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Introduction to Counting and Probability / LEVEL 4 · DIFFICULTY 4/5

Expectation and Identities

Use weighted averages and combinatorial explanations.

3 stages · 24 practice problems · two 6-question assessment forms

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  1. MINI QUEST CP 4.1Expected Net ValueRead the lesson
  2. MINI QUEST CP 4.2Binomial CoefficientsRead the lesson
  3. MINI QUEST CP 4.3Summing Combination CountsRead the lesson
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STAGE CP 4.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Expected Net Value

Useful preparation: Dependent Draws

Goal: Understand and apply expected net value.

Before you begin: Dependent Draws

Understand the idea

Expected value is a probability-weighted average over outcomes. A fixed cost is paid in every outcome, so it must be subtracted from the expected payout.

E(net)=Σpᵢ·payoutᵢ−cost

Choose and carry out a method

Multiply each payout by its probability, add the products, then subtract the entry cost. Negative values represent an average loss.

Check the reasoning

Expectation need not equal any single possible result. It describes the model’s long-run average over repeated trials.

WORKED EXAMPLE 1

A fair four-sided die pays $7 if it shows 1 and $0 otherwise. Each play costs $2. What is the expected net gain in dollars?

  1. Weight each payoff by its probability and subtract the certain cost.
  2. E(gross)=7/4; E(net)=7/4-2.
  3. The expected net gain is -1/4. An expectation need not be a possible single-play result.

-1/4

WORKED EXAMPLE 2

A fair four-sided die pays $8 if it shows 1 and $0 otherwise. Each play costs $3. What is the expected net gain in dollars?

  1. Weight each payoff by its probability and subtract the certain cost.
  2. E(gross)=8/4; E(net)=8/4-3.
  3. The expected net gain is -1. An expectation need not be a possible single-play result.

-1

Common pitfalls

Possible mix-up: Subtract the cost only when winning.

A fixed entry cost is paid in every outcome.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain why a game with a positive prize can still have negative expected net value.

Preview the eight practice prompts
  1. A fair four-sided die pays $10 if it shows 1 and $0 otherwise. Each play costs $5. What is the expected net gain in dollars?
  2. A fair four-sided die pays $11 if it shows 1 and $0 otherwise. Each play costs $6. What is the expected net gain in dollars?
  3. A fair four-sided die pays $12 if it shows 1 and $0 otherwise. Each play costs $7. What is the expected net gain in dollars?
  4. A fair four-sided die pays $13 if it shows 1 and $0 otherwise. Each play costs $8. What is the expected net gain in dollars?
  5. A fair four-sided die pays $14 if it shows 1 and $0 otherwise. Each play costs $9. What is the expected net gain in dollars?
  6. A fair four-sided die pays $15 if it shows 1 and $0 otherwise. Each play costs $10. What is the expected net gain in dollars?
  7. A random promotion awards $16 with probability 1/4 and nothing otherwise. Entering costs $11. What is the average net return per entry over many repetitions? New context
  8. A random promotion awards $17 with probability 1/4 and nothing otherwise. Entering costs $12. What is the average net return per entry over many repetitions? New context
Open stage CP 4.1 in the student workspace →

STAGE CP 4.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Binomial Coefficients

Useful preparation: Expected Net Value

Goal: Understand and apply binomial coefficients.

Before you begin: Expected Net Value

Understand the idea

Expanding a power of a sum means choosing one term from each factor. To obtain x^r, choose which r factors contribute their x term, then multiply their coefficients.

[x^r](1+2x)^n=C(n,r)2^r

Choose and carry out a method

For (1+2x)^n, choose r factors and multiply C(n,r) by 2^r. The other factors contribute one.

Check the reasoning

The exponent counts how many x terms were chosen. Forgetting their coefficients gives a subset count but not the coefficient of the expansion.

WORKED EXAMPLE 1

Find the coefficient of x³ in (1+2x)^6.

  1. Choose the three factors contributing x, then account for their coefficients.
  2. The coefficient is C(6,3)·2³.
  3. It equals 160; the same choice structure explains the box count.

160

WORKED EXAMPLE 2

Find the coefficient of x³ in (1+2x)^7.

  1. Choose the three factors contributing x, then account for their coefficients.
  2. The coefficient is C(7,3)·2³.
  3. It equals 280; the same choice structure explains the box count.

280

Common pitfalls

Possible mix-up: The coefficient is only C(n,r).

Each chosen 2x contributes a factor of two.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Give a two-color selection story for the same coefficient.

Preview the eight practice prompts
  1. Find the coefficient of x³ in (1+2x)^9.
  2. Find the coefficient of x³ in (1+2x)^10.
  3. Find the coefficient of x³ in (1+2x)^11.
  4. Find the coefficient of x³ in (1+2x)^12.
  5. Find the coefficient of x³ in (1+2x)^13.
  6. Find the coefficient of x³ in (1+2x)^14.
  7. 15 labeled boxes each may be empty or contain one of two colored markers. How many arrangements have exactly three occupied boxes? New context
  8. 16 labeled boxes each may be empty or contain one of two colored markers. How many arrangements have exactly three occupied boxes? New context
Open stage CP 4.2 in the student workspace →

STAGE CP 4.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Summing Combination Counts

Useful preparation: Binomial Coefficients

Goal: Understand and apply summing combination counts.

Before you begin: Binomial Coefficients

Understand the idea

Classifying a subset by its largest element splits all subsets into disjoint cases. The remaining elements are chosen from the smaller labels, producing a sum of combinations.

Σ from k=2 to n C(k,2)=C(n+1,3)

Choose and carry out a method

For three-element subsets, fix the largest label j and choose the two smaller labels. Sum over all allowed largest labels.

Check the reasoning

Each subset has exactly one largest element, so cases cannot overlap. Compare the sum with choosing three labels directly.

WORKED EXAMPLE 1

Evaluate C(2,2)+C(3,2)+…+C(6,2), where C(n,k) counts k-element subsets.

  1. Classify a three-element subset by its largest element.
  2. For largest label j, choose two smaller labels: C(j-1,2). Summing gives C(7,3).
  3. The total is 35; every team has exactly one largest label.

35

WORKED EXAMPLE 2

Evaluate C(2,2)+C(3,2)+…+C(7,2), where C(n,k) counts k-element subsets.

  1. Classify a three-element subset by its largest element.
  2. For largest label j, choose two smaller labels: C(j-1,2). Summing gives C(8,3).
  3. The total is 56; every team has exactly one largest label.

56

Common pitfalls

Possible mix-up: Use C(n,3) as the endpoint.

The final term C(n,2) represents largest label n+1.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain the identity without expanding any factorials.

Preview the eight practice prompts
  1. Evaluate C(2,2)+C(3,2)+…+C(9,2), where C(n,k) counts k-element subsets.
  2. Evaluate C(2,2)+C(3,2)+…+C(10,2), where C(n,k) counts k-element subsets.
  3. Evaluate C(2,2)+C(3,2)+…+C(11,2), where C(n,k) counts k-element subsets.
  4. Evaluate C(2,2)+C(3,2)+…+C(12,2), where C(n,k) counts k-element subsets.
  5. Evaluate C(2,2)+C(3,2)+…+C(13,2), where C(n,k) counts k-element subsets.
  6. Evaluate C(2,2)+C(3,2)+…+C(14,2), where C(n,k) counts k-element subsets.
  7. From labels 1 through 16, a team of three is classified by its largest label. How many teams are counted across all possible largest labels? New context
  8. From labels 1 through 17, a team of three is classified by its largest label. How many teams are counted across all possible largest labels? New context
Open stage CP 4.3 in the student workspace →
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