A teaching game chooses uniformly among four cards A, B, C, D, awarding 0, 1, 1, 4 points respectively. A random variable turns an experimental outcome into a number; expectation weights these numbers by their probabilities. It promises no individual game result. Use fractions, probability and independence.
From outcomes to a distribution
Let X be the score: X(A)=0, X(B)=X(C)=1, X(D)=4. Two outcomes give the same value 1, so its probability is twice that of one card.
| x, points | P(X=x) | x·P(X=x), points |
|---|---|---|
| 0 | 1/4 | 0 |
| 1 | 1/2 | 1/2 |
| 4 | 1/4 | 1 |
This variable is discrete: its possible values can be listed. Here there are finitely many; a discrete set can also be countably infinite. For a finite distribution:
p_(i) = P(X = x_(i)) ≥ 0, (∑)_(i)p_(i) = 1.
Check: 1/4+1/2+1/4=1. Replacing the final probability with 1/2 gives 5/4; calling it approximate does not validate that table. Distribution conditions.
An average with weights
E[X] = (∑)_(i)x_(i)p_(i) = 0 ⋅ (1)/(4) + 1 ⋅ (1)/(2) + 4 ⋅ (1)/(4) = 1.5.
E[X] means expectation of X; school books may use M(X). Probabilities are dimensionless, so expectation has units of points. It is not the average of the three distinct numbers (0+1+4)/3: value 1 occurs more often in the model. OpenStax
Distribution and expectation
X: A→0, B→1, C→1, D→4
Scoring 1.5 in one game is impossible here. If we independently repeat the same model many times, replacing the card, the average score approaches 1.5 in the law-of-large-numbers sense explained in A043. That is neither an exact result for any ten games nor a requirement that the next card compensate for the last.
Expectations add even with dependence
For finite-valued variables and constants a,b:
E[X + Y] = E[X] + E[Y], E[aX + b] = aE[X] + b.
Independence is unnecessary. Define Y=4−X on the same card. Large X means small Y, so they are dependent, yet X+Y always equals 4:
E[Y] = 4 − E[X] = 2.5, E[X + Y] = 4 = 1.5 + 2.5.
Why does this work? For each elementary outcome add X and Y, multiply by its probability, then group the X and Y terms separately. This uses distributivity rather than independence. Linearity of expectation, MIT.
For ten such awards the expected total is 10·1.5=15 points. Linearity gives an expected sum, without guaranteeing the sum itself. Name the variable and units, construct its distribution and check the probabilities sum to one.