coin case, P(X = 0) = 1/4, P(X = 1) = 2/4 and P(X = 2) = 1/4. These three probabilities form a probability mass function, often shortened to PMF: a table or rule giving the probability at every possible numerical value. Every entry is nonnegative, and the entries sum to 1 because each elementary outcome appears in exactly one group.

The grouping argument is more general than equal counts. For any output value x, add the probabilities of all elementary outcomes mapped to x. If the elementary outcomes are unequal in probability, add their weights; do not merely count them. Independence and fairness made each ordered coin pair have probability (1/2)(1/2) = 1/4 here. Neither assumption follows from seeing four written labels.

The expected value is the probability-weighted mean. In this case,

E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1 head.

Equivalently, average the four equally likely card values: (0 + 1 + 1 + 2)/4 = 1. Grouping the cards has not changed the mean because it has not lost their multiplicities. The variance measures average squared distance from that mean: (0 − 1)²/4 + (1 − 1)²/2 + (2 − 1)²/4 = 1/2. It describes the spread of the stated law, not an uncertainty estimate based on a collected sample.

A PMF graph uses numerical horizontal positions and probability heights. The total probability is the sum of the bar heights; these separated bars are not a continuous density whose areas must be integrated. Do not connect the tops and infer probabilities at impossible intermediate values. Label both axes so a viewer can recover the numerical meaning without relying on color.

Apply

Design a new value rule for the four coin outcomes. For instance, it might assign different fictional rewards to TT, a mixed pair, and HH. Produce three linked artifacts: the complete outcome mapping, the aggregated distribution, and its graph. Check that every outcome has exactly one assigned value, then compute the expectation in two ways—directly from elementary outcomes and from the aggregated PMF.

If the two means disagree, inspect the multiplicity of the mixed outcomes first. If probabilities do not sum to one, look for an omitted outcome or one counted in two groups. These are structural checks, not just arithmetic checks. Explain whether the expected value is itself attainable on a single repetition and why that does not decide whether the calculation is valid.

Continue

This lesson used a fully specified theoretical mechanism. The next constructs a distribution from observed frequencies. The same aggregation and weighted-mean arithmetic applies, but the interpretation changes: a finite observed dataset does not certify the true probabilities of future outcomes. Preserve that distinction when turning a table into a prediction.