đź“Š Expected Counts

Problems involving finding the expected number of occurrences using indicator variables and linearity of expectation.

🎯 Recognition Cues

  • Keywords: “expected number”, “average number”, “how many”, “count”, “occurrences”
  • Setup: Counting occurrences of events in multiple trials
  • Constraints: Independent or dependent events, fixed number of trials

đź“‹ Solution Template

  1. Identify the counting problem:

    • Independent events: Use linearity of expectation
    • Dependent events: Use conditional expectation
    • Fixed trials: Use indicator variables

  2. Apply the appropriate method:

    • Indicators: Define $I_i$ = 1 if event occurs in trial $i$, 0 otherwise
    • Linearity: $\mathbb{E}[\text{total count}] = \sum_i \mathbb{E}[I_i]$
    • Independence: $\mathbb{E}[I_i] = P(\text{event in trial } i)$

  3. Calculate the result:

    • Independent: $\mathbb{E}[\text{count}] = n \cdot P(\text{event})$
    • Dependent: Use conditional probability or casework

đź’ˇ Micro-Examples

Basic Expected Count

  • Problem: What’s the expected number of heads in 10 coin flips?
  • Solution: $10 \cdot \frac{1}{2} = 5$ heads

Indicator Variables

  • Problem: What’s the expected number of aces in a 5-card hand?
  • Solution: $5 \cdot \frac{4}{52} = \frac{5}{13}$ aces

Dependent Events

  • Problem: What’s the expected number of fixed points in a random permutation of ${1,2,3,4,5}$?
  • Solution: $5 \cdot \frac{1}{5} = 1$ fixed point

⚠️ Common Pitfalls & Variants

Pitfall: Assuming independence when events are dependent

  • Wrong: “Expected number of aces in 5 cards” = $5 \cdot \frac{4}{52}$ (assumes independence)
  • Right: This is actually correct! Linearity of expectation works even for dependent events.

Pitfall: Forgetting to use indicators

  • Wrong: “Expected number of heads in 10 flips” = Complex calculation
  • Right: Use indicators! $10 \cdot \frac{1}{2} = 5$

Pitfall: Confusing expected value with most likely value

  • Wrong: “Expected number of heads in 10 flips” = “Most likely number of heads”
  • Right: Expected value is 5, but the most likely value is also 5 (in this case).

🏆 AMC-Style Worked Example

Problem: What’s the expected number of runs in a sequence of 10 coin flips? (A run is a maximal sequence of consecutive identical outcomes.)

Solution:

  • Step 1: Let $I_i$ be 1 if there’s a run break between positions $i$ and $i+1$, 0 otherwise
  • Step 2: Number of runs = 1 + number of run breaks = 1 + $\sum_{i=1}^{9} I_i$
  • Step 3: $\mathbb{E}[\text{runs}] = 1 + \sum_{i=1}^{9} \mathbb{E}[I_i] = 1 + 9 \cdot P(\text{run break})$
  • Step 4: $P(\text{run break}) = P(\text{flip } i \neq \text{flip } i+1) = \frac{1}{2}$
  • Step 5: $\mathbb{E}[\text{runs}] = 1 + 9 \cdot \frac{1}{2} = 1 + 4.5 = 5.5$

Key insight: Use indicators to count complex events and apply linearity of expectation!

đź”— Related Topics

Next: Conditional Chains → Derangements