SME
Grade 11 Math · Problem type
Expected value of a game
Use this for “A spinner pays −2, 0, 5 or 20 points with chances 0.5, 0.3, 0.15, 0.05. What is the expected value?”
Chart
The bars add to 1.
E(X)
E(X) = −2 × 0.5 + 0 × 0.3 + 5 × 0.15 + 20 × 0.05
E(X) = 0.75
This is an example. Type your own values to start a new problem.
E(X)
=
x₁
×
p₁
+
x₂
×
p₂
+
x₃
×
p₃
+
x₄
×
p₄
=
E
Formulas
p₄
=
1
−
(
p₁
+
p₂
+
p₃
)
0.05 = 1 − (0.5 + 0.3 + 0.15)
E
=
x₁
×
p₁
+
x₂
×
p₂
+
x₃
×
p₃
+
x₄
×
p₄
0.75 = −2 × 0.5 + 0 × 0.3 + 5 × 0.15 + 20 × 0.05
Assumptions
•
X takes four values, each with its own probability.
•
The probabilities add to 1, so the last one is 1 minus the others.
•
E(X) is the average of X over many plays: each value times its probability, added.
•
For fewer outcomes, give the extra values probability 0.
Step-by-step
WE KNOW
x₄ = 20, x₁ = −2, x₂ = 0, x₃ = 5, p₁ = 0.5, p₂ = 0.3, p₃ = 0.15
FIND
probability of x₄ (p₄), expected value E(X) (E)
Step 1 · Find probability of x₄ (p₄)
Use p₄ = 1 − (p₁ + p₂ + p₃)
All the probabilities add to 1.
p₄
=
1
−
(0.5
+
0.3
+
0.15)
p₄
=
0.05
Step 2 · Find expected value E(X): E
Use E = x₁ × p₁ + x₂ × p₂ + x₃ × p₃ + x₄ × p₄
Weight each value by its probability, then add.
E
=
−2
×
0.5
+
0
×
0.3
+
5
×
0.15
+
20
×
0.05
E
=
−1
+
0.75
+
1
E
=
0.75
Check
Put the numbers back in. Both sides should match.
0.05 = 1 − (0.5 + 0.3 + 0.15) ✓
0.75 = −2 × 0.5 + 0 × 0.3 + 5 × 0.15 + 20 × 0.05 ✓
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Binomial: at least k successes
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